Dataset Viewer
problem
string | solution
string | answer
string | url
string | year
int64 |
|---|---|---|---|---|
What is the value of\[3+\frac{1}{3+\frac{1}{3+\frac13}}?\]
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We have\begin{align*} 3+\frac{1}{3+\frac{1}{3+\frac13}} &= 3+\frac{1}{3+\frac{1}{\left(\frac{10}{3}\right)}} \\ &= 3+\frac{1}{3+\frac{3}{10}} \\ &= 3+\frac{1}{\left(\frac{33}{10}\right)} \\ &= 3+\frac{10}{33} \\ &= \boxed{\textbf{(D)}\ \frac{109}{33}}. \end{align*}
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\frac{109}{33}
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_1
| 2,022 |
The sum of three numbers is $96.$ The first number is $6$ times the third number, and the third number is $40$ less than the second number. What is the absolute value of the difference between the first and second numbers?
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Let $x$ be the third number. It follows that the first number is $6x,$ and the second number is $x+40.$
We have\[6x+(x+40)+x=8x+40=96,\]from which $x=7.$
Therefore, the first number is $42,$ and the second number is $47.$ Their absolute value of the difference is $|42-47|=\boxed{\textbf{(E) } 5}.$
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5
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_2
| 2,022 |
The least common multiple of a positive integer $n$ and $18$ is $180$, and the greatest common divisor of $n$ and $45$ is $15$. What is the sum of the digits of $n$?
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Note that\begin{align*} 18 &= 2\cdot3^2, \\ 180 &= 2^2\cdot3^2\cdot5, \\ 45 &= 3^2\cdot5 \\ 15 &= 3\cdot5. \end{align*}Let $n = 2^a\cdot3^b\cdot5^c.$ It follows that:
From the least common multiple condition, we have\[\operatorname{lcm}(n,18) = \operatorname{lcm}(2^a\cdot3^b\cdot5^c,2\cdot3^2) = 2^{\max(a,1)}\cdot3^{\max(b,2)}\cdot5^{\max(c,0)} = 2^2\cdot3^2\cdot5,\]from which $a=2, b\in\{0,1,2\},$ and $c=1.$
From the greatest common divisor condition, we have\[\gcd(n,45) = \gcd(2^2\cdot3^b\cdot5,3^2\cdot5) = 2^{\min(2,0)}\cdot3^{\min(b,2)}\cdot5^{\min(1,1)} = 3\cdot5,\]from which $b=1.$
Together, we conclude that $n=2^2\cdot3\cdot5=60.$ The sum of its digits is $6+0=\boxed{\textbf{(B) } 6}.$
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6
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_4
| 2,022 |
The taxicab distance between points $(x_1, y_1)$ and $(x_2, y_2)$ in the coordinate plane is given by\[|x_1 - x_2| + |y_1 - y_2|.\]For how many points $P$ with integer coordinates is the taxicab distance between $P$ and the origin less than or equal to $20$?
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Let us consider the number of points for a certain $x$-coordinate. For any $x$, the viable points are in the range $[-20 + |x|, 20 - |x|]$. This means that our total sum is equal to\begin{align*} 1 + 3 + 5 + \cdots + 41 + 39 + 37 + \cdots + 1 &= (1 + 3 + 5 + \cdots + 39) + (1 + 3 + 5 + \cdots + 41) \\ & = 20^2 + 21^2 \\ & = 29^2 \\ &= \boxed{\textbf{(C)} \, 841}. \end{align*}
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841
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_5
| 2,022 |
A data set consists of $6$ (not distinct) positive integers: $1$, $7$, $5$, $2$, $5$, and $X$. The average (arithmetic mean) of the $6$ numbers equals a value in the data set. What is the sum of all positive values of $X$?
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First, note that $1+7+5+2+5=20$. There are $3$ possible cases:
Case 1: the mean is $5$.
$X = 5 \cdot 6 - 20 = 10$.
Case 2: the mean is $7$.
$X = 7 \cdot 6 - 20 = 22$.
Case 3: the mean is $X$.
$X= \frac{20+X}{6} \Rightarrow X=4$.
Therefore, the answer is $10+22+4=\boxed{\textbf{(D) }36}$.
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36
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_6
| 2,022 |
The infinite product\[\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots\]evaluates to a real number. What is that number?
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We can write $\sqrt[3]{10}$ as $10 ^ \frac{1}{3}$. Similarly, $\sqrt[3]{\sqrt[3]{10}} = (10 ^ \frac{1}{3}) ^ \frac{1}{3} = 10 ^ \frac{1}{3^2}$.
By continuing this, we get the form\[10 ^ \frac{1}{3} \cdot 10 ^ \frac{1}{3^2} \cdot 10 ^ \frac{1}{3^3} \cdots,\]which is\[10 ^ {\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots}.\]Using the formula for an infinite geometric series $S = \frac{a}{1-r}$, we get\[\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots = \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{1}{2}.\]Thus, our answer is $10 ^ \frac{1}{2} = \boxed{\textbf{(A) }\sqrt{10}}$.
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\sqrt{10}
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_8
| 2,022 |
On Halloween $31$ children walked into the principal's office asking for candy. They can be classified into three types: Some always lie; some always tell the truth; and some alternately lie and tell the truth. The alternaters arbitrarily choose their first response, either a lie or the truth, but each subsequent statement has the opposite truth value from its predecessor. The principal asked everyone the same three questions in this order.
"Are you a truth-teller?" The principal gave a piece of candy to each of the $22$ children who answered yes.
"Are you an alternater?" The principal gave a piece of candy to each of the $15$ children who answered yes.
"Are you a liar?" The principal gave a piece of candy to each of the $9$ children who answered yes.
How many pieces of candy in all did the principal give to the children who always tell the truth?
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Note that:
Truth-tellers would answer yes-no-no to the three questions in this order.
Liars would answer yes-yes-no to the three questions in this order.
Alternaters who responded truth-lie-truth would answer no-no-no to the three questions in this order.
Alternaters who responded lie-truth-lie would answer yes-yes-yes to the three questions in this order.
Suppose that there are $T$ truth-tellers, $L$ liars, and $A$ alternaters who responded lie-truth-lie.
The conditions of the first two questions imply that\begin{align*} T+L+A&=22, \\ L+A&=15. \end{align*}Subtracting the second equation from the first, we have $T=22-15=\boxed{\textbf{(A) } 7}.$
Remark
The condition of the third question is extraneous. However, we know that $A=9$ and $L=6,$ so there are $9$ alternaters who responded lie-truth-lie, $6$ liars, and $9$ alternaters who responded truth-lie-truth from this condition.
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7
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_9
| 2,022 |
How many ways are there to split the integers $1$ through $14$ into $7$ pairs such that in each pair, the greater number is at least $2$ times the lesser number?
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Clearly, the integers from $8$ through $14$ must be in different pairs, so are the integers from $1$ through $7.$ Note that $7$ must pair with $14.$
We pair the numbers $1,2,3,4,5,6$ with the numbers $8,9,10,11,12,13$ systematically:
$6$ can pair with either $12$ or $13.$
$5$ can pair with any of the three remaining numbers from $10,11,12,13.$
$1,2,3,4$ can pair with the other four remaining numbers from $8,9,10,11,12,13$ without restrictions.
Together, the answer is $2\cdot3\cdot4!=\boxed{\textbf{(E) } 144}.$
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144
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_10
| 2,022 |
What is the product of all real numbers $x$ such that the distance on the number line between $\log_6x$ and $\log_69$ is twice the distance on the number line between $\log_610$ and $1$?
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Let $a = 2 \cdot |\log_6 10 - 1| = |\log_6 9 - \log_6 x| = \left|\log_6 \frac{9}{x}\right|$.
$\pm a = \log_6 \frac{9}{x} \implies 6^{\pm a} = b^{\pm 1} = \frac{9}{x} \implies x = 9 \cdot b^{\pm 1}$
$9b^1 \cdot 9b^{-1} = \boxed{81}$.
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81
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_11
| 2,022 |
Let $M$ be the midpoint of $\overline{AB}$ in regular tetrahedron $ABCD$. What is $\cos(\angle CMD)$?
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Without loss of generality, let the edge-length of $ABCD$ be $2.$ It follows that $MC=MD=\sqrt3.$
Let $O$ be the center of $\triangle ABD,$ so $\overline{CO}\perp\overline{MOD}.$ Note that $MO=\frac13 MD=\frac{\sqrt{3}}{3}.$
In right $\triangle CMO,$ we have\[\cos(\angle CMD)=\frac{MO}{MC}=\boxed{\textbf{(B) } \frac13}.\]
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\frac{1}{3}
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_12
| 2,022 |
Let $\mathcal{R}$ be the region in the complex plane consisting of all complex numbers $z$ that can be written as the sum of complex numbers $z_1$ and $z_2$, where $z_1$ lies on the segment with endpoints $3$ and $4i$, and $z_2$ has magnitude at most $1$. What integer is closest to the area of $\mathcal{R}$?
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[asy] size(250); import TrigMacros; rr_cartesian_axes(-2,6,-2,6,complexplane=true, usegrid = true); Label f; f.p=fontsize(6); xaxis(-1,5,Ticks(f, 1.0)); yaxis(-1,5,Ticks(f, 1.0)); dot((3,0)); dot((0,4)); draw((0,4)--(3,0), blue); draw((0.8, 4.6)..(-.6,4.8)..(-.8, 3.4),red); draw((-.8, 3.4)--(2.2, -0.6), red); draw((2.2, -0.6)..(3.6,-0.8)..(3.8,0.6), red); draw((0.8, 4.6)--(3.8,0.6),red); draw((0.8, 4.6)--(-.8, 3.4),red+dashed); draw((2.2, -0.6)--(3.8,0.6),red+ dashed); draw((3,0)--(3,-1),Arrow); label("1",(3,0)--(3,-1),E); draw((0,4)--(-.6,4.8),Arrow); label("1",(0,4)--(-.6,4.8),SW); draw((1.5,2)--(2.3,2.6),Arrow); label("1",(1.5,2)--(2.3,2.6),SE); [/asy]
If $z$ is a complex number and $z = a + bi$, then the magnitude (length) of $z$ is $\sqrt{a^2 + b^2}$. Therefore, $z_1$ has a magnitude of 5. If $z_2$ has a magnitude of at most one, that means for each point on the segment given by $z_1$, the bounds of the region $\mathcal{R}$ could be at most 1 away. Alone the line, excluding the endpoints, a rectangle with a width of 2 and a length of 5, the magnitude, would be formed. At the endpoints, two semicircles will be formed with a radius of 1 for a total area of $\pi \approx 3$. Therefore, the total area is $5(2) + \pi \approx 10 + 3 = \boxed{\textbf{(A) } 13}$.
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13
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_13
| 2,022 |
What is the value of\[(\log 5)^{3}+(\log 20)^{3}+(\log 8)(\log 0.25)\]where $\log$ denotes the base-ten logarithm?
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Let $\text{log } 2 = x$. The expression then becomes\[(1+x)^3+(1-x)^3+(3x)(-2x)=\boxed{2}.\]
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2
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_14
| 2,022 |
The roots of the polynomial $10x^3 - 39x^2 + 29x - 6$ are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by $2$ units. What is the volume of the new box?
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Let $a$, $b$, $c$ be the three roots of the polynomial. The lengthened prism's volume is\[V = (a+2)(b+2)(c+2) = abc+2ac+2ab+2bc+4a+4b+4c+8 = abc + 2(ab+ac+bc) + 4(a+b+c) + 8.\]By Vieta's formulas, we know that a cubic polynomial $Ax^3+Bx^2+Cx+D$ with roots $a$, $b$, $c$ satisfies:\begin{alignat*}{8} a+b+c &= -\frac{B}{A} &&= \frac{39}{10}, \\ ab+ac+bc &= \hspace{2mm}\frac{C}{A} &&= \frac{29}{10}, \\ abc &= -\frac{D}{A} &&= \frac{6}{10}. \end{alignat*}We can substitute these into the expression, obtaining\[V = \frac{6}{10} + 2\left(\frac{29}{10}\right) + 4\left(\frac{39}{10}\right) + 8 = \boxed{\textbf{(D) } 30}.\]
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30
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_15
| 2,022 |
A $\emph{triangular number}$ is a positive integer that can be expressed in the form $t_n = 1+2+3+\cdots+n$, for some positive integer $n$. The three smallest triangular numbers that are also perfect squares are $t_1 = 1 = 1^2$, $t_8 = 36 = 6^2$, and $t_{49} = 1225 = 35^2$. What is the sum of the digits of the fourth smallest triangular number that is also a perfect square?
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We have $t_n = \frac{n (n+1)}{2}$. If $t_n$ is a perfect square, then it can be written as $\frac{n (n+1)}{2} = k^2$, where $k$ is a positive integer.
Thus, $n (n+1) = 2 k^2$. Rearranging, we get $(2n+1)^2-2(2k)^2=1$, a Pell equation (see https://artofproblemsolving.com/wiki/index.php/Pell_equation ). So $\frac{2n+1}{2k}$ must be a truncation of the continued fraction for $\sqrt{2}$:
\begin{eqnarray*} 1+\frac12&=&\frac{2\cdot1+1}{2\cdot1}\\ 1+\frac1{2+\frac1{2+\frac12}}&=&\frac{2\cdot8+1}{2\cdot6}\\ 1+\frac1{2+\frac1{2+\frac1{2+\frac1{2+\frac12}}}}&=&\frac{2\cdot49+1}{2\cdot35}\\ 1+\frac1{2+\frac1{2+\frac1{2+\frac1{2+\frac1{2+\frac1{2+\frac12}}}}}}&=&\frac{2\cdot288+1}{2\cdot204} \end{eqnarray*}
Therefore, $t_{288} = \frac{288\cdot289}{2} = 204^2 = 41616$, so the answer is $4+1+6+1+6=\boxed{\textbf{(D) 18}}$.
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18
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_16
| 2,022 |
Suppose $a$ is a real number such that the equation\[a\cdot(\sin{x}+\sin{(2x)}) = \sin{(3x)}\]has more than one solution in the interval $(0, \pi)$. The set of all such $a$ that can be written in the form\[(p,q) \cup (q,r),\]where $p, q,$ and $r$ are real numbers with $p < q< r$. What is $p+q+r$?
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We are given that $a\cdot(\sin{x}+\sin{(2x)})=\sin{(3x)}$
Using the sine double angle formula combine with the fact that $\sin{(3x)} = \sin{x}\cdot(4\cos^2{x}-1)$, which can be derived using sine angle addition with $\sin{(2x + x)}$, we have\[a\cdot(\sin{x}+2\sin{x}\cos{x})=\sin{x}\cdot(4\cos^2{x}-1)\]Since $\sin{x} \ne 0$ as it is on the open interval $(0, \pi)$, we can divide out $\sin{x}$ from both sides, leaving us with\[a\cdot(1+2\cos{x})=4\cos^2{x}-1\]Now, distributing $a$ and rearranging, we achieve the equation\[4\cos^2{x} - 2a\cos{x} - (1+a) = 0\]which is a quadratic in $\cos{x}$.
Applying the quadratic formula to solve for $\cos{x}$, we get\[\cos{x} =\frac{2a\pm\sqrt{4a^2+4(4)(1+a)}}{8}\]and expanding the terms under the radical, we get\[\cos{x} =\frac{2a\pm\sqrt{4a^2+16a+16}}{8}\]Factoring, since $4a^2+16a+16 = (2a+4)^2$, we can simplify our expression even further to\[\cos{x} =\frac{a\pm(a+2)}{4}\]
Now, solving for our two solutions, $\cos{x} = -\frac{1}{2}$ and $\cos{x} = \frac{a+1}{2}$.
Since $\cos{x} = -\frac{1}{2}$ yields a solution that is valid for all $a$, that being $x = \frac{2\pi}{3}$, we must now solve for the case where $\frac{a+1}{2}$ yields a valid value.
As $x\in (0, \pi)$, $\cos{x}\in (-1, 1)$, and therefore $\frac{a+1}{2}\in (-1, 1)$, and $a\in(-3,1)$.
There is one more case we must consider inside this interval though, the case where $\frac{a+1}{2} = -\frac{1}{2}$, as this would lead to a double root for $\cos{x}$, yielding only one valid solution for $x$. Solving for this case, $a \ne -2$.
Therefore, combining this fact with our solution interval, $a\in(-3, -2) \cup (-2, 1)$, so the answer is $-3-2+1 = \boxed{\textbf{(A) } {-}4}$.
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-4
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_17
| 2,022 |
Let $T_k$ be the transformation of the coordinate plane that first rotates the plane $k$ degrees counterclockwise around the origin and then reflects the plane across the $y$-axis. What is the least positive integer $n$ such that performing the sequence of transformations $T_1, T_2, T_3, \cdots, T_n$ returns the point $(1,0)$ back to itself?
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Let $P=(r,\theta)$ be a point in polar coordinates, where $\theta$ is in degrees.
Rotating $P$ by $k^{\circ}$ counterclockwise around the origin gives the transformation $(r,\theta)\rightarrow(r,\theta+k^{\circ}).$ Reflecting $P$ across the $y$-axis gives the transformation $(r,\theta)\rightarrow(r,180^{\circ}-\theta).$ Note that\begin{align*} T_k(P)&=(r,180^{\circ}-\theta-k^{\circ}), \\ T_{k+1}(T_k(P)) &= (r,\theta -1^{\circ}). \end{align*}We start with $(1,0^{\circ})$ in polar coordinates. For the sequence of transformations $T_1, T_2, T_3, \cdots, T_k,$ it follows that
After $T_1,$ we have $(1,179^{\circ}).$
After $T_2,$ we have $(1,-1^{\circ}).$
After $T_3,$ we have $(1,178^{\circ}).$
After $T_4,$ we have $(1,-2^{\circ}).$
After $T_5,$ we have $(1,177^{\circ}).$
After $T_6,$ we have $(1,-3^{\circ}).$
...
After $T_{2k-1},$ we have $(1,180^{\circ}-k^{\circ}).$
After $T_{2k},$ we have $(1,-k^{\circ}).$
The least such positive integer $k$ is $180.$ Therefore, the least such positive integer $n$ is $2k-1=\boxed{\textbf{(A) } 359}.$
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359
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_18
| 2,022 |
Suppose that $13$ cards numbered $1, 2, 3, \ldots, 13$ are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards $1, 2, 3$ are picked up on the first pass, $4$ and $5$ on the second pass, $6$ on the third pass, $7, 8, 9, 10$ on the fourth pass, and $11, 12, 13$ on the fifth pass. For how many of the $13!$ possible orderings of the cards will the $13$ cards be picked up in exactly two passes?
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For $1\leq k\leq 12,$ suppose that cards $1, 2, \ldots, k$ are picked up on the first pass. It follows that cards $k+1,k+2,\ldots,13$ are picked up on the second pass.
Once we pick the spots for the cards on the first pass, there is only one way to arrange all $\boldsymbol{13}$ cards.
For each value of $k,$ there are $\binom{13}{k}-1$ ways to pick the $k$ spots for the cards on the first pass: We exclude the arrangement[asy] size(11cm); draw((0,0)--(2,0)--(2,3)--(0,3)--cycle); label("1", (1,1.5)); draw((3,0)--(5,0)--(5,3)--(3,3)--cycle); label("2", (4,1.5)); draw((6,0)--(8,0)--(8,3)--(6,3)--cycle); label("3", (7,1.5)); draw((9,0)--(11,0)--(11,3)--(9,3)--cycle); label("4", (10,1.5)); draw((12,0)--(14,0)--(14,3)--(12,3)--cycle); label("5", (13,1.5)); draw((15,0)--(17,0)--(17,3)--(15,3)--cycle); label("6", (16,1.5)); draw((18,0)--(20,0)--(20,3)--(18,3)--cycle); label("7", (19,1.5)); draw((21,0)--(23,0)--(23,3)--(21,3)--cycle); label("8", (22,1.5)); draw((24,0)--(26,0)--(26,3)--(24,3)--cycle); label("9", (25,1.5)); draw((27,0)--(29,0)--(29,3)--(27,3)--cycle); label("10", (28,1.5)); draw((30,0)--(32,0)--(32,3)--(30,3)--cycle); label("11", (31,1.5)); draw((33,0)--(35,0)--(35,3)--(33,3)--cycle); label("12", (34,1.5)); draw((36,0)--(38,0)--(38,3)--(36,3)--cycle); label("13", (37,1.5)); [/asy]in which the cards are arranged such that the first pass consists of all $13$ cards.
Therefore, the answer is\[\sum_{k=1}^{12}\left[\binom{13}{k}-1\right] = \left[\sum_{k=1}^{12}\binom{13}{k}\right]-12 = \left[\sum_{k=0}^{13}\binom{13}{k}\right]-14 = 2^{13} - 14 = \boxed{\textbf{(D) } 8178}.\]
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8178
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_19
| 2,022 |
Isosceles trapezoid $ABCD$ has parallel sides $\overline{AD}$ and $\overline{BC},$ with $BC < AD$ and $AB = CD.$ There is a point $P$ in the plane such that $PA=1, PB=2, PC=3,$ and $PD=4.$ What is $\tfrac{BC}{AD}?$
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Consider the reflection $P^{\prime}$ of $P$ over the perpendicular bisector of $\overline{BC}$, creating two new isosceles trapezoids $DAPP^{\prime}$ and $CBPP^{\prime}$. Under this reflection, $P^{\prime}A=PD=4$, $P^{\prime}D=PA=1$, $P^{\prime}C=PB=2$, and $P^{\prime}B=PC=3$.
Since $DAPP'$ and $CBPP'$ are isosceles trapezoids, they are cyclic. Using Ptolemy's theorem on $DAPP'$, we get that $(PP')(AD) + (PA)(P'D) = (AP')(PD)$, so\[PP' \cdot AD + 1 \cdot 1 = 4 \cdot 4.\]Then, using Ptolemy's theorem again on $CBPP'$, we get that $(BC)(PP') + (BP)(CP') = (BP')(CP)$, so\[PP' \cdot BC + 2 \cdot 2 = 3 \cdot 3.\]Thus, $PP^{\prime}\cdot AD=15$ and $PP^{\prime}\cdot BC=5$; dividing these two equations and taking the reciprocal yields $\frac{BC}{AD}=\boxed{\textbf{(B) }\frac{1}{3}}$.[asy] size(300); pair A = (0,0); pair B = (1, 2); pair C = (2,2); pair D = (3,0); label("$A$", A, SW); label("$B$", B, NW); label("$C$", C, NE); label("$D$", D, SE); draw(A--B--C--D--cycle, blue); pair P = (0.8, 0.6); dot("$P$", P, NW); draw(P--A, magenta); draw(P--B, magenta); draw(P--C); draw(P--D); label("$1$", P--A, NW); label("$2$", P--B, E); label("$3$", P--C, NW); label("$4$", P--D, S); pair P1 = (2.2, 0.6); dot("$P'$", P1, NE); draw(P1--D, magenta); draw(P1--C, magenta); draw(P1--A); draw(P1--B); label("$1$", P1--D, NE); label("$2$", P1--C, E); label("$3$", P1--B, NE); label("$4$", P1--A, SE); draw(P--P1, dashed+magenta); [/asy]
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\frac{1}{3}
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_20
| 2,022 |
Let\[P(x) = x^{2022} + x^{1011} + 1.\]Which of the following polynomials is a factor of $P(x)$?
$\textbf{(A)} \, x^2 -x + 1 \qquad\textbf{(B)} \, x^2 + x + 1 \qquad\textbf{(C)} \, x^4 + 1 \qquad\textbf{(D)} \, x^6 - x^3 + 1 \qquad\textbf{(E)} \, x^6 + x^3 + 1$
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$P(x) = x^{2022} + x^{1011} + 1$ is equal to $\frac{x^{3033}-1}{x^{1011}-1}$ by difference of powers.
Therefore, the answer is a polynomial that divides $x^{3033}-1$ but not $x^{1011}-1$.
Note that any polynomial $x^m-1$ divides $x^n-1$ if and only if $m$ is a factor of $n$.
The prime factorizations of $1011$ and $3033$ are $3*337$ and $3^2*337$, respectively.
Hence, $x^9-1$ is a divisor of $x^{3033}-1$ but not $x^{1011}-1$.
By difference of powers, $x^9-1=(x^3-1)(x^6+x^3+1)$. Therefore, the answer is $\boxed{E}$.
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E
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_21
| 2,022 |
Let $c$ be a real number, and let $z_1$ and $z_2$ be the two complex numbers satisfying the equation $z^2 - cz + 10 = 0$. Points $z_1$, $z_2$, $\frac{1}{z_1}$, and $\frac{1}{z_2}$ are the vertices of (convex) quadrilateral $\mathcal{Q}$ in the complex plane. When the area of $\mathcal{Q}$ obtains its maximum possible value, $c$ is closest to which of the following?
$\textbf{(A) }4.5 \qquad\textbf{(B) }5 \qquad\textbf{(C) }5.5 \qquad\textbf{(D) }6\qquad\textbf{(E) }6.5$
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Because $c$ is real, $z_2 = \bar z_1$. We have\begin{align*} 10 & = z_1 z_2 \\ & = z_1 \bar z_1 \\ & = |z_1|^2 , \end{align*}where the first equality follows from Vieta's formula.
Thus, $|z_1| = \sqrt{10}$.
We have\begin{align*} c & = z_1 + z_2 \\ & = z_1 + \bar z_1 \\ & = 2 {\rm Re}(z_1), \end{align*}where the first equality follows from Vieta's formula.
Thus, ${\rm Re}(z_1) = \frac{c}{2}$.
We have\begin{align*} \frac{1}{z_1} & = \frac{1}{10}\cdot\frac{10}{z_1} \\ & = \frac{1}{10}\cdot\frac{z_1 z_2}{z_1} \\ & = \frac{z_2}{10} \\ & = \frac{\bar z_1}{10}. \end{align*}where the second equality follows from Vieta's formula.
We have\begin{align*} \frac{1}{z_2} & = \frac{1}{10}\cdot\frac{10}{z_2} \\ & = \frac{1}{10}\cdot\frac{z_1 z_2}{z_2} \\ & = \frac{z_1}{10}. \end{align*}where the second equality follows from Vieta's formula.
Therefore,\begin{align*} {\rm Area} \ Q & = \frac{1}{2} \left| {\rm Re}(z_1) \right| \cdot 2 \left| {\rm Im}(z_1) \right| \cdot \left( 1 - \frac{1}{10^2} \right) \\ & = \frac{1}{2} |c| \sqrt{10 - \frac{c^2}{4}} \left( 1 - \frac{1}{10^2} \right) \\ & = \frac{1 - \frac{1}{10^2}}{4} \sqrt{c^2 \left( 40 - c^2 \right)} \\ & \leq \frac{1 - \frac{1}{10^2}}{4} \cdot \frac{c^2 + \left( 40 - c^2 \right)}{2} \\ & = \frac{1 - \frac{1}{10^2}}{4} \cdot 20 , \end{align*}where the inequality follows from the AM-GM inequality, and it is augmented to an equality if and only if $c^2 = 40 - c^2$. Thus, $|c| = 2 \sqrt{5} \approx \boxed{\textbf{(A) 4.5}}$.
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A
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_22
| 2,022 |
Let $h_n$ and $k_n$ be the unique relatively prime positive integers such that\[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}=\frac{h_n}{k_n}.\]Let $L_n$ denote the least common multiple of the numbers $1, 2, 3, \ldots, n$. For how many integers with $1\le{n}\le{22}$ is $k_n<L_n$?
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We are given that\[\sum_{i=1}^{n}\frac1i = \frac{1}{L_n}\sum_{i=1}^{n}\frac{L_n}{i} = \frac{h_n}{k_n}.\]Since $k_n < L_n,$ we need $\gcd\left(\sum_{i=1}^{n}\frac{L_n}{i}, L_n\right)>1.$
For all primes $p$ such that $p\leq n,$ let $v_p(L_n)=e\geq1$ be the exponent of the largest power of $p$ that divides $L_n.$
It is clear that $L_n\equiv0\pmod{p},$ so we test whether $\sum_{i=1}^{n}\frac{L_n}{i}\equiv0\pmod{p}.$ Note that\[\sum_{i=1}^{n}\frac{L_n}{i} \equiv \sum_{i=1}^{\left\lfloor\tfrac{n}{p^e}\right\rfloor}\frac{L_n}{ip^e} \ (\operatorname{mod} \ p^e) \equiv \sum_{i=1}^{\left\lfloor\tfrac{n}{p^e}\right\rfloor}\frac{L_n}{ip^e} \ (\operatorname{mod} \ p).\]We construct the following table for $v_p(L_n)=e:$\[\begin{array}{c|c|l|c} \textbf{Case of }\boldsymbol{(p,e)} & \textbf{Interval of }\boldsymbol{n} & \hspace{22.75mm}\textbf{Sum} & \boldsymbol{\stackrel{?}{\equiv}0\ (\operatorname{mod} \ p)} \\ [0.5ex] \hline\hline & & & \\ [-2ex] (2,1) & [2,3] & L_n/2 & \\ (2,2) & [4,7] & L_n/4 & \\ (2,3) & [8,15] & L_n/8 & \\ (2,4) & [16,22] & L_n/16 & \\ [0.5ex] \hline & & & \\ [-2ex] (3,1) & [3,5] & L_n/3 & \\ & [6,8] & L_n/3 + L_n/6 & \checkmark \\ (3,2) & [9,17] & L_n/9 & \\ & [18,22] & L_n/9 + L_n/18 & \checkmark \\ [0.5ex] \hline & & & \\ [-2ex] (5,1) & [5,9] & L_n/5 & \\ & [10,14] & L_n/5 + L_n/10 & \\ & [15,19] & L_n/5 + L_n/10 + L_n/15 & \\ & [20,22] & L_n/5 + L_n/10 + L_n/15 + L_n/20 & \checkmark \\ [0.5ex] \hline & & & \\ [-2ex] (7,1) & [7,13] & L_n/7 & \\ & [14,20] & L_n/7 + L_n/14 & \\ & [21,22] & L_n/7 + L_n/14 + L_n/21 & \\ [0.5ex] \hline & & & \\ [-2ex] (11,1) & [11,21] & L_n/11 & \\ & \{22\} & L_n/11 + L_n/22 & \\ [0.5ex] \hline & & & \\ [-2ex] (13,1) & [13,22] & L_n/13 & \\ [0.5ex] \hline & & & \\ [-2ex] (17,1) & [17,22] & L_n/17 & \\ [0.5ex] \hline & & & \\ [-2ex] (19,1) & [19,22] & L_n/19 & \\ [0.5ex] \end{array}\]Note that:
If the Sum column has only one term, then it is never congruent to $0$ modulo $p.$
If $p$ and $q$ are positive integers such that $p\geq q,$ then $L_p$ is a multiple of $L_q.$ Therefore, for a specific case, if the sum is congruent to $0$ modulo $p$ for the smallest element in the interval of $n,$ then it is also congruent to $0$ modulo $p$ for all other elements in the interval of $n.$
Together, there are $\boxed{\textbf{(D) }8}$ such integers $n,$ namely\[6,7,8,18,19,20,21,22.\]
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8
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_23
| 2,022 |
How many strings of length $5$ formed from the digits $0$, $1$, $2$, $3$, $4$ are there such that for each $j \in \{1,2,3,4\}$, at least $j$ of the digits are less than $j$? (For example, $02214$ satisfies this condition because it contains at least $1$ digit less than $1$, at least $2$ digits less than $2$, at least $3$ digits less than $3$, and at least $4$ digits less than $4$. The string $23404$ does not satisfy the condition because it does not contain at least $2$ digits less than $2$.)
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For some $n$, let there be $n+1$ parking spaces counterclockwise in a circle. Consider a string of $n$ integers $c_1c_2 \ldots c_n$ each between $0$ and $n$, and let $n$ cars come into this circle so that the $i$th car tries to park at spot $c_i$, but if it is already taken then it instead keeps going counterclockwise and takes the next available spot. After this process, exactly one spot will remain empty.
Then the strings of $n$ numbers between $0$ and $n-1$ that contain at least $k$ integers $<k$ for $1 \leq k \leq n$ are exactly the set of strings that leave spot $n$ empty. Also note for any string $c_1c_2 \ldots c_n$, we can add $1$ to each $c_i$ (mod $n+1$) to shift the empty spot counterclockwise, meaning for each string there exists exactly one $j$ with $0 \leq j \leq n$ so that $(c_1+j)(c_2+j) \ldots (c_n+j)$ leaves spot $n$ empty. This gives there are $\frac{(n+1)^{n}}{n+1} = (n+1)^{n-1}$ such strings.
Plugging in $n = 5$ gives $\boxed{\textbf{(E) }1296}$ such strings.
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1296
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_24
| 2,022 |
A circle with integer radius $r$ is centered at $(r, r)$. Distinct line segments of length $c_i$ connect points $(0, a_i)$ to $(b_i, 0)$ for $1 \le i \le 14$ and are tangent to the circle, where $a_i$, $b_i$, and $c_i$ are all positive integers and $c_1 \le c_2 \le \cdots \le c_{14}$. What is the ratio $\frac{c_{14}}{c_1}$ for the least possible value of $r$?
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Suppose that with a pair $(a_i,b_i)$ the circle is an excircle. Then notice that the hypotenuse must be $(r-x)+(r-y)$, so it must be the case that\[a_i^2+b_i^2=(2r-a_i-b_i)^2.\]Similarly, if with a pair $(a_i,b_i)$ the circle is an incircle, the hypotenuse must be $(x-r)+(y-r)$, leading to the same equation.
Notice that this equation can be simplified through SFFT to\[(a_i-2r)(b_i-2r)=2r^2.\]Thus, we want the smallest $r$ such that this equation has at least $14$ distinct pairs $(a_i,b_i)$ for which this holds. The obvious choice to check is $r=6$. In this case, since $2r^2=2^3\cdot 3^2$ has $12$ positive factors, we get $12$ pairs, and we get another two if the factors are $-8,-9$ or vice versa. One can check that for smaller values of $r$, we do not even get close to $14$ possible pairs.
When $r=6$, the smallest possible $c$-value is clearly when the factors are negative. When this occurs, $a_i=4, b_i=3$ (or vice versa), so the mimimal $c$ is $5$. The largest possible $c$-value occurs when the largest of $a_i$ and $b_i$ are maximized. This occurs when the factors are $72$ and $1$, leading to $a_i=84, b_i=13$ (or vice-versa), leading to a maximal $c$ of $85$.
Hence the answer is $\frac{85}5=\boxed{17}$.
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17
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12A_Problems/Problem_25
| 2,022 |
Define $x\diamond y$ to be $|x-y|$ for all real numbers $x$ and $y.$ What is the value of\[(1\diamond(2\diamond3))-((1\diamond2)\diamond3)?\]
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We have\begin{align*} (1\diamond(2\diamond3))-((1\diamond2)\diamond3) &= |1-|2-3|| - ||1-2|-3| \\ &= |1-1| - |1-3| \\ &= 0-2 \\ &= \boxed{\textbf{(A)}\ {-}2}. \end{align*}
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-2
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_1
| 2,022 |
How many of the first ten numbers of the sequence $121, 11211, 1112111, \ldots$ are prime numbers?
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The $n$th term of this sequence is\[\sum_{k=n}^{2n}10^k + \sum_{k=0}^{n}10^k = 10^n\sum_{k=0}^{n}10^k + \sum_{k=0}^{n}10^k = \left(10^n+1\right)\sum_{k=0}^{n}10^k.\]It follows that the terms are\begin{align*} 121 &= 11\cdot11, \\ 11211 &= 101\cdot111, \\ 1112111 &= 1001\cdot1111, \\ & \ \vdots \end{align*}Therefore, there are $\boxed{\textbf{(A) } 0}$ prime numbers in this sequence.
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0
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_3
| 2,022 |
For how many values of the constant $k$ will the polynomial $x^{2}+kx+36$ have two distinct integer roots?
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Let $p$ and $q$ be the roots of $x^{2}+kx+36.$ By Vieta's Formulas, we have $p+q=-k$ and $pq=36.$
It follows that $p$ and $q$ must be distinct factors of $36.$ The possibilities of $\{p,q\}$ are\[\pm\{1,36\},\pm\{2,18\},\pm\{3,12\},\pm\{4,9\}.\]Each unordered pair gives a unique value of $k.$ Therefore, there are $\boxed{\textbf{(B) }8}$ values of $k,$ corresponding to $\mp37,\mp20,\mp15,\mp13,$ respectively.
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8
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_4
| 2,022 |
The point $(-1, -2)$ is rotated $270^{\circ}$ counterclockwise about the point $(3, 1)$. What are the coordinates of its new position?
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$(-1,-2)$ is $4$ units west and $3$ units south of $(3,1)$. Performing a counterclockwise rotation of $270^{\circ}$, which is equivalent to a clockwise rotation of $90^{\circ}$, the answer is $3$ units west and $4$ units north of $(3,1)$, or $\boxed{\textbf{(B)}\ (0,5)}$.
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(0, 5)
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_5
| 2,022 |
Consider the following $100$ sets of $10$ elements each:\begin{align*} &\{1,2,3,\ldots,10\}, \\ &\{11,12,13,\ldots,20\},\\ &\{21,22,23,\ldots,30\},\\ &\vdots\\ &\{991,992,993,\ldots,1000\}. \end{align*}How many of these sets contain exactly two multiples of $7$?
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We apply casework to this problem. The only sets that contain two multiples of seven are those for which:
The multiples of $7$ are $1\pmod{10}$ and $8\pmod{10}.$ That is, the first and eighth elements of such sets are multiples of $7.$
The first element is $1+10k$ for some integer $0\leq k\leq99.$ It is a multiple of $7$ when $k=2,9,16,\ldots,93.$
The multiples of $7$ are $2\pmod{10}$ and $9\pmod{10}.$ That is, the second and ninth elements of such sets are multiples of $7.$
The second element is $2+10k$ for some integer $0\leq k\leq99.$ It is a multiple of $7$ when $k=4,11,18,\ldots,95.$
The multiples of $7$ are $3\pmod{10}$ and $0\pmod{10}.$ That is, the third and tenth elements of such sets are multiples of $7.$
The third element is $3+10k$ for some integer $0\leq k\leq99.$ It is a multiple of $7$ when $k=6,13,20,\ldots,97.$
Each case has $\left\lfloor\frac{100}{7}\right\rfloor=14$ sets. Therefore, the answer is $14\cdot3=\boxed{\textbf{(B)}\ 42}.$
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42
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_6
| 2,022 |
Camila writes down five positive integers. The unique mode of these integers is $2$ greater than their median, and the median is $2$ greater than their arithmetic mean. What is the least possible value for the mode?
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Let $M$ be the median. It follows that the two largest integers are both $M+2.$
Let $a$ and $b$ be the two smallest integers such that $a<b.$ The sorted list is\[a,b,M,M+2,M+2.\]Since the median is $2$ greater than their arithmetic mean, we have $\frac{a+b+M+(M+2)+(M+2)}{5}+2=M,$ or\[a+b+14=2M.\]Note that $a+b$ must be even. We minimize this sum so that the arithmetic mean, the median, and the unique mode are minimized. Let $a=1$ and $b=3,$ from which $M=9$ and $M+2=\boxed{\textbf{(D)}\ 11}.$
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11
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_7
| 2,022 |
What is the graph of $y^4+1=x^4+2y^2$ in the coordinate plane?
$\textbf{(A) } \text{two intersecting parabolas} \qquad \textbf{(B) } \text{two nonintersecting parabolas} \qquad \textbf{(C) } \text{two intersecting circles} \qquad \\\\ \textbf{(D) } \text{a circle and a hyperbola} \qquad \textbf{(E) } \text{a circle and two parabolas}$
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Since the equation has even powers of $x$ and $y$, let $y'=y^2$ and $x' = x^2$. Then $y'^2 + 1 = x'^2 + 2y'$. Rearranging gives $y'^2 - 2y' + 1 = x'^2$, or $(y'-1)^2=x'^2$. There are two cases: $y' \leq 1$ or $y' > 1$.
If $y' \leq 1$, taking the square root of both sides gives $1 - y' = x'$, and rearranging gives $x' + y' = 1$. Substituting back in $x'=x^2$ and $y'=y^2$ gives us $x^2+y^2=1$, the equation for a circle.
Similarly, if $y' > 1$, we take the square root of both sides to get $y' - 1 = x'$, or $y' - x' = 1$, which is equivalent to $y^2 - x^2 = 1$, a hyperbola.
Hence, our answer is $\boxed{\textbf{(D)}\ \text{a circle and a hyperbola}}$.
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D
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_8
| 2,022 |
The sequence $a_0,a_1,a_2,\cdots$ is a strictly increasing arithmetic sequence of positive integers such that\[2^{a_7}=2^{27} \cdot a_7.\]What is the minimum possible value of $a_2$?
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We can rewrite the given equation as $2^{a_7-27}=a_7$. Hence, $a_7$ must be a power of $2$ and larger than $27$. The first power of 2 that is larger than $27$, namely $32$, does satisfy the equation: $2^{32 - 27} = 2^5 = 32$. In fact, this is the only solution; $2^{a_7-27}$ is exponential whereas $a_7$ is linear, so their graphs will not intersect again.
Now, let the common difference in the sequence be $d$. Hence, $a_0 = 32 - 7d$ and $a_2 = 32 - 5d$. To minimize $a_2$, we maxmimize $d$. Since the sequence contains only positive integers, $32 - 7d > 0$ and hence $d \leq 4$. When $d = 4$, $a_2 = \boxed{\textbf{(B)}\ 12}$.
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12
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_9
| 2,022 |
Regular hexagon $ABCDEF$ has side length $2$. Let $G$ be the midpoint of $\overline{AB}$, and let $H$ be the midpoint of $\overline{DE}$. What is the perimeter of $GCHF$?
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Let the center of the hexagon be $O$. $\triangle AOB$, $\triangle BOC$, $\triangle COD$, $\triangle DOE$, $\triangle EOF$, and $\triangle FOA$ are all equilateral triangles with side length $2$. Thus, $CO = 2$, and $GO = \sqrt{AO^2 - AG^2} = \sqrt{3}$. By symmetry, $\angle COG = 90^{\circ}$. Thus, by the Pythagorean theorem, $CG = \sqrt{2^2 + \sqrt{3}^2} = \sqrt{7}$. Because $CO = OF$ and $GO = OH$, $CG = HC = FH = GF = \sqrt{7}$. Thus, the solution to our problem is $\sqrt{7} + \sqrt{7} + \sqrt{7} + \sqrt{7} = \boxed{\textbf{(D)}\ 4\sqrt7}$.
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4\sqrt{7}
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_10
| 2,022 |
Let $f(n) = \left( \frac{-1+i\sqrt{3}}{2} \right)^n + \left( \frac{-1-i\sqrt{3}}{2} \right)^n$, where $i = \sqrt{-1}$. What is $f(2022)$?
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Converting both summands to exponential form,\begin{align*} -1 + i\sqrt{3} &= 2e^{\frac{2\pi i}{3}}, \\ -1 - i\sqrt{3} &= 2e^{-\frac{2\pi i}{3}} = 2e^{\frac{4\pi i}{3}}. \end{align*}Notice that the two terms in the problem are two of the third roots of unity (that is, both of them equal $1$ when raised to the power of $3$). When we replace the summands with their exponential form, we get\[f(n) = \left(e^{\frac{2\pi i}{3}}\right)^n + \left(e^{\frac{4\pi i}{3}}\right)^n.\]When we substitute $n = 2022$, we get\[f(2022) = \left(e^{\frac{2\pi i}{3}}\right)^{2022} + \left(e^{\frac{4\pi i}{3}}\right)^{2022}.\]We can rewrite $2022$ as $3 \cdot 674$, how does that help?\[f(2022) = \left(e^{\frac{2\pi i}{3}}\right)^{3 \cdot 674} + \left(e^{\frac{4\pi i}{3}}\right)^{3 \cdot 674} = \left(\left(e^{\frac{2\pi i}{3}}\right)^{3}\right)^{674} + \left(\left(e^{\frac{4\pi i}{3}}\right)^{3}\right)^{674} = 1^{674} + 1^{674} = \boxed{\textbf{(E)} \ 2}.\]Since any third root of unity must cube to $1$.
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2
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_11
| 2,022 |
Kayla rolls four fair $6$-sided dice. What is the probability that at least one of the numbers Kayla rolls is greater than $4$ and at least two of the numbers she rolls are greater than $2$?
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We will subtract from one the probability that the first condition is violated and the probability that only the second condition is violated, being careful not to double-count the probability that both conditions are violated.
For the first condition to be violated, all four dice must read $4$ or less, which happens with probability $\left( \frac23 \right)^4 = \frac{16}{81}$.
For the first condition to be met but the second condition to be violated, at least one of the dice must read greater than $4$, but less than two of the dice can read greater than $2$. Therefore, one of the four die must read $5$ or $6$, while the remaining three dice must read $2$ or less, which happens with probability ${4 \choose 1} \left(\frac13\right) \left(\frac13\right)^3 = 4 \cdot \frac13 \cdot \frac{1}{27} = \frac{4}{81}$.
Therefore, the overall probability of meeting both conditions is $1 - \frac{16}{81} - \frac{4}{81} = \boxed{\textbf{(D)}\ \frac{61}{81}}$.
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\frac{61}{81}
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_12
| 2,022 |
The graph of $y=x^2+2x-15$ intersects the $x$-axis at points $A$ and $C$ and the $y$-axis at point $B$. What is $\tan(\angle ABC)$?
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First, find $A=(-5,0)$, $B=(0,-15)$, and $C=(3,0)$. Create vectors $\overrightarrow{BA}$ and $\overrightarrow{BC}.$ These can be reduced to $\langle -1, 3 \rangle$ and $\langle 1, 5 \rangle$, respectively. Then, we can use the dot product to calculate the cosine of the angle (where $\theta=\angle ABC$) between them:
\begin{align*} \langle -1, 3 \rangle \cdot \langle 1, 5 \rangle = 15-1 &= \sqrt{10}\sqrt{26}\cos(\theta),\\ \implies \cos (\theta) &= \frac{7}{\sqrt{65}}. \end{align*}
Thus,\[\tan(\angle ABC) = \sqrt{\frac{65}{49}-1}= \boxed{\textbf{(E)}\ \frac{4}{7}}.\]
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\frac{4}{7}
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_14
| 2,022 |
One of the following numbers is not divisible by any prime number less than $10.$ Which is it?
$\textbf{(A) } 2^{606}-1 \qquad\textbf{(B) } 2^{606}+1 \qquad\textbf{(C) } 2^{607}-1 \qquad\textbf{(D) } 2^{607}+1\qquad\textbf{(E) } 2^{607}+3^{607}$
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For $\textbf{(A)}$ modulo $3,$\begin{align*} 2^{606} - 1 & \equiv (-1)^{606} - 1 \\ & \equiv 1 - 1 \\ & \equiv 0 . \end{align*}Thus, $2^{606} - 1$ is divisible by $3.$
For $\textbf{(B)}$ modulo $5,$\begin{align*} 2^{606} + 1 & \equiv 2^{{\rm Rem} ( 606, \phi(5) )} + 1 \\ & \equiv 2^{{\rm Rem} ( 606, 4 )} + 1 \\ & \equiv 2^2 + 1 \\ & \equiv 0 . \end{align*}Thus, $2^{606} + 1$ is divisible by $5.$
For $\textbf{(D)}$ modulo $3,$\begin{align*} 2^{607} + 1 & \equiv (-1)^{607} + 1 \\ & \equiv - 1 + 1 \\ & \equiv 0 . \end{align*}Thus, $2^{607} + 1$ is divisible by $3.$
For $\textbf{(E)}$ modulo $5,$\begin{align*} 2^{607} + 3^{607} & \equiv 2^{607} + (-2)^{607} \\ & \equiv 2^{607} - 2^{607} \\ & \equiv 0 . \end{align*}Thus, $2^{607} + 3^{607}$ is divisible by $5.$
Therefore, the answer is $\boxed{\textbf{(C) }2^{607} - 1}.$
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C
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_15
| 2,022 |
Suppose $x$ and $y$ are positive real numbers such that\[x^y=2^{64}\text{ and }(\log_2{x})^{\log_2{y}}=2^{7}.\]What is the greatest possible value of $\log_2{y}$?
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Take the base-two logarithm of both equations to get\[y\log_2 x = 64\quad\text{and}\quad (\log_2 y)(\log_2\log_2 x) = 7.\]Now taking the base-two logarithm of the first equation again yields\[\log_2 y + \log_2\log_2 x = 6.\]It follows that the real numbers $r:=\log_2 y$ and $s:=\log_2\log_2 x$ satisfy $r+s=6$ and $rs = 7$. Solving this system yields\[\{\log_2 y,\log_2\log_2 x\}\in\{3-\sqrt 2, 3 + \sqrt 2\}.\]Thus the largest possible value of $\log_2 y$ is $3+\sqrt 2 \implies \boxed{\textbf {(C)}}$.
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3 + \sqrt{2}
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https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_16
| 2,022 |
How many $4 \times 4$ arrays whose entries are $0$s and $1$s are there such that the row sums (the sum of the entries in each row) are $1, 2, 3,$ and $4,$ in some order, and the column sums (the sum of the entries in each column) are also $1, 2, 3,$ and $4,$ in some order? For example, the array\[\left[ \begin{array}{cccc} 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 0 \\ \end{array} \right]\]satisfies the condition.
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Note that the arrays and the sum configurations have one-to-one correspondence. Furthermore, the row sum configuration and the column sum configuration are independent of each other. Therefore, the answer is $(4!)^2=\boxed{\textbf{(D) }576}.$
Remark
For any given sum configuration, we can uniquely reconstruct the array it represents. Conversely, for any array, it is clear that we can determine the unique sum configuration associated with it. Therefore, this establishes a one-to-one correspondence between the arrays and the sum configurations.
|
576
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_17
| 2,022 |
In $\triangle{ABC}$ medians $\overline{AD}$ and $\overline{BE}$ intersect at $G$ and $\triangle{AGE}$ is equilateral. Then $\cos(C)$ can be written as $\frac{m\sqrt p}n$, where $m$ and $n$ are relatively prime positive integers and $p$ is a positive integer not divisible by the square of any prime. What is $m+n+p?$
|
Let $AG=AE=EG=2x$. Since $E$ is the midpoint of $\overline{AC}$, we must have $EC=2x$.
Since the centroid splits the median in a $2:1$ ratio, $GD=x$ and $BG=4x$.
Applying Law of Cosines on $\triangle ADC$ and $\triangle{}AGB$ yields $AB=\sqrt{28}x$ and $CD=BD=\sqrt{13}x$. Finally, applying Law of Cosines on $\triangle ABC$ yields $\cos(C)=\frac{5}{2\sqrt{13}}=\frac{5\sqrt{13}}{26}$. The requested sum is $5+13+26=44$.
|
44
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_19
| 2,022 |
Let $P(x)$ be a polynomial with rational coefficients such that when $P(x)$ is divided by the polynomial $x^2 + x + 1$, the remainder is $x+2$, and when $P(x)$ is divided by the polynomial $x^2+1$, the remainder is $2x+1$. There is a unique polynomial of least degree with these two properties. What is the sum of the squares of the coefficients of that polynomial?
|
Given that all the answer choices and coefficients are integers, we hope that $P(x)$ has positive integer coefficients.
Throughout this solution, we will express all polynomials in base $x$. E.g. $x^2 + x + 1 = 111_{x}$.
We are given:\[111a + 12 = 101b + 21 = P(x).\]We add $111$ and $101$ to each side and balance respectively:\[111(a - 1) + 123 = 101(b - 1) + 122 = P(x).\]We make the unit's digits equal:\[111(a - 1) + 123 = 101(b - 2) + 223 = P(x).\]We now notice that:\[111(a - 11) + 1233 = 101(b - 12) + 1233 = P(x).\]Therefore $a = 11_{x} = x + 1$, $b = 12_{x} = x + 2$, and $P(x) = 1233_{x} = x^3 + 2x^2 + 3x + 3$. $3$ is the minimal degree of $P(x)$ since there is no way to influence the $x$‘s digit in $101b + 21$ when $b$ is an integer. The desired sum is $1^2 + 2^2 +3^2+ 3^2 = \boxed{\textbf{(E)} \ 23}$
P.S. The four computational steps can be deduced through quick experimentation.
|
23
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_20
| 2,022 |
Let $S$ be the set of circles in the coordinate plane that are tangent to each of the three circles with equations $x^{2}+y^{2}=4$, $x^{2}+y^{2}=64$, and $(x-5)^{2}+y^{2}=3$. What is the sum of the areas of all circles in $S$?
|
[asy] import geometry; unitsize(0.5cm); void dc(pair x, pen p) { pair y = intersectionpoints(circle((0,0),8),(0,0)--1000*x)[0]; draw(circle(x, abs(x-y)),p+linewidth(2)); } pair O1 = (0,0),O2=(5,0),P1=intersectionpoints(circle(O1,5),circle(O2,3+sqrt(3)))[0],P2=intersectionpoints(circle(O1,3),circle(O2,5+sqrt(3)))[0],P3=intersectionpoints(circle(O1,5),circle(O2,3-sqrt(3)))[0],P4=intersectionpoints(circle(O1,3),circle(O2,5-sqrt(3)))[0]; draw(circle(O1,2)); draw(circle(O1,8)); draw(circle(O2,sqrt(3))); dc(P1,blue); dc(P2,red); dc(P3,mediumgreen); dc(P4,brown); [/asy]The circles match up as follows: Case $1$ is brown, Case $2$ is blue, Case $3$ is green, and Case 4 is red. Let $x^2 + y^2 = 64$ be circle $O$, $x^2 + y^2 = 4$ be circle $P$, and $(x-5)^2 + y^2 = 3$ be circle $Q$. All the circles in S are internally tangent to circle $O$. There are four cases with two circles belonging to each:
$*$ $P$ and $Q$ are internally tangent to $S$.
$*$ $P$ and $Q$ are externally tangent to $S$.
$*$ $P$ is externally and Circle $Q$ is internally tangent to $S$.
$*$ $P$ is internally and Circle $Q$ is externally tangent to $S$.
Consider Cases $1$ and $4$ together. Since circles $O$ and $P$ have the same center, the line connecting the center of $S$ and the center of $O$ will pass through the tangency point of both $S$ and $O$ and the tangency point of $S$ and $P$. This line will be the diameter of $S$ and have length $r_P + r_O = 10$. Therefore the radius of $S$ in these cases is $5$.
Consider Cases $2$ and $3$ together. Similarly to Cases $1$ and $4$, the line connecting the center of $S$ to the center of $O$ will pass through the tangency points. This time, however, the diameter of $S$ will have length $r_P-r_O=6$. Therefore, the radius of $S$ in these cases is $3$.
The set of circles $S$ consists of $8$ circles - $4$ of which have radius $5$ and $4$ of which have radius $3$. The total area of all circles in $S$ is $4(5^2\pi + 3^2\pi) = 136\pi \Rightarrow \boxed{\textbf{(E)}}$.
|
136 \pi
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_21
| 2,022 |
Ant Amelia starts on the number line at $0$ and crawls in the following manner. For $n=1,2,3,$ Amelia chooses a time duration $t_n$ and an increment $x_n$ independently and uniformly at random from the interval $(0,1).$ During the $n$th step of the process, Amelia moves $x_n$ units in the positive direction, using up $t_n$ minutes. If the total elapsed time has exceeded $1$ minute during the $n$th step, she stops at the end of that step; otherwise, she continues with the next step, taking at most $3$ steps in all. What is the probability that Amelia’s position when she stops will be greater than $1$?
|
Let $x$ and $y$ be random variables that are independently and uniformly distributed in the interval $(0,1).$ Note that\[P(x+y\leq 1)=\frac{\frac12\cdot1^2}{1^2}=\frac12,\]as shown below:[asy] /* Made by MRENTHUSIASM */ size(200); real xMin = -0.25; real xMax = 1.25; real yMin = -0.25; real yMax = 1.25; //Draws the horizontal ticks void horizontalTicks() { for (real i = 1; i < yMax; ++i) { draw((-1/32,i)--(1/32,i), black+linewidth(1)); } } //Draws the vertical ticks void verticalTicks() { for (real i = 1; i < xMax; ++i) { draw((i,-1/32)--(i,1/32), black+linewidth(1)); } } horizontalTicks(); verticalTicks(); label("$0$",(0,0),2*SW); label("$1$",(1,0),2*S); label("$1$",(0,1),2*W); fill((0,0)--(1,0)--(0,1)--cycle,yellow); draw((0,1)--(1,1)^^(1,0)--(1,1),dashed); draw((0,1)--(1,0)); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(8)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(8)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2)); [/asy]Let $x,y,$ and $z$ be random variables that are independently and uniformly distributed in the interval $(0,1).$ Note that\[P(x+y+z\leq 1)=\frac{\frac13\cdot\left(\frac12\cdot1^2\right)\cdot1}{1^3}=\frac16,\]as shown below:[asy] /* Made by MRENTHUSIASM */ size(200); import graph3; import solids; currentprojection=orthographic((0.3,0.1,0.1)); draw(surface((1,0,0)--(0,1,0)--(0,0,1)--cycle),yellow); draw(surface((1,0,0)--(0,1,0)--(0,0,0)--cycle),yellow); draw(surface((1,0,0)--(0,0,1)--(0,0,0)--cycle),yellow); draw(surface((0,1,0)--(0,0,1)--(0,0,0)--cycle),yellow); draw((0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--cycle,dashed); draw((0,1,0)--(1,1,0)--(1,0,0),dashed); draw((0,1,1)--(0,1,0)^^(1,1,1)--(1,1,0)^^(1,0,1)--(1,0,0),dashed); draw((-0.5,0,0)--(1.5,0,0),linewidth(1.25),EndArrow3(10)); draw((0,-0.5,0)--(0,1.5,0),linewidth(1.25),EndArrow3(10)); draw((0,0,-0.5)--(0,0,1.5),linewidth(1.25),EndArrow3(10)); draw((-0.1,0,1)--(0.1,0,1),linewidth(1)); draw((0,1,-0.1)--(0,1,0.1),linewidth(1)); draw((1,-0.1,0)--(1,0.1,0),linewidth(1)); label("$x$",(1.5,0,0),4*dir((1.5,0,0))); label("$y$",(0,1.5,0),2*dir((0,1.5,0))); label("$z$",(0,0,1.5),2*dir((0,0,1.5))); label("$0$",(0,0,0),2*dir((0,0.5,-0.5))); label("$1$",(1,0,0),4*dir((0,-1,0))); label("$1$",(0,1,0),4*dir((0,0,-1))); label("$1$",(0,0,1),5*dir((-1,0,0))); draw((1,0,0)--(0,1,0)--(0,0,1)--cycle); [/asy]We have two cases:
Amelia takes exactly $2$ steps.
We need $x_1+x_2>1$ and $t_1+t_2>1.$ So, the probability is\[P(x_1+x_2>1)\cdot P(t_1+t_2>1)=\left(1-\frac12\right)\cdot\left(1-\frac12\right)=\frac14.\]
Amelia takes exactly $3$ steps.
We need $x_1+x_2+x_3>1$ and $t_1+t_2\leq1.$ So, the probability is\[P(x_1+x_2+x_3>1)\cdot P(t_1+t_2\leq1)=\left(1-\frac16\right)\cdot\frac12=\frac{5}{12}.\]
Together, the answer is $\frac14 + \frac{5}{12} = \boxed{\textbf{(C) }\frac{2}{3}}.$
|
\frac{2}{3}
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_22
| 2,022 |
Let $x_0,x_1,x_2,\dotsc$ be a sequence of numbers, where each $x_k$ is either $0$ or $1$. For each positive integer $n$, define\[S_n = \sum_{k=0}^{n-1} x_k 2^k\]Suppose $7S_n \equiv 1 \pmod{2^n}$ for all $n \geq 1$. What is the value of the sum\[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}?\]
|
In binary numbers, we have\[S_n = (x_{n-1} x_{n-2} x_{n-3} x_{n-4} \ldots x_{2} x_{1} x_{0})_2.\]It follows that\[8S_n = (x_{n-1} x_{n-2} x_{n-3} x_{n-4} \ldots x_{2} x_{1} x_{0}000)_2.\]We obtain $7S_n$ by subtracting the equations:\[\begin{array}{clccrccccccr} & (x_{n-1} & x_{n-2} & x_{n-3} & x_{n-4} & \ldots & x_2 & x_1 & x_0 & 0 & 0 & 0 \ )_2 \\ -\quad & & & & (x_{n-1} & \ldots & x_5 & x_4 & x_3 & x_2 & x_1 & x_0)_2 \\ \hline & & & & & & & & & & & \\ [-2.5ex] & ( \ \ ?& ? & ? & 0 \ \ \ & \ldots & 0 & 0 & 0 & 0 & 0 & 1 \ )_2 \\ \end{array}\]We work from right to left:\begin{alignat*}{6} x_0=x_1=x_2=1 \quad &\implies \quad &x_3 &= 0& \\ \quad &\implies \quad &x_4 &= 1& \\ \quad &\implies \quad &x_5 &= 1& \\ \quad &\implies \quad &x_6 &= 0& \\ \quad &\implies \quad &x_7 &= 1& \\ \quad &\implies \quad &x_8 &= 1& \\ \quad &\quad \vdots & & & \end{alignat*}For all $n\geq3,$ we conclude that
$x_n=0$ if and only if $n\equiv 0\pmod{3}.$
$x_n=1$ if and only if $n\not\equiv 0\pmod{3}.$
Finally, we get $(x_{2019},x_{2020},x_{2021},x_{2022})=(0,1,1,0),$ from which\[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \boxed{\textbf{(A) } 6}.\]
|
6
|
https://artofproblemsolving.com/wiki/index.php/2022_AMC_12B_Problems/Problem_23
| 2,022 |
Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$. Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet?
|
This is a $d=st$ problem, so let $x$ be the time it takes to meet. We can write the following equation:\[12x+18x=45\]Solving gives us $x=1.5$. The $18x$ is Alicia so $18\times1.5=\boxed{\textbf{(E) 27}}$
|
27
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https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_1
| 2,023 |
Problem 2
The weight of $\frac{1}{3}$ of a large pizza together with $3 \frac{1}{2}$ cups of orange slices is the same as the weight of $\frac{3}{4}$ of a large pizza together with $\frac{1}{2}$ cup of orange slices. A cup of orange slices weighs $\frac{1}{4}$ of a pound. What is the weight, in pounds, of a large pizza?
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Use a system of equations. Let $x$ be the weight of a pizza and $y$ be the weight of a cup of orange slices. We have\[\frac{1}{3}x+\frac{7}{2}y=\frac{3}{4}x+\frac{1}{2}y.\]Rearranging, we get\begin{align*} \frac{5}{12}x&=3y, \\ x&=\frac{36}{5}y. \end{align*}Plugging in $\frac{1}{4}$ pounds for $y$ by the given gives $\frac{9}{5}=\boxed{\textbf{(A) }1\frac{4}{5}}.$
|
\frac{9}{5}
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_2
| 2,023 |
How many positive perfect squares less than $2023$ are divisible by $5$?
|
Since $\left \lfloor{\sqrt{2023}}\right \rfloor = 44$, there are $\left \lfloor{\frac{44}{5}}\right \rfloor = \boxed{\textbf{(A) 8}}$ perfect squares less than 2023 that are divisible by 5.
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8
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https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_3
| 2,023 |
How many digits are in the base-ten representation of $8^5 \cdot 5^{10} \cdot 15^5$?
|
Prime factorizing this gives us $2^{15}\cdot3^{5}\cdot5^{15}=10^{15}\cdot3^5=243\cdot10^{15}$.
$10^{15}$ has $16$ digits and $243$ = $2.43*10^{2}$ gives us $3$ more digits. $16+2=\text{\boxed{\textbf{(E) }18}}$
$2.43*10^{17}$ has $18$ digits
|
18
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https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_4
| 2,023 |
Janet rolls a standard $6$-sided die $4$ times and keeps a running total of the numbers she rolls. What is the probability that at some point, her running total will equal $3$?
|
There are $3$ cases where the running total will equal $3$: one roll; two rolls; or three rolls:
Case 1: The chance of rolling a running total of $3$, namely $(3)$ in exactly one roll is $\frac{1}{6}$.
Case 2: The chance of rolling a running total of $3$ in exactly two rolls, namely $(1, 2)$ and $(2, 1)$ is $\frac{1}{6}\cdot\frac{1}{6}\cdot2=\frac{1}{18}$.
Case 3: The chance of rolling a running total of 3 in exactly three rolls, namely $(1, 1, 1)$ is $\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{216}$.
Using the rule of sum we have $\frac{1}{6}+\frac{1}{18}+\frac{1}{216}=\boxed{\textbf{(B) }\frac{49}{216}}$.
|
\frac{49}{216}
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_5
| 2,023 |
Points $A$ and $B$ lie on the graph of $y=\log_{2}x$. The midpoint of $\overline{AB}$ is $(6, 2)$. What is the positive difference between the $x$-coordinates of $A$ and $B$?
|
Let $A(6+m,2+n)$ and $B(6-m,2-n)$, since $(6,2)$ is their midpoint. Thus, we must find $2m$. We find two equations due to $A,B$ both lying on the function $y=\log_{2}x$. The two equations are then $\log_{2}(6+m)=2+n$ and $\log_{2}(6-m)=2-n$. Now add these two equations to obtain $\log_{2}(6+m)+\log_{2}(6-m)=4$. By logarithm rules, we get $\log_{2}((6+m)(6-m))=4$. By raising 2 to the power of both sides, we obtain $(6+m)(6-m)=16$. We then get\[36-m^2=16 \rightarrow m^2=20 \rightarrow m=2\sqrt{5}\]. Since we're looking for $2m$, we obtain $(2)(2\sqrt{5})=\boxed{\textbf{(D) }4\sqrt{5}}$
|
4\sqrt{5}
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_6
| 2,023 |
A digital display shows the current date as an $8$-digit integer consisting of a $4$-digit year, followed by a $2$-digit month, followed by a $2$-digit date within the month. For example, Arbor Day this year is displayed as 20230428. For how many dates in $2023$ will each digit appear an even number of times in the 8-digital display for that date?
|
Do careful casework by each month. In the month and the date, we need a $0$, a $3$, and two digits repeated (which has to be $1$ and $2$ after consideration). After the casework, we get $\boxed{\textbf{(E)}~9}$. For curious readers, the numbers (in chronological order) are:
20230113
20230131
20230223
20230311
20230322
20231013
20231031
20231103
20231130
|
9
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_7
| 2,023 |
Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1$. If she scores an $11$ on each of the next three quizzes, her mean will increase by $2$. What is the mean of her quiz scores currently?
|
Let $a$ represent the amount of tests taken previously and $x$ the mean of the scores taken previously.
We can write the following equations:
\[\frac{ax+11}{a+1}=x+1\qquad (1)\]\[\frac{ax+33}{a+3}=x+2\qquad (2)\]
Multiplying equation $(1)$ by $(a+1)$ and solving, we get:\[ax+11=ax+a+x+1\]\[11=a+x+1\]\[a+x=10\qquad (3)\]
Multiplying equation $(2)$ by $(a+3)$ and solving, we get:\[ax+33=ax+2a+3x+6\]\[33=2a+3x+6\]\[2a+3x=27\qquad (4)\]
Solving the system of equations for $(3)$ and $(4)$, we find that $a=3$ and $x=\boxed{\textbf{(D) }7}$.
|
7
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_8
| 2,023 |
Positive real numbers $x$ and $y$ satisfy $y^3=x^2$ and $(y-x)^2=4y^2$. What is $x+y$?
|
Because $y^3=x^2$, set $x=a^3$, $y=a^2$ ($a\neq 0$). Put them in $(y-x)^2=4y^2$ we get $(a^2(a-1))^2=4a^4$ which implies $a^2-2a+1=4$. Solve the equation to get $a=3$ or $-1$. Since $x$ and $y$ are positive, $a=3$ and $x+y=3^3+3^2=\boxed{\textbf{(D)} 36}$.
|
36
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_10
| 2,023 |
What is the degree measure of the acute angle formed by lines with slopes $2$ and $\frac{1}{3}$?
|
Remind that $\text{slope}=\dfrac{\Delta y}{\Delta x}=\tan \theta$ where $\theta$ is the angle between the slope and $x$-axis. $k_1=2=\tan \alpha$, $k_2=\dfrac{1}{3}=\tan \beta$. The angle formed by the two lines is $\alpha-\beta$. $\tan(\alpha-\beta)=\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}=\dfrac{2-1/3}{1+2\cdot 1/3}=1$. Therefore, $\alpha-\beta=\boxed{\textbf{(C)} 45^\circ}$.
|
45
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https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_11
| 2,023 |
What is the value of\[2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3?\]
|
To solve this problem, we will be using difference of cube, sum of squares and sum of arithmetic sequence formulas.
\[2^3-1^3+4^3-3^3+6^3-5^3+...+18^3-17^3\]
$=(2-1)(2^2+1 \cdot 2+1^2)+(4-3)(4^2+4 \cdot 3+3^2)+(6-5)(6^2+6 \cdot 5+5^2)+...+(18-17)(18^2+18 \cdot 17+17^2)$
$=(2^2+1 \cdot 2+1^2)+(4^2+4 \cdot 3+3^2)+(6^2+6 \cdot 5+5^2)+...+(18^2+18 \cdot 17+17^2)$
$=1^2+2^2+3^2+4^2+5^2+6^2...+17^2+18^2+1 \cdot 2+3 \cdot 4+5 \cdot 6+...+17 \cdot 18$
$=\frac{18(18+1)(36+1)}{6}+1 \cdot 2+3 \cdot 4+5 \cdot 6+...+17 \cdot 18$
we could rewrite the second part as $\sum_{n=1}^{9}(2n-1)(2n)$
$(2n-1)(2n)=4n^2-2n$
$\sum_{n=1}^{9}4n^2=4(\frac{9(9+1)(18+1)}{6})$
$\sum_{n=1}^{9}-2n=-2(\frac{9(9+1)}{2})$
Hence,
$1 \cdot 2+3 \cdot 4+5 \cdot 6+...+17 \cdot 18 = 4(\frac{9(9+1)(18+1)}{6})-2(\frac{9(9+1)}{2})$
Adding everything up:
$2^3-1^3+4^3-3^3+6^3-5^3+...+18^3-17^3$
$=\frac{18(18+1)(36+1)}{6}+4(\frac{9(9+1)(18+1)}{6})-2(\frac{9(9+1)}{2})$
$=3(19)(37)+6(10)(19)-9(10)$
$=2109+1140-90$
$=\boxed{\textbf{(D) } 3159}$
|
3159
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_12
| 2,023 |
In a table tennis tournament, every participant played every other participant exactly once. Although there were twice as many right-handed players as left-handed players, the number of games won by left-handed players was $40\%$ more than the number of games won by right-handed players. (There were no ties and no ambidextrous players.) What is the total number of games played?
|
We know that the total amount of games must be the sum of games won by left and right handed players. Then, we can write $g = l + r$, and since $l = 1.4r$, $g = 2.4r$. Given that $r$ and $g$ are both integers, $g/2.4$ also must be an integer. From here we can see that $g$ must be divisible by 12, leaving only answers B and D. Now we know the formula for how many games are played in this tournament is $n(n-1)/2$, the sum of the first $n-1$ triangular numbers. Now, setting 36 and 48 equal to the equation will show that two consecutive numbers must have a product of 72 or 96. Clearly, $72=8*9$, so the answer is $\boxed{\textbf{(B) }36}$.
|
36
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https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_13
| 2,023 |
How many complex numbers satisfy the equation $z^5=\overline{z}$, where $\overline{z}$ is the conjugate of the complex number $z$?
|
When $z^5=\overline{z}$, there are two conditions: either $z=0$ or $z\neq 0$. When $z\neq 0$, since $|z^5|=|\overline{z}|$, $|z|=1$. $z^5\cdot z=z^6=\overline{z}\cdot z=|z|^2=1$. Consider the $r(\cos \theta +i\sin \theta)$ form, when $z^6=1$, there are 6 different solutions for $z$. Therefore, the number of complex numbers satisfying $z^5=\bar{z}$ is $\boxed{\textbf{(E)} 7}$.
|
7
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_14
| 2,023 |
Consider the set of complex numbers $z$ satisfying $|1+z+z^{2}|=4$. The maximum value of the imaginary part of $z$ can be written in the form $\tfrac{\sqrt{m}}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?
|
First, substitute in $z=a+bi$.
\[|1+(a+bi)+(a+bi)^2|=4\]\[|(1+a+a^2-b^2)+ (b+2ab)i|=4\]\[(1+a+a^2-b^2)^2+ (b+2ab)^2=16\]\[(1+a+a^2-b^2)^2+ b^2(1+4a+4a^2)=16\]
Let $p=b^2$ and $q=1+a+a^2$
\[(q-p)^2+ p(4q-3)=16\]\[p^2-2pq+q^2 + 4pq -3p=16\]
We are trying to maximize $b=\sqrt p$, so we'll turn the equation into a quadratic to solve for $p$ in terms of $q$.
\[p^2+(2q-3)p+(q^2-16)=0\]\[p=\frac{(-2q+3)\pm \sqrt{-12q+73}}{2}\]
We want to maximize $p$. Since $q$ is always negatively contributing to $p$'s value, we want to minimize $q$.
Due to the trivial inequality: $q=1+a+a^2=(a+\frac 12)^2+\frac{3}4 \geq \frac{3}4$
If we plug $q$'s minimum value in, we get that $p$'s maximum value is\[p=\frac{(-2(\frac 34)+3)+ \sqrt{-12(\frac 34)+73}}{2}=\frac{\frac 32+ 8}{2}=\frac{19}{4}\]
Then\[b=\frac{\sqrt{19}}{2}\]and\[m+n=\boxed{\textbf{(B)}~21}\]
|
21
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_16
| 2,023 |
Flora the frog starts at 0 on the number line and makes a sequence of jumps to the right. In any one jump, independent of previous jumps, Flora leaps a positive integer distance $m$ with probability $\frac{1}{2^m}$.
What is the probability that Flora will eventually land at 10?
|
Initially, the probability of landing at $10$ and landing past $10$ (summing the infinte series) are exactly the same. Landing before 10 repeats this initial condition, with a different irrelevant scaling factor. Therefore, the probability must be $\boxed{\textbf{(E)}~\frac12}$.
|
\frac{1}{2}
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_17
| 2,023 |
What is the product of all solutions to the equation\[\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023\]
|
For $\log_{7x}2023\cdot \log_{289x}2023=\log_{2023x}2023$, transform it into $\dfrac{\ln 289+\ln 7}{\ln 7 + \ln x}\cdot \dfrac{\ln 289+\ln 7}{\ln 289 + \ln x}=\dfrac{\ln 289+\ln 7}{\ln 289+\ln 7+\ln x}$. Replace $\ln x$ with $y$. Because we want to find the product of all solutions of $x$, it is equivalent to finding the exponential of the sum of all solutions of $y$. Change the equation to standard quadratic equation form, the term with 1 power of $y$ is canceled. By using Vieta, we see that since there does not exist a $by$ term, $\sum y=0$ and $\prod x=e^0=\boxed{\textbf{(C)} 1}$.
|
1
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_19
| 2,023 |
If $A$ and $B$ are vertices of a polyhedron, define the distance $d(A,B)$ to be the minimum number of edges of the polyhedron one must traverse in order to connect $A$ and $B$. For example, if $\overline{AB}$ is an edge of the polyhedron, then $d(A, B) = 1$, but if $\overline{AC}$ and $\overline{CB}$ are edges and $\overline{AB}$ is not an edge, then $d(A, B) = 2$. Let $Q$, $R$, and $S$ be randomly chosen distinct vertices of a regular icosahedron (regular polyhedron made up of 20 equilateral triangles). What is the probability that $d(Q, R) > d(R, S)$?
|
To find the total amount of vertices we first find the amount of edges, and that is $\frac{20 \times 3}{2}$. Next, to find the amount of vertices we can use Euler's characteristic, $V - E + F = 2$, and therefore the amount of vertices is $12$
So there are $P(12,3) = 1320$ ways to choose 3 distinct points.
Now, the furthest distance we can get from one point to another point in an icosahedron is 3. Which gives us a range of $1 \leq d(Q, R), d(R, S) \leq 3$
With some case work, we get two cases:
Case 1: $d(Q, R) = 3; d(R, S) = 1, 2$
Since we have only one way to choose Q, that is, the opposite point from R, we have one option for Q and any of the other points could work for S.
Then, we get $12 \times 1 \times 10 = 120$ (ways to choose R × ways to choose Q × ways to choose S)
Case 2: $d(Q, R) = 2; d(R, S) = 1$
We can visualize the icosahedron as 4 rows, first row with 1 vertex, second row with 5 vertices, third row with 5 vertices and fourth row with 1 vertex. We set R as the one vertex on the first row, and we have 12 options for R. Then, Q can be any of the 5 points on the third row and finally S can be one of the 5 points on the second row.
Therefore, we have $12 \times 5 \times 5 = 300$ (ways to choose R × ways to choose Q × ways to choose S)
Hence, $P(d(Q, R)>d(R, S)) = \frac{120+300}{1320} = \boxed{\textbf{(A) } \frac{7}{22}}$
|
\frac{7}{22}
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_21
| 2,023 |
Let $f$ be the unique function defined on the positive integers such that\[\sum_{d\mid n}d\cdot f\left(\frac{n}{d}\right)=1\]for all positive integers $n$. What is $f(2023)$?
|
First, we note that $f(1) = 1$, since the only divisor of $1$ is itself.
Then, let's look at $f(p)$ for $p$ a prime. We see that\[\sum_{d \mid p} d \cdot f\left(\frac{p}{d}\right) = 1\]\[1 \cdot f(p) + p \cdot f(1) = 1\]\[f(p) = 1 - p \cdot f(1)\]\[f(p) = 1-p\]Nice.
Now consider $f(p^k)$, for $k \in \mathbb{N}$.\[\sum_{d \mid p^k} d \cdot f\left(\frac{p^k}{d}\right) = 1\]\[1 \cdot f(p^k) + p \cdot f(p^{k-1}) + p^2 \cdot f(p^{k-2}) + \dotsc + p^k f(1) = 1\].
It can be (strongly) inductively shown that $f(p^k) = f(p) = 1-p$. Here's how.
We already showed $k=1$ works. Suppose it holds for $k = n$, then
\[1 \cdot f(p^n) + p \cdot f(p^{n-1}) + p^2 \cdot f(p^{n-2}) + \dotsc + p^n f(1) = 1 \implies f(p^m) = 1-p \; \forall \; m \leqslant n\]
For $k = n+1$, we have
\[1 \cdot f(p^{n+1}) + p \cdot f(p^{n}) + p^2 \cdot f(p^{n-1}) + \dotsc + p^{n+1} f(1) = 1\], then using $f(p^m) = 1-p \; \forall \; m \leqslant n$, we simplify to
\[1 \cdot f(p^{n+1}) + p \cdot (1-p) + p^2 \cdot (1-p) + \dotsc + p^n \cdot (1-p) + p^{n+1} f(1) = 1\]\[f(p^{n+1}) + \sum_{i=1}^n p^i (1-p) + p^{n+1} = 1\]\[f(p^{n+1}) + p(1 - p^n) + p^{n+1} = 1\]\[f(p^{n+1}) + p = 1 \implies f(p^{n+1}) = 1-p\].
Very nice! Now, we need to show that this function is multiplicative, i.e. $f(pq) = f(p) \cdot f(q)$ for $\textbf{distinct}$ $p,q$ prime. It's pretty standard, let's go through it quickly.\[\sum_{d \mid pq} d \cdot f\left(\frac{pq}{d}\right) = 1\]\[1 \cdot f(pq) + p \cdot f(q) + q \cdot f(p) + pq \cdot f(1) = 1\]Using our formulas from earlier, we have\[f(pq) + p(1-q) + q(1-p) + pq = 1 \implies f(pq) = 1 - p(1-q) - q(1-p) - pq = (1-p)(1-q) = f(p) \cdot f(q)\]
Great! We're almost done now. Let's actually plug in $2023 = 7 \cdot 17^2$ into the original formula.\[\sum_{d \mid 2023} d \cdot f\left(\frac{2023}{d}\right) = 1\]\[1 \cdot f(2023) + 7 \cdot f(17^2) + 17 \cdot f(7 \cdot 17) + 7 \cdot 17 \cdot f(17) + 17^2 \cdot f(7) + 7 \cdot 17^2 \cdot f(1) = 1\]Let's use our formulas! We know\[f(7) = 1-7 = -6\]\[f(17) = 1-17 = -16\]\[f(7 \cdot 17) = f(7) \cdot f(17) = (-6) \cdot (-16) = 96\]\[f(17^2) = f(17) = -16\]
So plugging ALL that in, we have\[f(2023) = 1 - \left(7 \cdot (-16) + 17 \cdot (-6) \cdot (-16) + 7 \cdot 17 \cdot (-16) + 17^2 \cdot (-6) + 7 \cdot 17^2\right)\]
which, be my guest simplifying, is $\boxed{\textbf{(B)} \ 96}$
|
96
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_22
| 2,023 |
How many ordered pairs of positive real numbers $(a,b)$ satisfy the equation\[(1+2a)(2+2b)(2a+b) = 32ab?\]
|
Using the AM-GM inequality on the two terms in each factor on the left-hand side, we get\[(1+2a)(2+2b)(2a+b) \ge 8\sqrt{2a \cdot 4b \cdot 2ab}= 32ab,\]This means the equality condition must be satisfied. Therefore, we must have $1 = 2a = b$, so the only solution is $\boxed{\textbf{(B) }1}$.
|
1
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_23
| 2,023 |
Let $K$ be the number of sequences $A_1$, $A_2$, $\dots$, $A_n$ such that $n$ is a positive integer less than or equal to $10$, each $A_i$ is a subset of $\{1, 2, 3, \dots, 10\}$, and $A_{i-1}$ is a subset of $A_i$ for each $i$ between $2$ and $n$, inclusive. For example, $\{\}$, $\{5, 7\}$, $\{2, 5, 7\}$, $\{2, 5, 7\}$, $\{2, 5, 6, 7, 9\}$ is one such sequence, with $n = 5$.What is the remainder when $K$ is divided by $10$?
|
Consider any sequence with $n$ terms. Every 10 number has such choices: never appear, appear the first time in the first spot, appear the first time in the second spot… and appear the first time in the $n$th spot, which means every number has $(n+1)$ choices to show up in the sequence. Consequently, for each sequence with length $n$, there are $(n+1)^{10}$ possible ways.
Thus, the desired value is $\sum_{i=1}^{10}(i+1)^{10}\equiv \boxed{\textbf{(C) } 5}\pmod{10}$
|
5
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_24
| 2,023 |
There is a unique sequence of integers $a_1, a_2, \cdots a_{2023}$ such that\[\tan2023x = \frac{a_1 \tan x + a_3 \tan^3 x + a_5 \tan^5 x + \cdots + a_{2023} \tan^{2023} x}{1 + a_2 \tan^2 x + a_4 \tan^4 x \cdots + a_{2022} \tan^{2022} x}\]whenever $\tan 2023x$ is defined. What is $a_{2023}?$
|
\begin{align*} \cos 2023 x + i \sin 2023 x &= (\cos x + i \sin x)^{2023}\\ &= \cos^{2023} x + \binom{2023}{1} \cos^{2022} x (i\sin x) + \binom{2023}{2} \cos^{2021} x (i \sin x)^{2} +\binom{2020}{3} \cos^{2020} x (i \sin x)^{3}\\ &+ \dots + \binom{2023}{2022} \cos x (i \sin x)^{2022} + (i \sin x)^{2023}\\ &= \cos^{2023} x + i \binom{2023}{1} \cos^{2022} x \sin x - \binom{2023}{2} \cos^{2021} x \sin^{2} x - i\binom{2023}{3} \cos^{2020} x \sin^{3} x + \dots\\ &- \binom{2023}{2022} \cos x \sin^{2022} x - i \sin^{2023} x\\ \end{align*}
By equating real and imaginary parts:
\[\cos 2023 x = \cos^{2023} x - \binom{2023}{2} \cos^{2021} x \sin^{2} x + \dots - \binom{2023}{2022} \cos x \sin^{2022} x\]
\[\sin 2023 x = \binom{2023}{1} \cos^{2022} x \sin x - \binom{2023}{3} \cos^{2020} x \sin^{3} x + \dots - \sin^{2023} x\]
\begin{align*} \tan2023x &= \frac{ \sin2023x }{ \cos2023x } = \frac{ \binom{2023}{1} \cos^{2022} x \sin x - \binom{2023}{3} \cos^{2020} x \sin^{3} x + \dots - \sin^{2023} x }{ \cos^{2023} x - \binom{2023}{2} \cos^{2021} x \sin^{2} x + \dots - \binom{2023}{2022} \cos x \sin^{2022} x }\\ &= \frac{ \binom{2023}{1} \frac{\cos^{2022} x \sin x}{\cos^{2023} x} - \binom{2023}{3} \frac{\cos^{2020} x \sin^{3} x}{\cos^{2023} x} + \dots - \frac{\sin^{2023} x}{\cos^{2023} x} }{ \frac{\cos^{2023} x}{\cos^{2023} x} - \binom{2023}{2} \frac{\cos^{2021} x \sin^{2} x}{\cos^{2023} x} + \dots - \binom{2023}{2022} \frac{\cos x \sin^{2022} x}{\cos^{2023} x} }\\ &= \frac{ \binom{2023}{1} \tan x - \binom{2023}{3} \tan^{3}x + \dots - \tan^{2023}x }{ 1 - \binom{2023}{2} \tan^{2}x + \dots - \binom{2023}{2022} \tan^{2022} x }\\ \end{align*}
\[a_{2023} = \boxed{\textbf{(C)}-1}\]
This problem is the same as problem 7.64 in the Art of Problem Solving textbook Precalculus chapter 7 that asks to prove $\tan{nx} = \frac{\binom{n}{1}\tan{x} - \binom{n}{3}\tan^{3}{x} + \binom{n}{5}\tan^{5}{x} - \binom{n}{7}\tan^{7}{x} + \dots}{1 - \binom{n}{2}\tan^{2}{x} + \binom{n}{4}\tan^{4}{x} - \binom{n}{6}\tan^{6}{x} + \dots}$
|
-1
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_25
| 2,023 |
Mrs. Jones is pouring orange juice into four identical glasses for her four sons. She fills the first three glasses completely but runs out of juice when the fourth glass is only $\frac{1}{3}$ full. What fraction of a glass must Mrs. Jones pour from each of the first three glasses into the fourth glass so that all four glasses will have the same amount of juice?
|
The first three glasses each have a full glass. Let's assume that each glass has "1 unit" of juice. It won't matter exactly how much juice everyone has because we're dealing with ratios, and that wouldn't affect our answer. The fourth glass has a glass that is one third. So the total amount of juice will be $1+1+1+\dfrac{1}{3} = \dfrac{10}{3}$. If we divide the total amount of juice by 4, we get $\dfrac{5}{6}$, which should be the amount of juice in each glass. This means that each of the first three glasses will have to contribute $1 - \dfrac{5}{6} = \boxed{\dfrac{1}{6}}$ to the fourth glass.
|
\frac{1}{6}
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_1
| 2,023 |
Carlos went to a sports store to buy running shoes. Running shoes were on sale, with prices reduced by $20\%$ on every pair of shoes. Carlos also knew that he had to pay a $7.5\%$ sales tax on the discounted price. He had $$43$ dollars. What is the original (before discount) price of the most expensive shoes he could afford to buy?
|
We can create the equation:\[0.8x \cdot 1.075 = 43\]using the information given. This is because x, the original price, got reduced by 20%, or multiplied by 0.8, and it also got multiplied by 1.075 on the discounted price. Solving that equation, we get\[\frac{4}{5} \cdot x \cdot \frac{43}{40} = 43\]\[\frac{4}{5} \cdot x \cdot \frac{1}{40} = 1\]\[\frac{1}{5} \cdot x \cdot \frac{1}{10} = 1\]\[x = \boxed{50}\]
|
50
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_2
| 2,023 |
A $3-4-5$ right triangle is inscribed in circle $A$, and a $5-12-13$ right triangle is inscribed in circle $B$. What is the ratio of the area of circle $A$ to the area of circle $B$?
|
Because the triangles are right triangles, we know the hypotenuses are diameters of circles $A$ and $B$. Thus, their radii are 2.5 and 6.5 (respectively). Square the two numbers and multiply $\pi$ to get $6.25\pi$ and $42.25\pi$ as the areas of the circles. Multiply 4 on both numbers to get $25\pi$ and $169\pi$. Cancel out the $\pi$, and lastly, divide, to get your answer $=\boxed{\textbf{(D) }\frac{25}{169}}.$
|
\frac{25}{169}
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_3
| 2,023 |
Jackson's paintbrush makes a narrow strip with a width of $6.5$ millimeters. Jackson has enough paint to make a strip $25$ meters long. How many square centimeters of paper could Jackson cover with paint?
|
$6.5$ millimeters is equal to $0.65$ centimeters. $25$ meters is $2500$ centimeters. The answer is $0.65 \times 2500$, so the answer is $\boxed{\textbf{(C) 1,625}}$.
|
1,625
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_4
| 2,023 |
You are playing a game. A $2 \times 1$ rectangle covers two adjacent squares (oriented either horizontally or vertically) of a $3 \times 3$ grid of squares, but you are not told which two squares are covered. Your goal is to find at least one square that is covered by the rectangle. A "turn" consists of you guessing a square, after which you are told whether that square is covered by the hidden rectangle. What is the minimum number of turns you need to ensure that at least one of your guessed squares is covered by the rectangle?
|
Notice that the $3\times3$ square grid has a total of $12$ possible $2\times1$ rectangles.
Suppose you choose the middle square for one of your turns. The middle square is covered by $4$ rectangles, and each of the remaining $8$ squares is covered by a maximum of $2$ uncounted rectangles. This means that the number of turns is at least $1+\frac{12-4}{2}=1+4=5$.
Now suppose you don't choose the middle square. The squares on the middle of the sides are covered by at most 3 uncounted rectangles, and the squares on the corners are covered by at most 2 uncounted rectangles. In this case, we see that the least number of turns needed to account for all 12 rectangles is $12\div 3=4.$ To prove that choosing only side squares indeed does cover all 12 rectangles, we need to show that the 3 rectangles per square that cover each side square do not overlap. Drawing the rectangles that cover one square, we see they form a $T$ shape and they do not cover any other side square. Hence, our answer is $4.$
[asy] draw((0,0)--(0.5,0)--(0.5,0.5)--(0,0.5)--(0,0)); draw((0,1)--(0.5,1)--(0.5,1.5)--(0,1.5)--(0,1)); draw((0.5,0.5)--(1,0.5)--(1,1)--(0.5,1)--(0.5,0.5)); draw((1,0)--(1.5,0)--(1.5,0.5)--(1,0.5)--(1,0)); draw((1,1)--(1.5,1)--(1.5,1.5)--(1,1.5)--(1,1)); draw((0,0.5)--(0.5,0.5)--(0.5,1)--(0,1)--(0,0.5)); draw((0.5,0)--(1,0)--(1,0.5)--(0.5,0.5)--(0.5,0)); draw((0.5,1)--(1,1)--(1,1.5)--(0.5,1.5)--(0.5,1)); draw((1,0.5)--(1.5,0.5)--(1.5,1)--(1,1)--(1,0.5)); filldraw((0,0.5)--(0.5,0.5)--(0.5,1)--(0,1)--(0,0.5)--cycle, red, black+linewidth(1)); filldraw((0,0)--(0.5,0)--(0.5,0.5)--(0,0.5)--(0,0)--cycle, red, black+linewidth(1)); filldraw((0,1)--(0.5,1)--(0.5,1.5)--(0,1.5)--(0,1)--cycle, red, black+linewidth(1)); filldraw((0.5,0.5)--(1,0.5)--(1,1)--(0.5,1)--(0.5,0.5)--cycle, red, black+linewidth(1)); [/asy]
|
4
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_5
| 2,023 |
When the roots of the polynomial
\[P(x) = (x-1)^1 (x-2)^2 (x-3)^3 \cdot \cdot \cdot \cdot (x-10)^{10}\]
are removed from the number line, what remains is the union of $11$ disjoint open intervals. On how many of these intervals is $P(x)$ positive?
|
The expressions to the power of even powers are always positive, so we don't need to care about those. We only need to care about $(x-1)^1(x-3)^3(x-5)^5(x-7)^7(x-9)^9$. We need 0, 2, or 4 of the expressions to be negative. The 9 through 10 interval and 10 plus interval make all of the expressions positive. The 5 through 6 and 6 through 7 intervals make two of the expressions negative. The 1 through 2 and 2 through 3 intervals make four of the expressions negative. There are $\boxed{\textbf{(C) 6}}$ intervals.
|
6
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_6
| 2,023 |
For how many integers $n$ does the expression\[\sqrt{\frac{\log (n^2) - (\log n)^2}{\log n - 3}}\]represent a real number, where log denotes the base $10$ logarithm?
|
We have\begin{align*} \sqrt{\frac{\log \left( n^2 \right) - \left( \log n \right)^2}{\log n - 3}} & = \sqrt{\frac{2 \log n - \left( \log n \right)^2}{\log n - 3}} \\ & = \sqrt{\frac{\left( \log n \right) \left( 2 - \log n\right)}{\log n - 3}} . \end{align*}
Because $n$ is an integer and $\log n$ is well defined, $n$ must be a positive integer.
Case 1: $n = 1$ or $10^2$.
The above expression is 0. So these are valid solutions.
Case 2: $n \neq 1, 10^2$.
Thus, $\log n > 0$ and $2 - \log n \neq 0$. To make the above expression real, we must have $2 < \log n < 3$. Thus, $100 < n < 1000$. Thus, $101 \leq n \leq 999$. Hence, the number of solutions in this case is 899.
Putting all cases together, the total number of solutions is $\boxed{\textbf{(E) 901}}$.
|
901
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_7
| 2,023 |
How many nonempty subsets $B$ of $\{0, 1, 2, 3, \cdots, 12\}$ have the property that the number of elements in $B$ is equal to the least element of $B$? For example, $B = \{4, 6, 8, 11\}$ satisfies the condition.
|
There is no way to have a set with 0. If a set is to have its lowest element as 1, it must have only 1 element: 1. If a set is to have its lowest element as 2, it must have 2, and the other element will be chosen from the natural numbers between 3 and 12, inclusive. To calculate this, we do $\binom{10}{1}$. If the set is the have its lowest element as 3, the other 2 elements will be chosen from the natural numbers between 4 and 12, inclusive. To calculate this, we do $\binom{9}{2}$. We can see a pattern emerge, where the top is decreasing by 1 and the bottom is increasing by 1. In other words, we have to add $1 + \binom{10}{1} + \binom{9}{2} + \binom{8}{3} + \binom{7}{4} + \binom{6}{5}$. This is $1+10+36+56+35+6 = \boxed{\textbf{(D) 144}}$.
|
144
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_8
| 2,023 |
What is the area of the region in the coordinate plane defined by
$| | x | - 1 | + | | y | - 1 | \le 1$?
|
First consider, $|x-1|+|y-1| \le 1.$ We can see that it is a square with a side length of $\sqrt{2}$ (diagonal $2$). The area of the square is $\sqrt{2}^2 = 2.$
Next, we insert an absolute value sign into the equation and get $|x-1|+||y|-1| \le 1.$ This will double the square reflecting over x-axis.
So now we have $2$ squares.
Finally, we add one more absolute value and obtain $||x|-1|+||y|-1| \le 1.$ This will double the squares as we reflect the $2$ squares we already have over the y-axis.
Concluding, we have $4$ congruent squares. Thus, the total area is $4\cdot2 =$ $\boxed{\text{(B) 8}}$
|
8
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_9
| 2,023 |
In the $xy$-plane, a circle of radius $4$ with center on the positive $x$-axis is tangent to the $y$-axis at the origin, and a circle with radius $10$ with center on the positive $y$-axis is tangent to the $x$-axis at the origin. What is the slope of the line passing through the two points at which these circles intersect?
|
The center of the first circle is $(4,0)$. The center of the second circle is $(0,10)$. Thus, the slope of the line that passes through these two centers is $- \frac{10}{4} = - \frac{5}{2}$.
Because this line is the perpendicular bisector of the line that passes through two intersecting points of two circles, the slope of the latter line is $\frac{-1}{- \frac{5}{2}} = \boxed{\textbf{(E) } \frac{2}{5}}$.
|
\frac{2}{5}
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_10
| 2,023 |
What is the maximum area of an isosceles trapezoid that has legs of length $1$ and one base twice as long as the other?
|
Let the trapezoid be $ABCD$ with $AD = BC = 1, \; AB = x, CD = 2x$. Extend $AD$ and $BC$ to meet at point $E$. Then, notice $\triangle ABE \sim \triangle DCE$ with side length ratio $1:2$ and $AE = BE = 1$. Thus, $[DCE] = 4 \cdot [ABE]$ and $[ABCD] = [DCE] - [ABE] = \frac{3}{4} \cdot [DCE]$.
The problem reduces to maximizing the area of $[DCE]$, an isosceles triangle with legs of length $2$. Analyzing the sine area formula, this is clearly maximized when $\angle DEC = 90^{\circ}$, so $[DCE] = 2$ and $[ABCD] = \frac{3}{4} \cdot 2 = \boxed{\frac{3}{2}}.$
|
\frac{3}{2}
|
https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_11
| 2,023 |
For complex number $u = a+bi$ and $v = c+di$ (where $i=\sqrt{-1}$), define the binary operation
$u \otimes v = ac + bdi$
Suppose $z$ is a complex number such that $z\otimes z = z^{2}+40$. What is $|z|$?
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let $z$ = $a+bi$.
$z \otimes z = a^{2}+b^{2}i$.
This is equal to $z^{2} + 40 = a^{2}-b^{2}+40+2abi$
Since the real values have to be equal to each other, $a^{2}-b^{2}+40 = a^{2}$. Simple algebra shows $b^{2} = 40$, so $b$ is $2\sqrt{10}$.
The imaginary components must also equal each other, meaning $b^{2} = 2ab$, or $b = 2a$. This means $a = \frac{b}{2} = \sqrt{10}$.
Thus, the magnitude of z is $\sqrt{a^{2}+b^{2}} = \sqrt{50} = 5\sqrt{2}$ $=\text{\boxed{\textbf{(E) }5\sqrt{2}}}$
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5\sqrt{2}
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https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_12
| 2,023 |
A rectangular box $\mathcal{P}$ has distinct edge lengths $a$, $b$, and $c$. The sum of the lengths of all $12$ edges of $\mathcal{P}$ is $13$, the areas of all $6$ faces of $\mathcal{P}$ is $\frac{11}{2}$, and the volume of $\mathcal{P}$ is $\frac{1}{2}$. What is the length of the longest interior diagonal connecting two vertices of $\mathcal{P}$?
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[asy] import geometry; pair A = (-3, 4); pair B = (-3, 5); pair C = (-1, 4); pair D = (-1, 5); pair AA = (0, 0); pair BB = (0, 1); pair CC = (2, 0); pair DD = (2, 1); draw(D--AA,dashed); draw(A--B); draw(A--C); draw(B--D); draw(C--D); draw(A--AA); draw(B--BB); draw(C--CC); draw(D--DD); // Dotted vertices dot(A); dot(B); dot(C); dot(D); dot(AA); dot(BB); dot(CC); dot(DD); draw(AA--BB); draw(AA--CC); draw(BB--DD); draw(CC--DD); label("a",midpoint(D--DD),E); label("b",midpoint(CC--DD),E); label("c",midpoint(AA--CC),S); [/asy]
We can create three equations using the given information.\[4a+4b+4c = 13\]\[2ab+2ac+2bc=\frac{11}{2}\]\[abc=\frac{1}{2}\]We also know that we want $\sqrt{a^2 + b^2 + c^2}$ because that is the length that can be found from using the Pythagorean Theorem. We cleverly notice that $a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab+ac+bc)$. We know that $a+b+c = \frac{13}{4}$ and $2(ab+ac+bc)=\dfrac{11}2$, so $a^2 + b^2 + c^2 = \left(\frac{13}{4}\right)^2 - \frac{11}{2} = \frac{169-88}{16} = \frac{81}{16}$. So our answer is $\sqrt{\frac{81}{16}} = \boxed{\textbf{(D)}~\tfrac94}$.
Interestingly, we don't use the fact that the volume is $\frac{1}{2}$.
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\frac{9}{4}
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https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_13
| 2,023 |
For how many ordered pairs $(a,b)$ of integers does the polynomial $x^3+ax^2+bx+6$ have $3$ distinct integer roots?
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Denote three roots as $r_1 < r_2 < r_3$. Following from Vieta's formula, $r_1r_2r_3 = -6$.
Case 1: All roots are negative.
We have the following solution: $\left( -3, -2, -1 \right)$.
Case 2: One root is negative and two roots are positive.
We have the following solutions: $\left( -3, 1, 2 \right)$, $\left( -2, 1, 3 \right)$, $\left( -1, 2, 3 \right)$, $\left( -1, 1, 6 \right)$.
Putting all cases together, the total number of solutions is $\boxed{\textbf{(A) 5}}$.
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5
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https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_14
| 2,023 |
Suppose $a$, $b$, and $c$ are positive integers such that\[\frac{a}{14}+\frac{b}{15}=\frac{c}{210}.\]Which of the following statements are necessarily true?
I. If $\gcd(a,14)=1$ or $\gcd(b,15)=1$ or both, then $\gcd(c,210)=1$.
II. If $\gcd(c,210)=1$, then $\gcd(a,14)=1$ or $\gcd(b,15)=1$ or both.
III. $\gcd(c,210)=1$ if and only if $\gcd(a,14)=\gcd(b,15)=1$.
$\textbf{(A)}~\text{I, II, and III}\qquad\textbf{(B)}~\text{I only}\qquad\textbf{(C)}~\text{I and II only}\qquad\textbf{(D)}~\text{III only}\qquad\textbf{(E)}~\text{II and III only}$
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We examine each of the conditions.
The first condition is false. A simple counterexample is $a=3$ and $b=5$. The corresponding value of $c$ is $115$. Since $\gcd(3,14)=1$, condition $I$ would imply that $\gcd(c,210)=1.$ However, $\gcd(115,210)$ is clearly not $1$ (they share a common factor of $5$). Condition $I$ is false so that we can rule out choices $A,B,$ and $C$.
We now decide between the two answer choices $D$ and $E$. What differs between them is the validity of condition $II$, so it suffices to check $II$ simply.
We look at statement $II$'s contrapositive to prove it. The contrapositive states that if $\gcd(a,14)\neq1$ and $\gcd(b,15)\neq1$, then $\gcd(c,210)\neq1.$ In other words, if $a$ shares some common factor that is not $1$ with $14$ and $b$ shares some common factor that is not $1$ with $15$, then $c$ also shares a common factor that is not $1$ with $210$. Let's say that $a=a'\cdot n$, where $a'$ is a factor of $14$ not equal to $1$. (So $a'$ is the common factor.)
We can rewrite the given equation as $15a+14b=c\implies15(a'n)+14b=c.$ We can express $14$ as $a'\cdot n'$, for some positive integer $n'$ (this $n'$ can be $1$). We can factor $a'$ out to get $a'(15n+bn')=c.$
Since all values in this equation are integers, $c$ must be divisible by $a'$. Since $a'$ is a factor of $14$, $a'$ must also be a factor of $210$, a multiple of $14$. Therefore, we know that $c$ shares a common factor with $210$ (which is $a'$), so $\gcd(c,210)\neq1$. This is what $II$ states, so therefore $II$ is true.
Thus, our answer is $\boxed{\textbf{(E) }\text{II and III only}}.$
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E
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https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_15
| 2,023 |
In the state of Coinland, coins have values $6,10,$ and $15$ cents. Suppose $x$ is the value in cents of the most expensive item in Coinland that cannot be purchased using these coins with exact change. What is the sum of the digits of $x?$
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This problem asks to find largest $x$ that cannot be written as\[6 a + 10 b + 15 c = x, \hspace{1cm} (1)\]
where $a, b, c \in \Bbb Z_+$.
Denote by $r \in \left\{ 0, 1 \right\}$ the remainder of $x$ divided by 2. Modulo 2 on Equation (1), we get By using modulus $m \in \left\{ 2, 3, 5 \right\}$ on the equation above, we get $c \equiv r \pmod{2}$.
Following from Chicken McNugget's Theorem, we have that any number that is no less than $(3-1)(5-1) = 8$ can be expressed in the form of $3a + 5b$ with $a, b \in \Bbb Z_+$.
Therefore, all even numbers that are at least equal to $2 \cdot 8 + 15 \cdot 0 = 16$ can be written in the form of Equation (1) with $a, b, c \in \Bbb Z_+$. All odd numbers that are at least equal to $2 \cdot 8 + 15 \cdot 1 = 31$ can be written in the form of Equation (1) with $a, b, c \in \Bbb Z_+$.
The above two cases jointly imply that all numbers that are at least 30 can be written in the form of Equation (1) with $a, b, c \in \Bbb Z_+$.
Next, we need to prove that 29 cannot be written in the form of Equation (1) with $a, b, c \in \Bbb Z_+$.
Because 29 is odd, we must have $c \equiv 1 \pmod{2}$. Because $a, b, c \in \Bbb Z_+$, we must have $c = 1$. Plugging this into Equation (1), we get $3 a + 5 b = 7$. However, this equation does not have non-negative integer solutions.
All analysis above jointly imply that the largest $x$ that has no non-negative integer solution to Equation (1) is 29. Therefore, the answer is $2 + 9 = \boxed{\textbf{(D) 11}}$.
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11
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https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_16
| 2,023 |
Triangle $ABC$ has side lengths in arithmetic progression, and the smallest side has length $6.$ If the triangle has an angle of $120^\circ,$ what is the area of $ABC$?
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The length of the side opposite to the angle with $120^\circ$ is longest. We denote its value as $x$.
Because three side lengths form an arithmetic sequence, the middle-valued side length is $\frac{x + 6}{2}$.
Following from the law of cosines, we have\begin{align*} 6^2 + \left( \frac{x + 6}{2} \right)^2 - 2 \cdot 6 \cdot \frac{x + 6}{2} \cdot \cos 120^\circ = x^2 . \end{align*}
By solving this equation, we get $x = 14$. Thus, $\frac{x + 6}{2} = 10$.
Therefore, the area of the triangle is\begin{align*} \frac{1}{2} 6 \cdot 10 \cdot \sin 120^\circ = \boxed{\textbf{(E) } 15 \sqrt{3}} . \end{align*}
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15\sqrt{3}
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https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_17
| 2,023 |
Last academic year Yolanda and Zelda took different courses that did not necessarily administer the same number of quizzes during each of the two semesters. Yolanda's average on all the quizzes she took during the first semester was $3$ points higher than Zelda's average on all the quizzes she took during the first semester. Yolanda's average on all the quizzes she took during the second semester was $18$ points higher than her average for the first semester and was again $3$ points higher than Zelda's average on all the quizzes Zelda took during her second semester. Which one of the following statements cannot possibly be true?
$\textbf{(A)}$ Yolanda's quiz average for the academic year was $22$ points higher than Zelda's.
$\textbf{(B)}$ Zelda's quiz average for the academic year was higher than Yolanda's.
$\textbf{(C)}$ Yolanda's quiz average for the academic year was $3$ points higher than Zelda's.
$\textbf{(D)}$ Zelda's quiz average for the academic year equaled Yolanda's.
$\textbf{(E)}$ If Zelda had scored $3$ points higher on each quiz she took, then she would have had the same average for the academic year as Yolanda.
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Denote by $A_i$ the average of person with initial $A$ in semester $i \in \left\{1, 2 \right\}$ Thus, $Y_1 = Z_1 + 3$, $Y_2 = Y_1 + 18$, $Y_2 = Z_2 + 3$.
Denote by $A_{12}$ the average of person with initial $A$ in the full year. Thus, $Y_{12}$ can be any number in $\left( Y_1 , Y_2 \right)$ and $Z_{12}$ can be any number in $\left( Z_1 , Z_2 \right)$.
Therefore, the impossible solution is $\boxed{\textbf{(A)}~\text{Yolanda's quiz average for the academic year was 22 points higher than Zelda's.}}$
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A
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https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_18
| 2,023 |
Each of $2023$ balls is randomly placed into one of $3$ bins. Which of the following is closest to the probability that each of the bins will contain an odd number of balls?
$\textbf{(A) } \frac{2}{3} \qquad\textbf{(B) } \frac{3}{10} \qquad\textbf{(C) } \frac{1}{2} \qquad\textbf{(D) } \frac{1}{3} \qquad\textbf{(E) } \frac{1}{4}$
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Because each bin will have an odd number, they will have at least one ball. So we can put one ball in each bin prematurely. We then can add groups of 2 balls into each bin, meaning we now just have to spread 1010 pairs over 3 bins. This will force every bin to have an odd number of balls. Using stars and bars, we find that this is equal to $\binom{1012}{2}$. This is equal to $\frac{1012\cdot1011}{2}$. The total amount of ways would also be found using stars and bars. That would be $\binom{2023+3-1}{3-1} = \binom{2025}{2}$. Dividing our two quantities, we get $\frac{1012 \cdot 1011 \cdot 2}{2 \cdot 2025 \cdot 2024}$. We can roughly cancel $\frac{1012 \cdot 1011}{2025 \cdot 2024}$ to get $\frac{1}{4}$. The 2 in the numerator and denominator also cancels out, so we're left with $\boxed{\frac{1}{4}}$.
Note: My solution does not provide an exact probability, but is a good estimate, which is how this problem was designed. Most of the solutions on this page will give a decent estimate. This is due to the fact that the binomial expansions are symmetric for even vs odd. Try experimenting to determine why this is true.
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E
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https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_19
| 2,023 |
Cyrus the frog jumps $2$ units in a direction, then $2$ more in another direction. What is the probability that he lands less than $1$ unit away from his starting position?
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2023AMC12BP20.png
Let Cyrus's starting position be $S$. WLOG, let the place Cyrus lands at for his first jump be $O$. From $O$, Cyrus can reach all the points on $\odot O$. The probability that Cyrus will land less than $1$ unit away from $S$ is $\frac{4 \alpha }{ 2 \pi}$.
\[\sin \alpha = \frac{ \frac12 }{2} = \frac14, \quad \alpha = \arcsin \frac14\]
Therefore, the answer is
\[\frac{4 \arcsin \frac14 }{ 2 \pi} = \boxed{\textbf{(E) } \frac{2 \arcsin \frac{1}{4}}{\pi}}\]
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\frac{2 \arcsin \frac{1}{4}}{\pi}
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https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_20
| 2,023 |
A lampshade is made in the form of the lateral surface of the frustum of a right circular cone. The height of the frustum is $3\sqrt3$ inches, its top diameter is $6$ inches, and its bottom diameter is $12$ inches. A bug is at the bottom of the lampshade and there is a glob of honey on the top edge of the lampshade at the spot farthest from the bug. The bug wants to crawl to the honey, but it must stay on the surface of the lampshade. What is the length in inches of its shortest path to the honey?
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We augment the frustum to a circular cone. Denote by $O$ the apex of the cone. Denote by $A$ the bug and $B$ the honey.
By using the numbers given in this problem, the height of the cone is $6 \sqrt{3}$. Thus, $OA = 12$ and $OB = 6$.
We unfold the lateral face. So we get a circular sector. The radius is 12 and the length of the arc is $12\pi$. Thus, the central angle of this circular sector is $180^\circ$.
Because $A$ and $B$ are opposite in the original frustum, in the unfolded circular cone, $\angle AOB = \frac{180^\circ}{2} = 90^\circ$.
Notice that a feasible path between $A$ and $B$ can only fall into the region with the range of radii between $OB = 6$ and $OA = 12$. Therefore, we cannot directly connect $A$ and $B$ and must make a detour. Denote by $AC$ a tangent to the circular sector with radius 6 that meets it at point $C$. Therefore, the shortest path between $A$ and $B$ consists of a segment $AC$ and an arc from $C$ to $B$.
Because $OA = 12$, $OC = 6$ and $\angle OCA = 90^\circ$, we have $AC = \sqrt{OA^2 - OC^2} = 6 \sqrt{3}$ and $\angle AOC = 60^\circ$. This implies $\angle COB = \angle AOB - \angle AOC = 30^\circ$. Therefore, the length of the arc between $C$ and $B$ is $OB \cdot \pi \cdot \frac{\angle COB}{180^\circ} = \pi$. Therefore, the shortest distance between $A$ and $B$ is $\boxed{\textbf{(E) } 6 \sqrt{3} + \pi}$.
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6\sqrt{3} + \pi
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https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_21
| 2,023 |
A real-valued function $f$ has the property that for all real numbers $a$ and $b,$\[f(a + b) + f(a - b) = 2f(a) f(b).\]Which one of the following cannot be the value of $f(1)?$
$\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } -1 \qquad \textbf{(D) } 2 \qquad \textbf{(E) } -2$
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Substituting $a = b$ we get\[f(2a) + f(0) = 2f(a)^2\]Substituting $a= 0$ we find\[2f(0) = 2f(0)^2 \implies f(0) \in \{0, 1\}.\]This gives\[f(2a) = 2f(a)^2 - f(0) \geq 0-1\]Plugging in $a = \frac{1}{2}$ implies $f(1) \geq -1$, so answer choice $\boxed{\textbf{(E) -2}}$ is impossible.
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E
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https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_22
| 2,023 |
When $n$ standard six-sided dice are rolled, the product of the numbers rolled can be any of $936$ possible values. What is $n$?
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We start by trying to prove a function of $n$, and then we can apply the function and equate it to $936$ to find the value of $n$.
It is helpful to think of this problem in the format $(1+2+3+4+5+6) \cdot (1+2+3+4+5+6) \dots$. Note that if we represent the scenario in this manner, we can think of picking a $1$ for one factor and then a $5$ for another factor to form their product - this is similar thinking to when we have the factorized form of a polynomial. Unfortunately this is not quite accurate to the problem because we can reach the same product in many ways: for example for $n=2$, $4$ can be reached by picking $1$ and $4$ or $2$ and $2$. However, this form gives us insights that will be useful later in the problem.
Note that there are only $3$ primes in the set $\{1,2,3,4,5,6\}$: $2,3,$ and $5$. Thus if we're forming the product of possible values of a dice roll, the product has to be written in the form $2^h \cdot 3^i \cdot 5^j$ (the choice of variables will become clear later), for integer nonnegative values $h,i,j$. So now the problem boils down to how many distinct triplets $(h,i,j)$ can be formed by taking the product of $n$ dice values.
We start our work on representing $j$: the powers of $5$, because it is the simplest in this scenario because there is only one factor of $5$ in the set. Because of this, having $j$ fives in our prime factorization of the product is equivalent to picking $j$ factors from the polynomial $(1+\dots + 6) \cdots$ and choosing each factor to be a $5$. Now that we've selected $j$ factors, there are $n-j$ factors remaining to choose our powers of $3$ and $2$.
Suppose our prime factorization of this product contains $i$ powers of $3$. These powers of $3$ can either come from a $3$ factor or a $6$ factor, but since both $3$ and $6$ contain only one power of $3$, this means that a product with $i$ powers of $3$ corresponds directly to picking $i$ factors from the polynomial, each of which is either $3$ or $6$ (but this distinction doesn't matter when we consider only the powers of $3$.
Now we can reframe the problem again. Our method will be as follows: Suppose we choose an arbitrary pair $(i,j)$ that match the requirements, corresponding to the number of $3$'s and the number of $5$'s our product will have. Then how many different $h$ values for the powers of $2$ are possible?
In the $i+j$ factors we have already chosen, we obviously can't have any factors of $2$ in the $j$ factors with $5$. However, we can have a factor of $2$ pairing with factors of $3$, if we choose a $6$. The maximal possible power of $2$ in these $i$ factors is thus $2^i$, which occurs when we pick every factor to be $6$.
We now have $n-i-j$ factors remaining, and we want to allocate these to solely powers of $2$. For each of these factors, we can choose either a $1,2,$ or $4$. Therefore the maximal power of $2$ achieved in these factors is when we pick $4$ for all of them, which is equivalent to $2^{2\cdot (n-i-j)}$.
Now if we multiply this across the total $n$ factors (or $n$ dice) we have a total of $2^{2n-2i-2j} \cdot 2^i = 2^{2n-i-2j}$, which is the maximal power of $2$ attainable in the product for a pair $(i,j)$. Now note that every power of $2$ below this power is attainable: we can simply just take away a power of $2$ from an existing factor by dividing by $2$. Therefore the powers of $2$, and thus the $h$ value ranges from $h=0$ to $h=2n-i-2j$, so there are a total of $2n+1-i-2j$ distinct values for $h$ for a given pair $(i,j)$.
Now to find the total number of distinct triplets, we must sum this across all possible $i$s and $j$s. Lets take note of our restrictions on $i,j$: the only restriction is that $i+j \leq n$, since we're picking factors from $n$ dice.
\[\sum_{i+j\leq n}^{} 2n+1-i-2j = \sum_{i+j \leq n}^{} 2n+1 - \sum_{i+j \leq n}^{} i+2j\]
We start by calculating the first term. $2n+1$ is constant, so we just need to find out how many pairs there are such that $i+j \leq n$. Set $i$ to $0$: $j$ can range from $0$ to $n$, then set $i$ to $1$: $j$ can range from $0$ to $n-1$, etc. The total number of pairs is thus $n+1+n+n-1+\dots+1 = \frac{(n+1)(n+2)}{2}$. Therefore the left summation evaluates to\[\frac{(2n+1)(n+1)(n+2)}{2}\]
Now we calculate $\sum_{i+j \leq n}^{} i+2j$. This simplifies to $\sum_{i+j \leq n}^{} i + 2 \cdot \sum_{i+j \leq n}^{} j$. Note that because $i+j = n$ is symmetric with respect to $i,j$, the sum of $i$ in all of the pairs will be equal to the sum of $j$ in all of the pairs. Thus this is equal to calculating $3 \cdot \sum_{i+j \leq n}^{} i$.
In the pairs, $i=1$ appears for $j$ ranging between $0$ and $n-1$ so the sum here is $1 \cdot (n)$. Similarly $i=2$ appears for $j$ ranging from $0$ to $n-2$, so the sum is $2 \cdot (n-1)$. If we continue the pattern, the sum overall is $(n)+2 \cdot (n-1) + 3 \cdot (n-2) + \dots + (n) \cdot 1$. We can rearrange this as $((n)+(n-1)+ \dots + 1) + ((n-1)+(n-2)+ \dots + 1)+ ((n-2)+(n-3)+ \dots + 1) + \dots + 1)$
\[= \frac{(n)(n+1)}{2} + \frac{(n-1)(n)}{2}+ \dots + 1\]
We can write this in easier terms as $\sum_{k=0}^{n} \frac{(k)(k+1)}{2} = \frac{1}{2} \cdot \sum_{k=0}^{n} k^2+k$\[=\frac{1}{2} \cdot( \sum_{k=0}^{n} k^2 + \sum_{k=0}^{n} k)\]\[= \frac{1}{2} \cdot ( \frac{(n)(n+1)(2n+1)}{6} + \frac{(n)(n+1)}{2})\]\[= \frac{1}{2} \cdot ( \frac{(n)(n+1)(2n+1)}{6} + \frac{3n(n+1)}{6}) = \frac{1}{2} \cdot \frac{n(n+1)(2n+4)}{6}\]\[= \frac{n(n+1)(n+2)}{6}\]
We multiply this by $3$ to obtain that\[\sum_{i+j \leq n}^{} i+2j = \frac{n(n+1)(n+2)}{2}\]
Thus our final answer for the number of distinct triplets $(h,i,j)$ is:\[\sum_{i+j\leq n}^{} 2n+1-i-2j = \frac{(2n+1)(n+1)(n+2)}{2} - \frac{n(n+1)(n+2)}{2}\]\[= \frac{(n+1)(n+2)}{2} \cdot (2n+1-n) = \frac{(n+1)(n+2)}{2} \cdot (n+1)\]\[= \frac{(n+1)^2(n+2)}{2}\]
Now most of the work is done. We set this equal to $936$ and prime factorize. $936 = 12 \cdot 78 = 2^3 \cdot 3^2 \cdot 13$, so $(n+1)^2(n+2) = 936 \cdot 2 = 2^4 \cdot 3^2 \cdot 13$. Clearly $13$ cannot be anything squared and $2^4 \cdot 3^2$ is a perfect square, so $n+2 = 13$ and $n = 11 = \boxed{A}$
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11
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https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_23
| 2,023 |
Suppose that $a$, $b$, $c$ and $d$ are positive integers satisfying all of the following relations.
\[abcd=2^6\cdot 3^9\cdot 5^7\]\[\text{lcm}(a,b)=2^3\cdot 3^2\cdot 5^3\]\[\text{lcm}(a,c)=2^3\cdot 3^3\cdot 5^3\]\[\text{lcm}(a,d)=2^3\cdot 3^3\cdot 5^3\]\[\text{lcm}(b,c)=2^1\cdot 3^3\cdot 5^2\]\[\text{lcm}(b,d)=2^2\cdot 3^3\cdot 5^2\]\[\text{lcm}(c,d)=2^2\cdot 3^3\cdot 5^2\]
What is $\text{gcd}(a,b,c,d)$?
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Denote by $\nu_p (x)$ the number of prime factor $p$ in number $x$.
We index Equations given in this problem from (1) to (7).
First, we compute $\nu_2 (x)$ for $x \in \left\{ a, b, c, d \right\}$.
Equation (5) implies $\max \left\{ \nu_2 (b), \nu_2 (c) \right\} = 1$. Equation (2) implies $\max \left\{ \nu_2 (a), \nu_2 (b) \right\} = 3$. Equation (6) implies $\max \left\{ \nu_2 (b), \nu_2 (d) \right\} = 2$. Equation (1) implies $\nu_2 (a) + \nu_2 (b) + \nu_2 (c) + \nu_2 (d) = 6$.
Therefore, all above jointly imply $\nu_2 (a) = 3$, $\nu_2 (d) = 2$, and $\left( \nu_2 (b), \nu_2 (c) \right) = \left( 0 , 1 \right)$ or $\left( 1, 0 \right)$.
Second, we compute $\nu_3 (x)$ for $x \in \left\{ a, b, c, d \right\}$.
Equation (2) implies $\max \left\{ \nu_3 (a), \nu_3 (b) \right\} = 2$. Equation (3) implies $\max \left\{ \nu_3 (a), \nu_3 (c) \right\} = 3$. Equation (4) implies $\max \left\{ \nu_3 (a), \nu_3 (d) \right\} = 3$. Equation (1) implies $\nu_3 (a) + \nu_3 (b) + \nu_3 (c) + \nu_3 (d) = 9$.
Therefore, all above jointly imply $\nu_3 (c) = 3$, $\nu_3 (d) = 3$, and $\left( \nu_3 (a), \nu_3 (b) \right) = \left( 1 , 2 \right)$ or $\left( 2, 1 \right)$.
Third, we compute $\nu_5 (x)$ for $x \in \left\{ a, b, c, d \right\}$.
Equation (5) implies $\max \left\{ \nu_5 (b), \nu_5 (c) \right\} = 2$. Equation (2) implies $\max \left\{ \nu_5 (a), \nu_5 (b) \right\} = 3$. Thus, $\nu_5 (a) = 3$.
From Equations (5)-(7), we have either $\nu_5 (b) \leq 1$ and $\nu_5 (c) = \nu_5 (d) = 2$, or $\nu_5 (b) = 2$ and $\max \left\{ \nu_5 (c), \nu_5 (d) \right\} = 2$.
Equation (1) implies $\nu_5 (a) + \nu_5 (b) + \nu_5 (c) + \nu_5 (d) = 7$. Thus, for $\nu_5 (b)$, $\nu_5 (c)$, $\nu_5 (d)$, there must be two 2s and one 0.
Therefore,\begin{align*} {\rm gcd} (a,b,c,d) & = \Pi_{p \in \{ 2, 3, 5\}} p^{\min\{ \nu_p (a), \nu_p(b) , \nu_p (c), \nu_p(d) \}} \\ & = 2^0 \cdot 3^1 \cdot 5^0 \\ & = \boxed{\textbf{(C) 3}}. \end{align*}
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3
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https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_24
| 2,023 |
A regular pentagon with area $\sqrt{5}+1$ is printed on paper and cut out. The five vertices of the pentagon are folded into the center of the pentagon, creating a smaller pentagon. What is the area of the new pentagon?
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Pentagon 2023 12B Q25 dissmo.png
Let the original pentagon be $ABCDE$ centered at $O$. The dashed lines represent the fold lines. WLOG, let's focus on vertex $A$.
Since $A$ is folded onto $O$, $AM = MO$ where $M$ is the intersection of $AO$ and the creaseline between $A$ and $O$. Note that the inner pentagon is regular, and therefore similar to the original pentagon, due to symmetry.
Because of their similarity, the ratio of the inner pentagon's area to that of the outer pentagon can be represented by
$\left(\frac{OM}{ON}\right)^{2} = \left(\frac{\frac{OA}{2}}{OA\sin (\angle OAE)}\right)^{2} = \frac{1}{4\sin^{2}54}$
Option 1: Knowledge
Remember that $\sin54 = \frac{1+\sqrt5}{4}$.
Option 2: Angle Identities
$\cos54 = \sin36$
$4\cos^{3}18-3\cos18 = 2\sin18\cos18$
$4(1-\sin^{2}18)-3-2\sin18=0$
$4\sin^{2}18+2\sin18-1=0$
$\sin18 = \frac{-1+\sqrt5}{4}$
$\sin54 = \cos36 = 1-2\sin^{2}18 = \frac{1+\sqrt5}{4}$
$\sin^{2}54 =\frac{3+\sqrt5}{8}$
Let the inner pentagon be $Z$.
$[Z] = \frac{1}{4\sin^{2}54}[ABCDE]$
$= \frac{2(1+\sqrt5)}{3+\sqrt5}$
$= \sqrt5-1$
So the answer is $\boxed{B}$
|
\sqrt{5}-1
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https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_25
| 2,023 |
What is the value of $9901\cdot101-99\cdot10101?$
|
We have\begin{align*} 9901\cdot101-99\cdot10101 &= (10000-99)\cdot101-99\cdot(10000+101) \\ &= 10000\cdot101-99\cdot101-99\cdot10000-99\cdot101 \\ &= (10000\cdot101-99\cdot10000)-2\cdot(99\cdot101) \\ &= 2\cdot10000-2\cdot9999 \\ &= \boxed{\textbf{(A) }2}. \end{align*}
|
2
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https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_1
| 2,024 |
Define $\blacktriangledown(a) = \sqrt{a - 1}$ and $\blacktriangle(a) = \sqrt{a + 1}$ for all real numbers $a$. What is the value of\[\frac{\blacktriangledown(20 + \blacktriangle(2024))}{\blacktriangledown(\blacktriangle(24))}~?\]
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The value of the expression is\[\frac{\sqrt{20+\sqrt{2024+1}-1}}{\sqrt{\sqrt{24+1}-1}}=\frac{\sqrt{20+\sqrt{2025}-1}}{\sqrt{\sqrt{25}-1}}=\frac{\sqrt{20+45-1}}{\sqrt{5-1}}=\frac{\sqrt{64}}{\sqrt{4}}=\frac{8}{2}=\boxed{\textbf{(C)}~4}.\]
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4
|
https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_2
| 2,024 |
What is the least value of $n$ such that $n!$ is a multiple of $2024$?
|
Note that $2024=2^3\cdot11\cdot23$ in the prime factorization. Since $23!$ is a multiple of $2^3, 11,$ and $23,$ we conclude that $23!$ is a multiple of $2024.$ Therefore, we have $n=\boxed{\textbf{(D) } 23}.$
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23
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https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_4
| 2,024 |
A data set containing $20$ numbers, some of which are $6$, has mean $45$. When all the 6s are removed, the data set has mean $66$. How many 6s were in the original data set?
|
Because the set has $20$ numbers and mean $45$, the sum of the terms in the set is $45\cdot 20=900$.
Let there be $s$ sixes in the set.
Then, the mean of the set without the sixes is $\frac{900-6s}{20-s}$. Equating this expression to $66$ and solving yields $s=7$, so we choose answer choice $\boxed{\textbf{(D) }7}$.
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7
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https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_5
| 2,024 |
The product of three integers is $60$. What is the least possible positive sum of the three integers?
|
We notice that the optimal solution involves two negative numbers and a positive number. Thus we may split $60$ into three factors and choose negativity. We notice that $10\cdot6\cdot1=10\cdot(-6)\cdot(-1)=60$, and trying other combinations does not yield lesser results so the answer is $10-6-1=\boxed{\textbf{(B) }3}$.
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3
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https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_6
| 2,024 |
In $\Delta ABC$, $\angle ABC = 90^\circ$ and $BA = BC = \sqrt{2}$. Points $P_1, P_2, \dots, P_{2024}$ lie on hypotenuse $\overline{AC}$ so that $AP_1= P_1P_2 = P_2P_3 = \dots = P_{2023}P_{2024} = P_{2024}C$. What is the length of the vector sum\[\overrightarrow{BP_1} + \overrightarrow{BP_2} + \overrightarrow{BP_3} + \dots + \overrightarrow{BP_{2024}}?\]
|
Let us find an expression for the $x$- and $y$-components of $\overrightarrow{BP_i}$. Note that $AP_1+P_1P_2+\dots+P_{2023}P_{2024}+P_{2024}C=AC=2$, so $AP_1=P_1P_2=\dots=P_{2023}P_{2024}=P_{2024}C=\dfrac2{2025}$. All of the vectors $\overrightarrow{AP_1},\overrightarrow{P_1P_2},$ and so on up to $\overrightarrow{P_{2024}C}$ are equal; moreover, they equal $\textbf v=\left\langle\dfrac1{\sqrt2}\cdot\dfrac2{2025},-\dfrac1{\sqrt2}\cdot\dfrac2{2025}\right\rangle=\left\langle\dfrac{\sqrt2}{2025},-\dfrac{\sqrt2}{2025}\right\rangle$.
We now note that $\overrightarrow{AP_i}=i\textbf v=\left\langle\dfrac{i\sqrt2}{2025},-\dfrac{i\sqrt2}{2025}\right\rangle$ ($i$ copies of $\textbf v$ added together). Furthermore, note that $\overrightarrow{BP_i}=\overrightarrow{BA}+\overrightarrow{AP_i}=\left\langle0,\sqrt2\right\rangle+\left\langle\dfrac{i\sqrt2}{2025},-\dfrac{i\sqrt2}{2025}\right\rangle=\left\langle\dfrac{i\sqrt2}{2025},\sqrt2-\dfrac{i\sqrt2}{2025}\right\rangle.$
We want $\sum_{i=1}^{2024}\overrightarrow{BP_i}$'s length, which can be determined from the $x$- and $y$-components. Note that the two values should actually be the same - in this problem, everything is symmetric with respect to the line $x=y$, so the magnitudes of the $x$- and $y$-components should be identical. The $x$-component is easier to calculate.
\[\sum_{i=1}^{2024}\left(\overrightarrow{BP_i}\right)_x=\sum_{i=1}^{2024}\dfrac{i\sqrt2}{2025}=\dfrac{\sqrt2}{2025}\sum_{i=1}^{2024}i=\dfrac{\sqrt2}{2025}\cdot\dfrac{2024\cdot2025}2=1012\sqrt2.\]
One can similarly evaulate the $y$-component and obtain an identical answer; thus, our desired length is $\sqrt{\left(1012\sqrt2\right)^2+\left(1012\sqrt2\right)^2}=\sqrt{4\cdot1012^2}=\boxed{\textbf{(D) }2024}$.
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2024
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https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_7
| 2,024 |
How many angles $\theta$ with $0\le\theta\le2\pi$ satisfy $\log(\sin(3\theta))+\log(\cos(2\theta))=0$?
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Note that this is equivalent to $\sin(3\theta)\cos(2\theta)=1$, which is clearly only possible when $\sin(3\theta)=\cos(2\theta)=\pm1$. (If either one is between $1$ and $-1$, the other one must be greater than $1$ or less than $-1$ to offset the product, which is impossible for sine and cosine.) They cannot be both $-1$ since we cannot take logarithms of negative numbers, so they are both $+1$. Then $3\theta$ is $\dfrac\pi2$ more than a multiple of $2\pi$ and $2\theta$ is a multiple of $2\pi$, so $\theta$ is $\dfrac\pi6$ more than a multiple of $\dfrac23\pi$ and also a multiple of $\pi$. However, a multiple of $\dfrac23\pi$ will always have a denominator of $1$ or $3$, and never $6$; it can thus never add with $\dfrac\pi6$ to form an integral multiple of $\pi$. Thus, there are $\boxed{\textbf{(A) }0}$ solutions.
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0
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https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_8
| 2,024 |
Let $M$ be the greatest integer such that both $M+1213$ and $M+3773$ are perfect squares. What is the units digit of $M$?
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Let $M+1213=P^2$ and $M+3773=Q^2$ for some positive integers $P$ and $Q.$ We subtract the first equation from the second, then apply the difference of squares:\[(Q+P)(Q-P)=2560.\]Note that $Q+P$ and $Q-P$ have the same parity, and $Q+P>Q-P.$
We wish to maximize both $P$ and $Q,$ so we maximize $Q+P$ and minimize $Q-P.$ It follows that\begin{align*} Q+P&=1280, \\ Q-P&=2, \end{align*}from which $(P,Q)=(639,641).$
Finally, we get $M=P^2-1213=Q^2-3773\equiv1-3\equiv8\pmod{10},$ so the units digit of $M$ is $\boxed{\textbf{(E) }8}.$
|
8
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https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_9
| 2,024 |
Let $\alpha$ be the radian measure of the smallest angle in a $3{-}4{-}5$ right triangle. Let $\beta$ be the radian measure of the smallest angle in a $7{-}24{-}25$ right triangle. In terms of $\alpha$, what is $\beta$?
|
From the question,\[\tan\alpha=\frac{3}{4}, \space \tan\beta=\frac{7}{24}\]\[\tan(\alpha+\beta)= \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\]\[\tan(\alpha+\beta)= \frac{\frac{3}{4}+\frac{7}{24}}{1-\frac{3}{4} \cdot \frac{7}{24}}\]\[\tan(\alpha+\beta)=\frac{4}{3}\]\[\alpha+\beta=\tan^{-1}\left(\frac{4}{3}\right)\]\[\alpha+\beta=\frac{\pi}{2}-\tan^{-1}\left(\frac{3}{4}\right)\]\[\alpha+\beta=\frac{\pi}{2}-\alpha\]\[\beta=\boxed{\textbf{(C) }\frac{\pi}{2}-2\alpha}\]
|
\frac{\pi}{2}-2\alpha
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https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_10
| 2,024 |
There are exactly $K$ positive integers $b$ with $5 \leq b \leq 2024$ such that the base-$b$ integer $2024_b$ is divisible by $16$ (where $16$ is in base ten). What is the sum of the digits of $K$?
|
$2b^3+2b+4\equiv 0\pmod{16}\implies b^3+b+2\equiv 0\pmod 8$, if $b$ even then $b+2\equiv 0\pmod 8\implies b\equiv 6\pmod 8$. If $b$ odd then $b^2\equiv 1\pmod 8\implies b^3+b+2\equiv 2b+2\pmod 8$ so $2b+2\equiv 0\pmod 8\implies b+1\equiv 0\pmod 4\implies b\equiv 3,7\pmod 8$. Now $8\mid 2024$ so $\frac38\cdot 2024=759$ but one of the answers we got from that, $3$, is too small, so $759 - 1 = 758\implies\boxed{\textbf{(D) }20}$.
|
20
|
https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_11
| 2,024 |
The first three terms of a geometric sequence are the integers $a,\,720,$ and $b,$ where $a<720<b.$ What is the sum of the digits of the least possible value of $b?$
|
For a geometric sequence, we have $ab=720^2=2^8 3^4 5^2$, and we can test values for $b$. We find that $b=768$ and $a=675$ works, and we can test multiples of $5$ in between the two values. Finding that none of the multiples of 5 divide $720^2$ besides $720$ itself, we know that the answer is $7+6+8=\boxed{\textbf{(E) } 21}$.
|
21
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https://artofproblemsolving.com/wiki/index.php/2024_AMC_12A_Problems/Problem_12
| 2,024 |
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