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0 | 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 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_1 |
1 | 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 | 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 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_11 |
3 | 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:
$2^3-1^3+4^3-3^3+6^3-5^3+...+18^3-17^3$
$=\frac{18(18+1)(36+1)}{6}+4(\frac{9(9+1)(18+1)}{6})-2(\frac{9(9+1)}{2})$
$=3(19)(37)+6(10)(19)-9(10)$
$=2109+1140-90$
$=\boxed{\textbf{(D) } 3159}$
| 3,159 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_12 |
4 | 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 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_13 |
5 | 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 |
7 | 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
\[m+n=\boxed{\textbf{(B)}~21}\]
- CherryBerry
| 21 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_16 |
8 | 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? Write the answer as a simplified fraction $\frac{m}{n}$, find $m+n$ | 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 $\boxed{\textbf{(E)}~\frac12}$.
| 3 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_17 |
10 | 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 |
11 | 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? The answer can be written in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m-n$? | 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 $\frac{9}{5}=\boxed{\textbf{(A) }1\frac{4}{5}}.$
| 4 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_2 |
12 | Rows 1, 2, 3, 4, and 5 of a triangular array of integers are shown below.
1
1 1
1 3 1
1 5 5 1
1 7 11 7 1
Each row after the first row is formed by placing a 1 at each end of the row, and each interior entry is 1 greater than the sum of the two numbers diagonally above it in the previous row. What is the units digits of the sum of the 2023 numbers in the 2023rd row? | First, let $R(n)$ be the sum of the $n$th row. Now, with some observation and math instinct, we can guess that $R(n) = 2^n - n$.
Now we try to prove it by induction,
$R(1) = 2^n - n = 2^1 - 1 = 1$ (works for base case)
$R(k) = 2^k - k$
$R(k+1) = 2^{k+1} - (k + 1) = 2(2^k) - k - 1$
By definition from the question, the next row is always$:$
Double the sum of last row (Imagine each number from last row branches off toward left and right to the next row), plus # of new row, minus 2 (minus leftmost and rightmost's 1)
Which gives us $:$
$2(2^k - k) + (k + 1) - 2 = 2(2^k) - k - 1$
Hence, proven
Last, simply substitute $n = 2023$, we get $R(2023) = 2^{2023} - 2023$
Last digit of $2^{2023}$ is $8$, $8-3 = \boxed{\textbf{(C) } 5}$
| 5 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_20 |
13 | 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). Find the probability that $d(Q, R) > d(R, S)$. The final answer can be written in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$? | 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, $P(d(Q, R)>d(R, S)) = \frac{120+300}{1320} = \boxed{\textbf{(A) } \frac{7}{22}}$
| 29 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_21 |
14 | 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 $\boxed{\textbf{(B)} \ 96}$
$\color{magenta} zoomanTV$
| 96 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_22 |
15 | 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 |
16 | 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 |
17 | 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 |
18 | 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.
| 8 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_3 |
19 | 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 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_4 |
20 | 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 $3$? The final answer can be written in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$? | There are $3$ cases where the running total will equal $3$: one roll; two rolls; or three rolls:
The chance of rolling a running total of $3$, namely $(3)$ in exactly one roll is $\frac{1}{6}$.
The chance of rolling a running total of $3$ in exactly two rolls, namely $(1, 2)$ and $(2, 1)$ is $\frac{1}{6}\cdot\frac{1}{6}\cdot2=\frac{1}{18}$.
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 $\frac{1}{6}+\frac{1}{18}+\frac{1}{216}=\boxed{\textbf{(B) }\frac{49}{216}}$.
| 265 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_5 |
21 | 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 $B$? The final answer can be written in the form $m \sqrt{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$? | 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 $(2)(2\sqrt{5})=\boxed{\textbf{(D) }4\sqrt{5}}$
| 9 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_6 |
22 | 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:
| 9 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12A_Problems/Problem_7 |
23 | 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 |
25 | 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 final answer can be written in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$? | 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.
| 7 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_1 |
26 | 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 final answer can be written in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$? | 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 $\frac{-1}{- \frac{5}{2}} = \boxed{\textbf{(E) } \frac{2}{5}}$.
| 7 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_10 |
27 | Calculate the maximum area of an isosceles trapezoid that has legs of length $1$ and one base twice as long as the other. The final answer can be written in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m^2+n^2$? | 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 $[ABCD] = \frac{3}{4} \cdot 2 = \boxed{\frac{3}{2}}.$
-PIDay
| 13 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_11 |
28 | 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 $|z|^2$? | 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 $\sqrt{a^{2}+b^{2}} = \sqrt{50} = 5\sqrt{2}$
$=\text{\boxed{\textbf{(E) }5\sqrt{2}}}$
| 50 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_12 |
29 | A rectangular box $P$ has distinct edge lengths $a$, $b$, and $c$. The sum of the lengths of all $12$ edges of $P$ is $13$, the areas of all $6$ faces of $P$ is $\frac{11}{2}$, and the volume of $P$ is $\frac{1}{2}$. Find the length of the longest interior diagonal connecting two vertices of $P$. The final answer can be written in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$? | [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 $\frac{1}{2}$.
| 13 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_13 |
30 | For how many ordered pairs $(a,b)$ of integers does the polynomial $x^3+ax^2+bx+6$ have $3$ distinct integer roots? | 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
$\boxed{\textbf{(A) 5}}$.
| 5 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_14 |
32 | 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 $x?$ | 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 , 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 $2 + 9 = \boxed{\textbf{(D) 11}}$.
| 11 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_16 |
33 | Triangle $ABC$ has side lengths in arithmetic progression, and the smallest side has length $6.$ If the triangle has an angle of $120^\circ,$ Find the area of $ABC$. The final answer can be simplified in the form $m \sqrt{n}$, where $m$ and $n$ are positive integers and $n$ without square factore. What is $m+n$? | 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*}
| 18 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_17 |
36 | 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 |
40 | When $n$ standard six-sided dice are rolled, the product of the numbers rolled can be any of $936$ possible values. What is $n$? | 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 $n = 11 = \boxed{A}$
| 11 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_23 |
41 | 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 $\text{gcd}(a,b,c,d)$? | 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*}
| 3 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_24 |
43 | A $3-4-5$ right triangle is inscribed in circle $A$, and a $5-12-13$ right triangle is inscribed in circle $B$. Find the ratio of the area of circle $A$ to the area of circle $B$. The final answer can be written in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$? | 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 $=\boxed{\textbf{(D) }\frac{25}{169}}.$
| 194 | https://artofproblemsolving.com/wiki/index.php/2023_AMC_12B_Problems/Problem_3 |
44 | 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 |
45 | 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 |
46 | When the roots of the polynomial
\[P(x) = (x-1)^1 (x-2)^2 (x-3)^3 \cdot \cdot \cdot (x-10)^{10}\]
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 |
47 | 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 |
48 | 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 |
49 | 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 |
math-ai/amc23 with solutions, scraped from their original URL.
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