problem
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Find the sum of all positive integers $n$ such that, given an unlimited supply of stamps of denominations $5,n,$ and $n+1$ cents, $91$ cents is the greatest postage that cannot be formed. | \(\boxed{71}\) | Obviously $n\le 90$. We see that the problem's condition is equivalent to: 96 is the smallest number that can be formed which is 1 mod 5, and 92, 93, 94 can be formed (95 can always be formed). Now divide this up into cases. If $n\equiv 0\pmod{5}$, then 91 can be formed by using $n+1$ and some 5's, so there are no solutions for this case. If $n\equiv 1\pmod{5}$, then 91 can be formed by using $n$ and some 5's, so there are no solutions for this case either.
For $n\equiv 2\pmod{5}$, $2n+2$ is the smallest value that can be formed which is 1 mod 5, so $2n+2=96$ and $n=47$. We see that $92=45+47$, $93=48+45$, and $94=47+47$, so $n=47$ does work. If $n\equiv 3\pmod{5}$, then the smallest value that can be formed which is 1 mod 5 is $2n$, so $2n=96$ and $n=48$. We see that $94=49+45$ and $93=48+45$, but 92 cannot be formed, so there are no solutions for this case. If $n\equiv 4\pmod{5}$, then we can just ignore $n+1$ since it is a multiple of 5, meaning that the Chicken McNuggest theorem is a both necessary and sufficient condition, and it states that $5n-n-5=91$ meaning $4n=96$ and $n=24$. Hence, the only two $n$ that work are $n=24$ and $n=47$, so our answer is $24+47=\boxed{071}$. -Stormersyle
Video solution by Dr. Osman Nal: | -7,934,588,176,340,929,000 |
A positive integer $N$ has base-eleven representation $\underline{a}\kern 0.1em\underline{b}\kern 0.1em\underline{c}$ and base-eight representation $\underline1\kern 0.1em\underline{b}\kern 0.1em\underline{c}\kern 0.1em\underline{a},$ where $a,b,$ and $c$ represent (not necessarily distinct) digits. Find the least such $N$ expressed in base ten. | \(\boxed{621}\) | From the given information, $121a+11b+c=512+64b+8c+a \implies 120a=512+53b+7c$. Since $a$, $b$, and $c$ have to be positive, $a \geq 5$. Since we need to minimize the value of $n$, we want to minimize $a$, so we have $a = 5$. Then we know $88=53b+7c$, and we can see the only solution is $b=1$, $c=5$. Finally, $515_{11} = 621_{10}$, so our answer is $\boxed{621}$.
~ JHawk0224 | -8,975,637,807,720,872,000 |
Six cards numbered $1$ through $6$ are to be lined up in a row. Find the number of arrangements of these six cards where one of the cards can be removed leaving the remaining five cards in either ascending or descending order. | \(\boxed{52}\) | Similar to above, a $1-1$ correspondence between ascending and descending is established by subtracting each number from $7$.
We note that the given condition is equivalent to "cycling" $123456$ for a contiguous subset of it. For example,
$12(345)6 \rightarrow 125346, 124536$
It's not hard to see that no overcount is possible, and that the cycle is either $1$ "right" or $1$ "left." Therefore, we consider how many elements we flip by. If we flip $1$ or $2$ such elements, then there is one way to cycle them. Otherwise, we have $2$ ways. Therefore, the total number of ascending is $1 + 5 + 2(4 + 3 + 2 + 1) = 26$, and multiplying by two gives $\boxed{052}.$
~awang11 | 794,600,372,110,184,800 |
A club consisting of $11$ men and $12$ women needs to choose a committee from among its members so that the number of women on the committee is one more than the number of men on the committee. The committee could have as few as $1$ member or as many as $23$ members. Let $N$ be the number of such committees that can be formed. Find the sum of the prime numbers that divide $N.$ | \(\boxed{81}\) | Let $k$ be the number of women selected. Then, the number of men not selected is $11-(k-1)=12-k$. Note that the sum of the number of women selected and the number of men not selected is constant at $12$. Each combination of women selected and men not selected corresponds to a committee selection. Since choosing 12 individuals from the total of 23 would give $k$ women and $12-k$ men, the number of committee selections is $\binom{23}{12}$. The answer is $\boxed{081}$. ~awang11's sol | -8,896,810,917,364,741,000 |
Let $S$ be the set of positive integer divisors of $20^9.$ Three numbers are chosen independently and at random with replacement from the set $S$ and labeled $a_1,a_2,$ and $a_3$ in the order they are chosen. The probability that both $a_1$ divides $a_2$ and $a_2$ divides $a_3$ is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m.$ | \(\boxed{77}\) | Similar to before, we calculate that there are $190^3$ ways to choose $3$ factors with replacement. Then, we figure out the number of triplets ${a,b,c}$ and ${d,f,g}$, where $a$, $b$, and $c$ represent powers of $2$ and $d$, $f$, and $g$ represent powers of $5$, such that the triplets are in non-descending order. The maximum power of $2$ is $18$, and the maximum power of $5$ is $9$. Using the Hockey Stick identity, we figure out that there are $\dbinom{12}{3}$ ways to choose $d$, $f$ and $g$, and $\dbinom{21}{3}$ ways to choose $a$, $b$, and $c$. Therefore, the probability of choosing $3$ factors which satisfy the conditions is \[\frac{\dbinom{21}{3} \cdot \dbinom{12}{3}}{190^3}.\] This simplifies to $\frac{77}{1805}$, therefore $m =$ $\boxed{077}$. | -5,794,982,965,406,523,000 |
For integers $a,b,c$ and $d,$ let $f(x)=x^2+ax+b$ and $g(x)=x^2+cx+d.$ Find the number of ordered triples $(a,b,c)$ of integers with absolute values not exceeding $10$ for which there is an integer $d$ such that $g(f(2))=g(f(4))=0.$ | \(\boxed{510}\) | Either $f(2)=f(4)$ or not. If it is, note that Vieta's forces $a = -6$. Then, $b$ can be anything. However, $c$ can also be anything, as we can set the root of $g$ (not equal to $f(2) = f(4)$) to any integer, producing a possible integer value of $d$. Therefore there are $21^2 = 441$ in this case*. If it isn't, then $f(2),f(4)$ are the roots of $g$. This means by Vieta's, that:
\[f(2)+f(4) = -c \in [-10,10]\] \[20 + 6a + 2b \in [-10,10]\] \[3a + b \in [-15,-5].\]
Solving these inequalities while considering that $a \neq -6$ to prevent $f(2) = f(4)$, we obtain $69$ possible tuples and adding gives $441+69=\boxed{510}$. ~awang11 | -7,783,187,565,554,845,000 |
Let $n$ be the least positive integer for which $149^n-2^n$ is divisible by $3^3\cdot5^5\cdot7^7.$ Find the number of positive integer divisors of $n.$ | \(\boxed{270}\) | Lifting the Exponent shows that \[v_3(149^n-2^n) = v_3(n) + v_3(147) = v_3(n)+1\] so thus, $3^2$ divides $n$. It also shows that \[v_7(149^n-2^n) = v_7(n) + v_7(147) = v_7(n)+2\] so thus, $7^5$ divides $n$.
Now, setting $n = 4c$ (necessitated by $149^n \equiv 2^n \pmod 5$ in order to set up LTE), we see \[v_5(149^{4c}-2^{4c}) = v_5(149^{4c}-16^{c})\] and since $149^{4} \equiv 1 \pmod{25}$ and $16^1 \equiv 16 \pmod{25}$ then $v_5(149^{4c}-2^{4c})=1+v_5(c)$ meaning that we have that by LTE, $5^4 | c$ and $4 \cdot 5^4$ divides $n$.
Since $3^2$, $7^5$ and $4\cdot 5^4$ all divide $n$, the smallest value of $n$ working is their LCM, also $3^2 \cdot 7^5 \cdot 4 \cdot 5^4 = 2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5$. Thus the number of divisors is $(2+1)(2+1)(4+1)(5+1) = \boxed{270}$.
~kevinmathz
clarified by another user | 3,211,575,271,358,981,000 |
Let $P(x)$ be a quadratic polynomial with complex coefficients whose $x^2$ coefficient is $1.$ Suppose the equation $P(P(x))=0$ has four distinct solutions, $x=3,4,a,b.$ Find the sum of all possible values of $(a+b)^2.$ | \(\boxed{85}\) | Either $P(3) = P(4)$ or not. We first see that if $P(3) = P(4)$ it's easy to obtain by Vieta's that $(a+b)^2 = 49$. Now, take $P(3) \neq P(4)$ and WLOG $P(3) = P(a), P(4) = P(b)$. Now, consider the parabola formed by the graph of $P$. It has vertex $\frac{3+a}{2}$. Now, say that $P(x) = x^2 - (3+a)x + c$. We note $P(3)P(4) = c = P(3)\left(4 - 4a + \frac{8a - 1}{2}\right) \implies a = \frac{7P(3) + 1}{8}$. Now, we note $P(4) = \frac{7}{2}$ by plugging in again. Now, it's easy to find that $a = -2.5, b = -3.5$, yielding a value of $36$. Finally, we add $49 + 36 = \boxed{085}$. ~awang11, charmander3333
Remark: We know that $c=\frac{8a-1}{2}$ from $P(3)+P(4)=3+a$. | 926,079,944,055,805,700 |
Let $\triangle ABC$ be an acute triangle with circumcircle $\omega,$ and let $H$ be the intersection of the altitudes of $\triangle ABC.$ Suppose the tangent to the circumcircle of $\triangle HBC$ at $H$ intersects $\omega$ at points $X$ and $Y$ with $HA=3,HX=2,$ and $HY=6.$ The area of $\triangle ABC$ can be written in the form $m\sqrt{n},$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n.$ | \(\boxed{58}\) | Let $O$ be circumcenter of $ABC,$ let $R$ be circumradius of $ABC,$ let $\omega'$ be the image of circle $\omega$ over line $BC$ (the circumcircle of $HBC$).
Let $P$ be the image of the reflection of $H$ over line $BC, P$ lies on circle $\omega.$ Let $M$ be the midpoint of $XY.$ Then $P$ lies on $\omega, OA = O'H, OA || O'H.$
(see here: https://brilliant.org/wiki/triangles-orthocenter/ or https://en.wikipedia.org/wiki/Altitude_(triangle) Russian)
$P$ lies on $\omega \implies OA = OP = R,$
$OA = O'H, OA || O'H \implies M$ lies on $OA.$
We use properties of crossing chords and get \[AH \cdot HP = XH \cdot HY = 2 \cdot 6 \implies HP = 4, AP = AH + HP = 7.\] We use properties of radius perpendicular chord and get \[MH = \frac{XH + HY}{2} – HY = 2.\] We find \[\sin OAH =\frac{MH}{AH} = \frac{2}{3} \implies \cos OAH = \frac{\sqrt{5}}{3}.\] We use properties of isosceles $\triangle OAP$ and find $\hspace{5mm}R = \frac{AP}{2\cos OAP} = \frac{7}{2\frac {\sqrt{5}}{3}} = \frac{21}{2\sqrt{5}}.$
We use $OM' = \frac{AH}{2} = \frac {3}{2}$ and find $\hspace{25mm} \frac{BC}{2} = \sqrt{R^2 – OM'^2} = 3 \sqrt {\frac {11}{5}}.$
The area of $ABC$ \[[ABC]=\frac{BC}{2} \cdot (AH + HD) = 3\cdot \sqrt{55} \implies 3+55 = \boldsymbol{\boxed{058}}.\] [email protected], vvsss | 4,685,868,726,477,670,000 |
Find the number of positive integers less than $1000$ that can be expressed as the difference of two integral powers of $2.$ | \(\boxed{50}\) | We look for all positive integers of the form $2^a-2^b<1000,$ where $0\leq b<a.$ Performing casework on $a,$ we can enumerate all possibilities in the table below: \[\begin{array}{c|c} & \\ [-2.25ex] \boldsymbol{a} & \boldsymbol{b} \\ \hline & \\ [-2ex] 1 & 0 \\ 2 & 0,1 \\ 3 & 0,1,2 \\ 4 & 0,1,2,3 \\ 5 & 0,1,2,3,4 \\ 6 & 0,1,2,3,4,5 \\ 7 & 0,1,2,3,4,5,6 \\ 8 & 0,1,2,3,4,5,6,7 \\ 9 & 0,1,2,3,4,5,6,7,8 \\ 10 & \xcancel{0},\xcancel{1},\xcancel{2},\xcancel{3},\xcancel{4},5,6,7,8,9 \\ [0.5ex] \end{array}\] As indicated by the X-marks, the ordered pairs $(a,b)=(10,0),(10,1),(10,2),(10,3),(10,4)$ generate $2^a-2^b>1000,$ which are invalid.
Note that each of the remaining ordered pairs generates one unique desired positive integer.
We prove this statement as follows:
The positive integers generated for each value of $a$ are clearly different.
For all integers $k$ such that $1\leq k\leq9,$ the largest positive integer generated for $a=k$ is $1$ less than the smallest positive integer generated for $a=k+1.$
Together, we have justified our claim in bold. The answer is \[1+2+3+4+5+6+7+8+9+5=\boxed{050}.\]
~MRENTHUSIASM | 8,707,660,538,450,936,000 |
Find the number of ways $66$ identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile. | \(\boxed{331}\) | Suppose we have $1$ coin in the first pile. Then $(1, 2, 63), (1, 3, 62), \ldots, (1, 32, 33)$ all work for a total of $31$ piles. Suppose we have $2$ coins in the first pile, then $(2, 3, 61), (2, 4, 60), \ldots, (2, 31, 33)$ all work, for a total of $29$. Continuing this pattern until $21$ coins in the first pile, we have the sum \begin{align*} 31+29+28+26+25+\cdots+4+2+1 &= (31+28+25+22+\cdots+1)+(29+26+23+\cdots+2) \\ &= 176+155 \\ &= \boxed{331}. \end{align*} | 1,429,113,754,250,875,600 |
Find the number of pairs $(m,n)$ of positive integers with $1\le m<n\le 30$ such that there exists a real number $x$ satisfying \[\sin(mx)+\sin(nx)=2.\] | \(\boxed{63}\) | We know that the range of $sin$ is between $-1$ and $1$.
Thus, the only way for the sum to be $2$ is for $sin$ of $mx$ and $nx$ to both be $1$.
The $sin$ of $(90+360k)$ is equal to 1.
Assuming $mx$ and $nm$ are both positive, m and n could be $1,5,9,13,17,21,25,29$. There are $8$ ways, so $\dbinom{8}{2}$.
If both are negative, m and n could be $3,7,11,15,19,23,27$. There are $7$ ways, so $\dbinom{7}{2}$.
However, the pair $(1,5)$ could also be $(2, 10)$ and so on. The same goes for some other pairs.
In total there are $14$ of these extra pairs.
The answer is $28+21+14 = \boxed{063}$
Remark
The graphs of $r\leq\sin(m\theta)+\sin(n\theta)$ and $r=2$ are shown here in Desmos: https://www.desmos.com/calculator/busxadywja
Move the sliders around for $1\leq m \leq 29$ and $2\leq m+1\leq n\leq30$ to observe the geometric representation generated by each pair $(m,n).$
~MRENTHUSIASM (inspired by TheAMCHub)
~mathproblemsolvingskills
~Mathematical Dexterity | -3,245,935,582,349,820,400 |
There are real numbers $a, b, c,$ and $d$ such that $-20$ is a root of $x^3 + ax + b$ and $-21$ is a root of $x^3 + cx^2 + d.$ These two polynomials share a complex root $m + \sqrt{n} \cdot i,$ where $m$ and $n$ are positive integers and $i = \sqrt{-1}.$ Find $m+n.$ | \(\boxed{330}\) | We plug -20 into the equation obtaining $(-20)^3-20a+b$, likewise, plugging -21 into the second equation gets $(-21)^3+441c+d$.
Both equations must have 3 solutions exactly, so the other two solutions must be $m + \sqrt{n} \cdot i$ and $m - \sqrt{n} \cdot i$.
By Vieta's, the sum of the roots in the first equation is $0$, so $m$ must be $10$.
Next, using Vieta's theorem on the second equation, you get $x1x2+x2x3+x1x3 = 0$. However, since we know that the sum of the roots with complex numbers are 20, we can factor out the terms with -21, so $-21*(20)+(m^2+n)=0$.
Given that $m$ is $10$, then $n$ is equal to $320$.
Therefore, the answer to the equation is $\boxed{330}$ | -3,636,804,197,637,068,300 |
Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients $2$ and $-2,$ respectively. The graphs of both polynomials pass through the two points $(16,54)$ and $(20,53).$ Find $P(0) + Q(0).$ | \(\boxed{116}\) | Let $R(x)=P(x)+Q(x).$ Since the $x^2$-terms of $P(x)$ and $Q(x)$ cancel, we conclude that $R(x)$ is a linear polynomial.
Note that \begin{alignat*}{8} R(16) &= P(16)+Q(16) &&= 54+54 &&= 108, \\ R(20) &= P(20)+Q(20) &&= 53+53 &&= 106, \end{alignat*} so the slope of $R(x)$ is $\frac{106-108}{20-16}=-\frac12.$
It follows that the equation of $R(x)$ is \[R(x)=-\frac12x+c\] for some constant $c,$ and we wish to find $R(0)=c.$
We substitute $x=20$ into this equation to get $106=-\frac12\cdot20+c,$ from which $c=\boxed{116}.$
~MRENTHUSIASM | 8,239,188,321,179,411,000 |
Let $w = \dfrac{\sqrt{3} + i}{2}$ and $z = \dfrac{-1 + i\sqrt{3}}{2},$ where $i = \sqrt{-1}.$ Find the number of ordered pairs $(r,s)$ of positive integers not exceeding $100$ that satisfy the equation $i \cdot w^r = z^s.$ | \(\boxed{834}\) | We rewrite $w$ and $z$ in polar form: \begin{align*} w &= e^{i\cdot\frac{\pi}{6}}, \\ z &= e^{i\cdot\frac{2\pi}{3}}. \end{align*} The equation $i \cdot w^r = z^s$ becomes \begin{align*} e^{i\cdot\frac{\pi}{2}} \cdot \left(e^{i\cdot\frac{\pi}{6}}\right)^r &= \left(e^{i\cdot\frac{2\pi}{3}}\right)^s \\ e^{i\left(\frac{\pi}{2}+\frac{\pi}{6}r\right)} &= e^{i\left(\frac{2\pi}{3}s\right)} \\ \frac{\pi}{2}+\frac{\pi}{6}r &= \frac{2\pi}{3}s+2\pi k \\ 3+r &= 4s+12k \\ 3+r &= 4(s+3k). \end{align*} for some integer $k.$
Since $4\leq 3+r\leq 103$ and $4\mid 3+r,$ we conclude that \begin{align*} 3+r &\in \{4,8,12,\ldots,100\}, \\ s+3k &\in \{1,2,3,\ldots,25\}. \end{align*} Note that the values for $s+3k$ and the values for $r$ have one-to-one correspondence.
We apply casework to the values for $s+3k:$
$s+3k\equiv0\pmod{3}$
There are $8$ values for $s+3k,$ so there are $8$ values for $r.$ It follows that $s\equiv0\pmod{3},$ so there are $33$ values for $s.$
There are $8\cdot33=264$ ordered pairs $(r,s)$ in this case.
$s+3k\equiv1\pmod{3}$
There are $9$ values for $s+3k,$ so there are $9$ values for $r.$ It follows that $s\equiv1\pmod{3},$ so there are $34$ values for $s.$
There are $9\cdot34=306$ ordered pairs $(r,s)$ in this case.
$s+3k\equiv2\pmod{3}$
There are $8$ values for $s+3k,$ so there are $8$ values for $r.$ It follows that $s\equiv2\pmod{3},$ so there are $33$ values for $s.$
There are $8\cdot33=264$ ordered pairs $(r,s)$ in this case.
Together, the answer is $264+306+264=\boxed{834}.$
~MRENTHUSIASM | 5,274,167,565,660,867,000 |
Let $a,b,c,d,e,f,g,h,i$ be distinct integers from $1$ to $9.$ The minimum possible positive value of \[\dfrac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i}\] can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | \(\boxed{289}\) | To minimize a positive fraction, we minimize its numerator and maximize its denominator. It is clear that $\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i} \geq \frac{1}{7\cdot8\cdot9}.$
If we minimize the numerator, then $a \cdot b \cdot c - d \cdot e \cdot f = 1.$ Note that $a \cdot b \cdot c \cdot d \cdot e \cdot f = (a \cdot b \cdot c) \cdot (a \cdot b \cdot c - 1) \geq 6! = 720,$ so $a \cdot b \cdot c \geq 28.$ It follows that $a \cdot b \cdot c$ and $d \cdot e \cdot f$ are consecutive composites with prime factors no other than $2,3,5,$ and $7.$ The smallest values for $a \cdot b \cdot c$ and $d \cdot e \cdot f$ are $36$ and $35,$ respectively. So, we have $\{a,b,c\} = \{2,3,6\}, \{d,e,f\} = \{1,5,7\},$ and $\{g,h,i\} = \{4,8,9\},$ from which $\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i} = \frac{1}{288}.$
If we do not minimize the numerator, then $a \cdot b \cdot c - d \cdot e \cdot f > 1.$ Note that $\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i} \geq \frac{2}{7\cdot8\cdot9} > \frac{1}{288}.$
Together, we conclude that the minimum possible positive value of $\frac{a \cdot b \cdot c - d \cdot e \cdot f}{g \cdot h \cdot i}$ is $\frac{1}{288}.$ Therefore, the answer is $1+288=\boxed{289}.$
~MRENTHUSIASM ~jgplay | 2,207,720,615,262,049,800 |
Let $S$ be the set of all rational numbers that can be expressed as a repeating decimal in the form $0.\overline{abcd},$ where at least one of the digits $a,$ $b,$ $c,$ or $d$ is nonzero. Let $N$ be the number of distinct numerators obtained when numbers in $S$ are written as fractions in lowest terms. For example, both $4$ and $410$ are counted among the distinct numerators for numbers in $S$ because $0.\overline{3636} = \frac{4}{11}$ and $0.\overline{1230} = \frac{410}{3333}.$ Find the remainder when $N$ is divided by $1000.$ | \(\boxed{392}\) | $0.abcd=\frac{\overline{abcd}}{9999}$, $9999=9\times 11\times 101$.
Then we need to find the number of positive integers less than $10000$ that can meet the requirement. Suppose the number is $x$.
Case $1$: $(9999, x)=1$. Clearly $x$ satisfies. \[\varphi \left( 9999 \right) =9999\times \left( 1-\frac{1}{3} \right) \times \left( 1-\frac{1}{11} \right) \times \left( 1-\frac{1}{101} \right)=6000\]
Case $2$: $3|x$ but $x$ is not a multiple of $11$ or $101.$ Then the least value of $abcd$ is $9x$, so that $x\le 1111$, $334$ values from $3$ to $1110.$
Case $3$: $11|x$ but $x$ is not a multiple of $3$ or $101$. Then the least value of $abcd$ is $11x$, so that $x\le 909$, $55$ values from $11$ to $902$.
Case $4$: $101|x$. None.
Case $5$: $3, 11|x$. Then the least value of $abcd$ is $11x$, $3$ values from $33$ to $99.$
To sum up, the answer is \[6000+334+55+3=\boxed{392}\] mod $1000$. | -1,672,954,175,228,689,200 |
Given $\triangle ABC$ and a point $P$ on one of its sides, call line $\ell$ the $\textit{splitting line}$ of $\triangle ABC$ through $P$ if $\ell$ passes through $P$ and divides $\triangle ABC$ into two polygons of equal perimeter. Let $\triangle ABC$ be a triangle where $BC = 219$ and $AB$ and $AC$ are positive integers. Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AC},$ respectively, and suppose that the splitting lines of $\triangle ABC$ through $M$ and $N$ intersect at $30^\circ.$ Find the perimeter of $\triangle ABC.$ | \(\boxed{459}\) | We wish to solve the Diophantine equation $a^2+ab+b^2=3^2 \cdot 73^2$. It can be shown that $3|a$ and $3|b$, so we make the substitution $a=3x$ and $b=3y$ to obtain $x^2+xy+y^2=73^2$ as our new equation to solve for.
Notice that $r^2+r+1=(r-\omega)(r-{\omega}^2)$, where $\omega=e^{i\frac{2\pi}{3}}$. Thus, \[x^2+xy+y^2 = y^2((x/y)^2+(x/y)+1) = y^2 (\frac{x}{y}-\omega)(\frac{x}{y}-{\omega}^2) = (x-y\omega)(x-y{\omega}^2).\]
Note that $8^2+1^2+8 \cdot 1=73$. Thus, $(8-\omega)(8-{\omega}^2)=73$. Squaring both sides yields \begin{align} (8-\omega)^2(8-{\omega}^2)^2&=73^2\\ (63-17\omega)(63-17{\omega}^2)&=73^2. \end{align} Thus, by $(2)$, $(63, 17)$ is a solution to $x^2+xy+y^2=73^2$. This implies that $a=189$ and $b=51$, so our final answer is $189+51+219=\boxed{459}$.
~ Leo.Euler | 8,432,783,925,725,289,000 |
Let $x,$ $y,$ and $z$ be positive real numbers satisfying the system of equations: \begin{align*} \sqrt{2x-xy} + \sqrt{2y-xy} &= 1 \\ \sqrt{2y-yz} + \sqrt{2z-yz} &= \sqrt2 \\ \sqrt{2z-zx} + \sqrt{2x-zx} &= \sqrt3. \end{align*} Then $\left[ (1-x)(1-y)(1-z) \right]^2$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | \(\boxed{33}\) | Let $1-x=a;1-y=b;1-z=c$, rewrite those equations
$\sqrt{(1-a)(1+b)}+\sqrt{(1+a)(1-b)}=1$;
$\sqrt{(1-b)(1+c)}+\sqrt{(1+b)(1-c)}=\sqrt{2}$
$\sqrt{(1-a)(1+c)}+\sqrt{(1-c)(1+a)}=\sqrt{3}$
square both sides, get three equations:
$2ab-1=2\sqrt{(1-a^2)(1-b^2)}$
$2bc=2\sqrt{(1-b^2)(1-c^2)}$
$2ac+1=2\sqrt{(1-c^2)(1-a^2)}$
Getting that $a^2+b^2-ab=\frac{3}{4}$
$b^2+c^2=1$
$a^2+c^2+ac=\frac{3}{4}$
Subtract first and third equation, getting $(b+c)(b-c)=a(b+c)$, $a=b-c$
Put it in first equation, getting $b^2-2bc+c^2+b^2-b(b-c)=b^2+c^2-bc=\frac{3}{4}$, $bc=\frac{1}{4}$
Since $a^2=b^2+c^2-2bc=\frac{1}{2}$, the final answer is $\frac{1}{4}*\frac{1}{4}*\frac{1}{2}=\frac{1}{32}$ the final answer is $\boxed{033}$
~bluesoul | -5,829,392,112,625,381,000 |
Adults made up $\frac5{12}$ of the crowd of people at a concert. After a bus carrying $50$ more people arrived, adults made up $\frac{11}{25}$ of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived. | \(\boxed{154}\) | Let $x$ be the number of people at the party before the bus arrives. We know that $x\equiv 0\pmod {12}$, as $\frac{5}{12}$ of people at the party before the bus arrives are adults. Similarly, we know that $x + 50 \equiv 0 \pmod{25}$, as $\frac{11}{25}$ of the people at the party are adults after the bus arrives. $x + 50 \equiv 0 \pmod{25}$ can be reduced to $x \equiv 0 \pmod{25}$, and since we are looking for the minimum amount of people, $x$ is $300$. That means there are $350$ people at the party after the bus arrives, and thus there are $350 \cdot \frac{11}{25} = \boxed{154}$ adults at the party.
~eamo | -6,134,327,127,445,261,000 |
Twenty distinct points are marked on a circle and labeled $1$ through $20$ in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles formed whose vertices are among the original $20$ points. | \(\boxed{72}\) | Let $a$, $b$, and $c$ be the vertex of a triangle that satisfies this problem, where $a > b > c$. \[a - b = p_1\] \[b - c = p_2\] \[a - c = p_3\]
$p_3 = a - c = a - b + b - c = p_1 + p_2$. Because $p_3$ is the sum of two primes, $p_1$ and $p_2$, $p_1$ or $p_2$ must be $2$. Let $p_1 = 2$, then $p_3 = p_2 + 2$. There are only $8$ primes less than $20$: $2, 3, 5, 7, 11, 13, 17, 19$. Only $3, 5, 11, 17$ plus $2$ equals another prime. $p_2 \in \{ 3, 5, 11, 17 \}$.
Once $a$ is determined, $a = b+2$ and $b = c + p_2$. There are $18$ values of $b$ where $b+2 \le 20$, and $4$ values of $p_2$. Therefore the answer is $18 \cdot 4 = \boxed{072}$
~isabelchen | -1,362,941,814,015,018,500 |
Positive real numbers $b \not= 1$ and $n$ satisfy the equations \[\sqrt{\log_b n} = \log_b \sqrt{n} \qquad \text{and} \qquad b \cdot \log_b n = \log_b (bn).\] The value of $n$ is $\frac{j}{k},$ where $j$ and $k$ are relatively prime positive integers. Find $j+k.$ | \(\boxed{881}\) | Denote $x = \log_b n$. Hence, the system of equations given in the problem can be rewritten as \begin{align*} \sqrt{x} & = \frac{1}{2} x , \\ bx & = 1 + x . \end{align*} Solving the system gives $x = 4$ and $b = \frac{5}{4}$. Therefore, \[n = b^x = \frac{625}{256}.\] Therefore, the answer is $625 + 256 = \boxed{881}$.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | -1,899,228,719,008,460,000 |
A plane contains $40$ lines, no $2$ of which are parallel. Suppose that there are $3$ points where exactly $3$ lines intersect, $4$ points where exactly $4$ lines intersect, $5$ points where exactly $5$ lines intersect, $6$ points where exactly $6$ lines intersect, and no points where more than $6$ lines intersect. Find the number of points where exactly $2$ lines intersect. | \(\boxed{607}\) | In this solution, let $\boldsymbol{n}$-line points be the points where exactly $n$ lines intersect. We wish to find the number of $2$-line points.
There are $\binom{40}{2}=780$ pairs of lines. Among them:
The $3$-line points account for $3\cdot\binom32=9$ pairs of lines.
The $4$-line points account for $4\cdot\binom42=24$ pairs of lines.
The $5$-line points account for $5\cdot\binom52=50$ pairs of lines.
The $6$-line points account for $6\cdot\binom62=90$ pairs of lines.
It follows that the $2$-line points account for $780-9-24-50-90=\boxed{607}$ pairs of lines, where each pair intersect at a single point.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
~MRENTHUSIASM
~IceMatrix | -3,998,887,712,230,138,400 |
The sum of all positive integers $m$ such that $\frac{13!}{m}$ is a perfect square can be written as $2^a3^b5^c7^d11^e13^f,$ where $a,b,c,d,e,$ and $f$ are positive integers. Find $a+b+c+d+e+f.$ | \(\boxed{012}\) | We first rewrite $13!$ as a prime factorization, which is $2^{10}\cdot3^5\cdot5^2\cdot7\cdot11\cdot13.$
For the fraction to be a square, it needs each prime to be an even power. This means $m$ must contain $7\cdot11\cdot13$. Also, $m$ can contain any even power of $2$ up to $2^{10}$, any odd power of $3$ up to $3^{5}$, and any even power of $5$ up to $5^{2}$. The sum of $m$ is \[(2^0+2^2+2^4+2^6+2^8+2^{10})(3^1+3^3+3^5)(5^0+5^2)(7^1)(11^1)(13^1) =\] \[1365\cdot273\cdot26\cdot7\cdot11\cdot13 = 2\cdot3^2\cdot5\cdot7^3\cdot11\cdot13^4.\] Therefore, the answer is $1+2+1+3+1+4=\boxed{012}$.
~chem1kall | -655,636,288,345,638,700 |
Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$ | \(\boxed{106}\) | WLOG, let $P$ be on minor arc $\overarc {AB}$. Let $r$ and $O$ be the radius and center of the circumcircle respectively, and let $\theta = \angle AOP$.
By the Pythagorean Theorem, the area of the square is $2r^2$. We can use the Law of Cosines on isosceles triangles $\triangle AOP, \, \triangle COP, \, \triangle BOP, \, \triangle DOP$ to get
\begin{align*} PA^2 &= 2r^2(1 - \cos \theta), \\ PC^2 &= 2r^2(1 - \cos (180 - \theta)) = 2r^2(1 + \cos \theta), \\ PB^2 &= 2r^2(1 - \cos (90 - \theta)) = 2r^2(1 - \sin \theta), \\ PD^2 &= 2r^2(1 - \cos (90 + \theta)) = 2r^2(1 + \sin \theta). \end{align*}
Taking the products of the first two and last two equations, respectively, \[56^2 = (PA \cdot PC)^2 = 4r^4(1 - \cos \theta)(1 + \cos \theta) = 4r^4(1 - \cos^2 \theta) = 4r^4 \sin^2 \theta,\] and \[90^2 = (PB \cdot PD)^2 = 4r^4(1 - \sin \theta)(1 + \sin \theta) = 4r^4(1 - \sin^2 \theta) = 4r^4 \cos^2 \theta.\] Adding these equations, \[56^2 + 90^2 = 4r^4,\] so \[2r^2 = \sqrt{56^2+90^2} = 2\sqrt{28^2+45^2} = 2\sqrt{2809} = 2 \cdot 53 = \boxed{106}.\] ~OrangeQuail9 | 812,196,417,709,542,900 |
Alice knows that $3$ red cards and $3$ black cards will be revealed to her one at a time in random order. Before each card is revealed, Alice must guess its color. If Alice plays optimally, the expected number of cards she will guess correctly is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | \(\boxed{051}\) | Denote by $N_{i,j}$ the optimal expected number of cards that Alice guesses correctly, where the number of cards are $i$ and $j \ge i.$
If $i = 0$ then Alice guesses correctly all cards, so $N_{0,j} = j.$
If $j = i$ then Alice guesses next card with probability $\frac {1}{2} \implies N_{i,i} = \frac {1}{2} + N_{i-1,i}.$
If $j = i+1$ then Alice guesses next card with probability $\frac {i+1}{2i+1} \implies N_{i,i+1} = \frac {i+1}{2i+1} (1+ N_{i,i}) + \frac{i}{2i+1} N_{i-1,i+1}.$
If $j = i+2$ then Alice guesses next card with probability $\frac {i+2}{2i+2} \implies N_{i,i+2} = \frac {i+2}{2i+2} (1+ N_{i,i+1}) + \frac{i}{2i+2} N_{i-1,i+2}.$
One can find consistently: $N_{1,1} = \frac {1}{2} + N_{0,1} = \frac {3}{2},$ \[N_{1,2} = \frac {2}{3} (1 + N_{1,1}) + \frac {1}{3} N_{0,2} = \frac {7}{3}.\] \[N_{2,2} = \frac {1}{2} + N_{1,2} = \frac {17}{6}.\] \[N_{1,3} = \frac {3}{4} (1 + N_{1,2}) + \frac {1}{4} N_{0,3} = \frac {13}{4}.\] \[N_{2,3} = \frac {3}{5} (1 + N_{2,2}) + \frac {2}{5} N_{1,3} = \frac {18}{5}.\] \[N_{3,3} = \frac {1}{2} + N_{2,3} = \frac {41}{10}.\] Therefore, the answer is $41 + 10 = \boxed{\textbf{(051) }}$.
[email protected], vvsss | -3,001,802,648,747,907,600 |
Find the number of cubic polynomials $p(x) = x^3 + ax^2 + bx + c,$ where $a, b,$ and $c$ are integers in $\{-20,-19,-18,\ldots,18,19,20\},$ such that there is a unique integer $m \not= 2$ with $p(m) = p(2).$ | \(\boxed{738}\) | $p(x)-p(2)$ is a cubic with at least two integral real roots, therefore it has three real roots, which are all integers.
There are exactly two distinct roots, so either $p(x)=p(2)+(x-2)^2(x-m)$ or $p(x)=p(2)+(x-2)(x-m)^2$, with $m\neq 2$.
In the first case $p(x)=x^3-(4+m)x^2+(4+4m)x-4m+p(2)$, with $|4+4m|\leq 20$ (which entails $|4+m|\leq 20$), so $m$ can be $-6,-5,-4,-3,-2,-1,0,1, (\textbf{not 2}!), 3,4$ and $-4m+p(2)$ can be any integer from $-20$ to $20$, giving $410$ polynomials (since the coefficients are given by linear functions of $m$ and thus are distinct).
In the second case $p(x)=x^3-(2+2m)x^2+(4m+m^2)x-2m^2+p(2)$, and $m$ can be $-6,-5,-4,-3,-2,-1,0,1$ and $-4m+p(2)$ can be any integer from $-20$ to $20$, giving $328$ polynomials.
The total is $\boxed{738}$.
~EVIN-
~MathProblemSolvingSkills.com | -1,163,425,622,183,980,000 |
Find the number of subsets of $\{1,2,3,\ldots,10\}$ that contain exactly one pair of consecutive integers. Examples of such subsets are $\{\mathbf{1},\mathbf{2},5\}$ and $\{1,3,\mathbf{6},\mathbf{7},10\}.$ | \(\boxed{235}\) | The problem is the same as laying out a line of polynomoes to cover spots $0,1,...10$: 1 triomino ($RGG$), $n$ dominoes ($RG$), and $8-2n$ monominoes ($R$). The $G$ spots cover the members of the subset. The total number spots is 11, because one $R$ spot always covers the 0, and the other spots cover 1 through 10.
There are 5 ways to choose polyomino sets, and many ways to order each set:
$R + RG + RGG =$ Polyominoes $\rightarrow$ Orderings
$0 + 4 + 1 = 5 \rightarrow 5! / 0!4!1! = ~~~5$
$2 + 3 + 1 = 6 \rightarrow 6! / 2!3!1! = ~60$
$4 + 2 + 1 = 7 \rightarrow 7! / 4!2!1! = 105$
$6 + 1 + 1 = 8 \rightarrow 8! / 6!1!1! = ~56$
$8 + 0 + 1 = 9 \rightarrow 9! / 8!0!1! = ~~~9$
The sum is $\boxed{235}$.
~BraveCobra22aops | 2,355,908,779,465,259,500 |
The numbers of apples growing on each of six apple trees form an arithmetic sequence where the greatest number of apples growing on any of the six trees is double the least number of apples growing on any of the six trees. The total number of apples growing on all six trees is $990.$ Find the greatest number of apples growing on any of the six trees. | \(\boxed{220}\) | Let the terms in the sequence be defined as \[a_1, a_2, ..., a_6.\]
Since this is an arithmetic sequence, we have $a_1+a_6=a_2+a_5=a_3+a_4.$ So, \[\sum_{i=1}^6 a_i=3(a_1+a_6)=990.\] Hence, $(a_1+a_6)=330.$ And, since we are given that $a_6=2a_1,$ we get $3a_1=330\implies a_1=110$ and $a_6=\boxed{220}.$
~Kiran | -311,779,547,040,070,500 |
Recall that a palindrome is a number that reads the same forward and backward. Find the greatest integer less than $1000$ that is a palindrome both when written in base ten and when written in base eight, such as $292 = 444_{\text{eight}}.$ | \(\boxed{585}\) | Assuming that such palindrome is greater than $777_8 = 511,$ we conclude that the palindrome has four digits when written in base $8.$ Let such palindrome be \[(\underline{ABBA})_8 = 512A + 64B + 8B + A = 513A + 72B.\]
It is clear that $A=1,$ so we repeatedly add $72$ to $513$ until we get palindromes less than $1000:$ \begin{align*} 513+72\cdot0 &= 513, \\ 513+72\cdot1 &= \boxed{585}, \\ 513+72\cdot2 &= 657, \\ 513+72\cdot3 &= 729, \\ 513+72\cdot4 &= 801, \\ 513+72\cdot5 &= 873, \\ 513+72\cdot6 &= 945, \\ 513+72\cdot7 &= 1017. \\ \end{align*}
~MRENTHUSIASM
~MathProblemSolvingSkills.com | -2,381,908,026,295,517,700 |
Let $x,y,$ and $z$ be real numbers satisfying the system of equations \begin{align*} xy + 4z &= 60 \\ yz + 4x &= 60 \\ zx + 4y &= 60. \end{align*} Let $S$ be the set of possible values of $x.$ Find the sum of the squares of the elements of $S.$ | \(\boxed{273}\) | We first subtract the second equation from the first, noting that they both equal $60$. \begin{align*} xy+4z-yz-4x&=0 \\ 4(z-x)-y(z-x)&=0 \\ (z-x)(4-y)&=0 \end{align*}
Case 1: Let $y=4$.
The first and third equations simplify to: \begin{align*} x+z&=15 \\ xz&=44 \end{align*} from which it is apparent that $x=4$ and $x=11$ are solutions.
Case 2: Let $x=z$.
The first and third equations simplify to: \begin{align*} xy+4x&=60 \\ x^2+4y&=60 \end{align*}
We subtract the following equations, yielding: \begin{align*} x^2+4y-xy-4x&=0 \\ x(x-4)-y(x-4)&=0 \\ (x-4)(x-y)&=0 \end{align*}
We thus have $x=4$ and $x=y$, substituting in $x=y=z$ and solving yields $x=6$ and $x=-10$.
Then, we just add the squares of the solutions (make sure not to double count the $4$), and get \[4^2+11^2+(-6)^2+10^2=16+121+36+100=\boxed{273}.\] ~SAHANWIJETUNGA | -4,818,128,078,214,237,000 |
Let $S$ be the set of all positive rational numbers $r$ such that when the two numbers $r$ and $55r$ are written as fractions in lowest terms, the sum of the numerator and denominator of one fraction is the same as the sum of the numerator and denominator of the other fraction. The sum of all the elements of $S$ can be expressed in the form $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$ | \(\boxed{719}\) | Denote $r = \frac{a}{b}$, where $\left( a, b \right) = 1$. We have $55 r = \frac{55a}{b}$. Suppose $\left( 55, b \right) = 1$, then the sum of the numerator and the denominator of $55r$ is $55a + b$. This cannot be equal to the sum of the numerator and the denominator of $r$, $a + b$. Therefore, $\left( 55, b \right) \neq 1$.
Case 1: $b$ can be written as $5c$ with $\left( c, 11 \right) = 1$.
Thus, $55r = \frac{11a}{c}$.
Because the sum of the numerator and the denominator of $r$ and $55r$ are the same, \[ a + 5c = 11a + c . \]
Hence, $2c = 5 a$.
Because $\left( a, b \right) = 1$, $\left( a, c \right) = 1$. Thus, $a = 2$ and $c = 5$. Therefore, $r = \frac{a}{5c} = \frac{2}{25}$.
Case 2: $b$ can be written as $11d$ with $\left( d, 5 \right) = 1$.
Thus, $55r = \frac{5a}{c}$.
Because the sum of the numerator and the denominator of $r$ and $55r$ are the same, \[ a + 11c = 5a + c . \]
Hence, $2a = 5 c$.
Because $\left( a, b \right) = 1$, $\left( a, c \right) = 1$. Thus, $a = 5$ and $c = 2$. Therefore, $r = \frac{a}{11c} = \frac{5}{22}$.
Case 3: $b$ can be written as $55 e$.
Thus, $55r = \frac{a}{c}$.
Because the sum of the numerator and the denominator of $r$ and $55r$ are the same, \[ a + 55c = a + c . \]
Hence, $c = 0$. This is infeasible. Thus, there is no solution in this case.
Putting all cases together, $S = \left\{ \frac{2}{25}, \frac{5}{22} \right\}$. Therefore, the sum of all numbers in $S$ is \[ \frac{2}{25} + \frac{5}{22} = \frac{169}{550} . \]
Therefore, the answer is $169 + 550 = \boxed{\textbf{(719) }}$.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | 2,004,866,186,478,964,500 |
Consider the L-shaped region formed by three unit squares joined at their sides, as shown below. Two points $A$ and $B$ are chosen independently and uniformly at random from inside the region. The probability that the midpoint of $\overline{AB}$ also lies inside this L-shaped region can be expressed as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ [asy] unitsize(2cm); draw((0,0)--(2,0)--(2,1)--(1,1)--(1,2)--(0,2)--cycle); draw((0,1)--(1,1)--(1,0),dashed); [/asy] | \(\boxed{035}\) | Consider this diagram:
First, the one of points must be in the uppermost box and the other in the rightmost box. This happens with probability 2/3*1/3=2/9.
We need the midpoints of the $x$ coordinates to be greater than $1$ but less than $2.$
We need the midpoints of the $y$ coordinates to be greater than $1$ but less than $2.$
Thus, we set up: \[1<\frac{x_1+x_2}{2}<2\] \[1<\frac{y_1+y_2}{2}<2\]
Using geometric probability, as shown below, we get that the probability of happening is 1/2. This is for both. So we have probability 1/2*1/2=1/4.
Thus, the desired is 2/9*1/4=1/18 => 1-1/18=17/18.
mathboy282 | 4,533,259,266,860,700,700 |
Circles $\omega_1$ and $\omega_2$ intersect at two points $P$ and $Q,$ and their common tangent line closer to $P$ intersects $\omega_1$ and $\omega_2$ at points $A$ and $B,$ respectively. The line parallel to $AB$ that passes through $P$ intersects $\omega_1$ and $\omega_2$ for the second time at points $X$ and $Y,$ respectively. Suppose $PX=10,$ $PY=14,$ and $PQ=5.$ Then the area of trapezoid $XABY$ is $m\sqrt{n},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n.$ | \(\boxed{033}\) | Notice that line $\overline{PQ}$ is the radical axis of circles $\omega_1$ and $\omega_2$. By the radical axis theorem, we know that the tangents of any point on line $\overline{PQ}$ to circles $\omega_1$ and $\omega_2$ are equal. Therefore, line $\overline{PQ}$ must pass through the midpoint of $\overline{AB}$, call this point M. In addition, we know that $AM=MB=6$ by circle properties and midpoint definition.
Then, by Power of Point,
\begin{align*} AM^2&=MP*MQ \\ 36&=MP(MP+5) \\ MP&=4 \\ \end{align*}
Call the intersection point of line $\overline{A\omega_1}$ and $\overline{XY}$ be C, and the intersection point of line $\overline{B\omega_2}$ and $\overline{XY}$ be D. $ABCD$ is a rectangle with segment $MP=4$ drawn through it so that $AM=MB=6$, $CP=5$, and $PD=7$. Dropping the altitude from $M$ to $\overline{XY}$, we get that the height of trapezoid $XABY$ is $\sqrt{15}$. Therefore the area of trapezoid $XABY$ is
\begin{align*} \frac{1}{2}\cdot(12+24)\cdot(\sqrt{15})=18\sqrt{15} \end{align*}
Giving us an answer of $\boxed{033}$.
~Danielzh | -4,849,982,330,447,947,000 |
Let $N$ be the number of ways to place the integers $1$ through $12$ in the $12$ cells of a $2 \times 6$ grid so that for any two cells sharing a side, the difference between the numbers in those cells is not divisible by $3.$ One way to do this is shown below. Find the number of positive integer divisors of $N.$ \[\begin{array}{|c|c|c|c|c|c|} \hline \,1\, & \,3\, & \,5\, & \,7\, & \,9\, & 11 \\ \hline \,2\, & \,4\, & \,6\, & \,8\, & 10 & 12 \\ \hline \end{array}\] | \(\boxed{144}\) | We replace the numbers which are 0 mod(3) to 0, 1 mod(3) to 1, and 2 mod(3) to 2.
Then, the problem is equivalent to arranging 4 0's,4 1's, and 4 2's into the grid (and then multiplying by $4!^3$ to account for replacing the remainder numbers with actual numbers) such that no 2 of the same numbers are adjacent.
Then, the numbers which are vertically connected must be different. 2 1s must be connected with 2 2s, and 2 1s must be connected with 2 2s (vertically), as if there are less 1s connected with either, then there will be 2s or 3s which must be connected within the same number. For instance if we did did: - 12 x 3 - 13 x 1 Then we would be left with 23 x1, and 2 remaining 3s which aren't supposed to be connected, so it is impossible. Similar logic works for why all 1s can't be connected with all 2s.
Thus, we are left with the problem of re-arranging 2x 12 pairs, 2x 13 pairs, 2x 23 pairs. Notice that the pairs can be re-arranged horizontally in any configuration, but when 2 pairs are placed adjacent there is only 1 configuration for the rightmost pair to be set after the leftmost one has been placed.
We have $\frac{6!}{2!2!2!}=90$ ways to horizontally re-arrange the pairs, with 2 ways to set the initial leftmost column. Thus, there are 180 ways to arrange the pairs. Accounting for the initial simplification of the problem with 1-12 -> 0-3, we obtain the answer is:
\[2488320=2^{11}\cdot3^5\cdot5^1\]
The number of divisors is: $12\cdot6\cdot2=\boxed{144}.$ ~SAHANWIJETUNGA | -5,721,514,728,047,152,000 |
Find the number of collections of $16$ distinct subsets of $\{1,2,3,4,5\}$ with the property that for any two subsets $X$ and $Y$ in the collection, $X \cap Y \not= \emptyset.$ | \(\boxed{081}\) | Denote the $A$ as $\{ 1,2,3,4,5 \}$ and the collection of subsets as $S$.
Case 1: There are only sets of size $3$ or higher in $S$: Any two sets in $S$ must have at least one element common to both of them (since $3+3>5$). Since there are $16$ subsets of $A$ that have size $3$ or higher, there is only one possibility for this case.
Case 2: There are only sets of size $2$ or higher in $S$: Firstly, there cannot be a no size $2$ set $S$, or it will be overcounting the first case.
If there is only one such size $2$ set, there are $10$ ways to choose it. That size $2$ set, say $X$, cannot be in $S$ with $Y = A/X$ (a three element set). Thus, there are only $15$ possible size $3$ subsets that can be in $S$, giving us $10$ for this case.
If there are two sets with size $2$ in $S$, we can choose the common elements of these two subsets in $5$ ways, giving us a total of $5\cdot 6 = 30$.
If there are three sets with size $2$, they can either share one common element, which can be done in $5 \cdot 4 = 20$ ways, or they can share pairwise common elements (sort of like a cycle), which can be done $\binom{5}{2} = 10$ ways. In total, we have $30$ possibilities.
If there are four sets with size $2$, they all have to share one common element, which can be done in $5\cdot 1$ ways.
Thus, summing everything up, this will give us $75$ possible sets $S$
Case 3: There is a set with size $1$ in $S$: Notice that be at most one size $1$ subset. There are $5$ ways to choose that single element set. Say it's $\{ 1\}$. All other subsets in $S$ must have a $1$ in them, but there are only $2^4-1 = 15$ of them. Thus this case yields $5\cdot 1 = 5$ possibilities.
Thus, the total number of sets $S$ would be $1+75+5 = \boxed{\textbf{(081)}}$.
~sml1809 | 3,690,800,589,982,402,000 |
In $\triangle ABC$ with side lengths $AB = 13,$ $BC = 14,$ and $CA = 15,$ let $M$ be the midpoint of $\overline{BC}.$ Let $P$ be the point on the circumcircle of $\triangle ABC$ such that $M$ is on $\overline{AP}.$ There exists a unique point $Q$ on segment $\overline{AM}$ such that $\angle PBQ = \angle PCQ.$ Then $AQ$ can be written as $\frac{m}{\sqrt{n}},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | \(\boxed{247}\) | We use the law of Cosine and get \[AB^2 = AM^2 + BM^2 - 2 AM \cdot BM \cos \angle AMB,\] \[AC^2 = AM^2 + CM^2 + 2 AM \cdot CM \cos \angle AMB \implies\] \[AM^2 = \frac {AB^2 + AC^2}{2}- BM^2 = \sqrt{148} \approx 12.\] We use the power of point $M$ with respect circumcircle $\triangle ABC$ and get \[AM \cdot MP = BM \cdot CM = BM^2 \implies\] \[PM = \frac {49}{\sqrt {148}} \approx \frac {48}{12} \approx 4 < AM.\] It is clear that if $Q = P,$ then $\angle PBQ = \angle PCQ = 0 \implies$
if $Q$ is symmetric to $P$ with respect $M$ then $\angle PBQ = \angle PCQ.$
There exists a unique point $Q$ on segment $\overline{AM}, PM < AM \implies$ \[PQ = AM - PM = \frac{99}{\sqrt{148}} \implies \boxed{247}.\] [email protected], vvsss | 6,721,571,186,367,917,000 |
Let $A$ be an acute angle such that $\tan A = 2 \cos A.$ Find the number of positive integers $n$ less than or equal to $1000$ such that $\sec^n A + \tan^n A$ is a positive integer whose units digit is $9.$ | \(\boxed{167}\) | \[\tan A = 2 \cos A \implies \sin A = 2 \cos^2 A \implies \sin^2 A + \cos^2 A = 4 \cos^4 A + \cos^2 A = 1\] \[\implies \cos^2 A = \frac {\sqrt {17} - 1}{8}.\] \[c_n = \sec^n A + \tan^n A = \frac {1}{\cos^n A} + 2^n \cos^n A = (4\cos^2 A +1)^{\frac {n}{2}}+(4 \cos^2 A)^{\frac {n}{2}} =\] \[= \left(\frac {\sqrt {17} + 1}{2}\right)^{\frac {n}{2}}+ \left(\frac {\sqrt {17} - 1}{2}\right)^{\frac {n}{2}}.\]
It is clear, that $c_n$ is not integer if $n \ne 4k, k > 0.$
Denote $x = \frac {\sqrt {17} + 1}{2}, y = \frac {\sqrt {17} - 1}{2} \implies$ \[x \cdot y = 4, x + y = \sqrt{17}, x - y = 1 \implies x^2 + y^2 = (x - y)^2 + 2xy = 9 = c_4.\]
\[c_8 = x^4 + y^4 = (x^2 + y^2)^2 - 2x^2 y^2 = 9^2 - 2 \cdot 16 = 49.\] \[c_{4k+4} = x^{4k+4} + y^{4k+4} = (x^{4k} + y^{4k})(x^2+y^2)- (x^2 y^2)(x^{4k-2}+y^{4k-2}) = 9 c_{4k}- 16 c_{4k – 4} \implies\] \[c_{12} = 9 c_8 - 16 c_4 = 9 \cdot 49 - 16 \cdot 9 = 9 \cdot 33 = 297.\] \[c_{16} = 9 c_{12} - 16 c_8 = 9 \cdot 297 - 16 \cdot 49 = 1889.\] \[c_{12m + 4} \pmod{10} = 9 \cdot c_{12m} \pmod{10} - 16 \pmod{10} \cdot c_{12m - 4} \pmod{10} =\] \[= (9 \cdot 7 - 6 \cdot 9) \pmod{10} = (3 - 4) \pmod{10} = 9.\] \[c_{12m + 8}\pmod{10} = 9 \cdot c_{12m+4} \pmod{10} - 16 \pmod{10} \cdot c_{12m } \pmod{10} =\] \[= (9 \cdot 9 - 6 \cdot 7) \pmod{10} = (1 - 2)\pmod{10} = 9.\] \[c_{12m + 12} \pmod{10} = 9 \cdot c_{12m + 8} \pmod{10} - 16 \pmod{10} \cdot c_{12m + 4} \pmod{10} =\] \[= (9 \cdot 9 - 6 \cdot 9) \pmod{10} = (1 - 4) \pmod{10} = 7 \implies\]
The condition is satisfied iff $n = 12 k + 4$ or $n = 12k + 8.$
If $n \le N$ then the number of possible n is $\left\lfloor \frac{N}{4} \right\rfloor - \left\lfloor \frac{N}{12} \right\rfloor.$
For $N = 1000$ we get $\left\lfloor \frac{1000}{4} \right\rfloor - \left\lfloor \frac{1000}{12} \right\rfloor = 250 - 83 = \boxed{167}.$
[email protected], vvsss
~MathProblemSolvingSkills.com | 3,405,713,459,275,372,000 |
For each positive integer $n$ let $a_n$ be the least positive integer multiple of $23$ such that $a_n \equiv 1 \pmod{2^n}.$ Find the number of positive integers $n$ less than or equal to $1000$ that satisfy $a_n = a_{n+1}.$ | \(\boxed{363}\) | Observe that if $a_{n-1} - 1$ is divisible by $2^n$, $a_n = a_{n-1}$. If not, $a_n = a_{n-1} + 23 \cdot 2^{n-1}$.
This encourages us to let $b_n = \frac{a_n - 1}{2^n}$. Rewriting the above equations, we have \[b_n = \begin{cases} \frac{b_{n-1}}{2} & \text{if } 2 \text{ } \vert \text{ } b_{n-1} \\ \frac{b_{n-1}+23}{2} &\text{if } 2 \not\vert \text{ } b_{n-1} \end{cases}\] The first few values of $b_n$ are $11, 17, 20, 10, 5, 14, 7, 15, 19, 21,$ and $22$. We notice that $b_{12} = b_1 = 11$, and thus the sequence is periodic with period $11$.
Note that $a_n = a_{n+1}$ if and only if $b_n$ is even. This occurs when $n$ is congruent to $0, 3, 4$ or $6$ mod $11$, giving four solutions for each period.
From $1$ to $1001$ (which is $91 \times 11$), there are $91 \times 4 = 364$ values of $n$. We subtract $1$ from the total since $1001$ satisfies the criteria but is greater than $1000$ to get a final answer of $\fbox{363}$ . ~Bxiao31415
(small changes by bobjoebilly and IraeVid13) | 5,793,692,957,040,758,000 |
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