problem stringlengths 10 5.15k | answer stringlengths 0 1.22k | solution stringlengths 0 11.1k | difficulty float64 0.75 2.02k | difficulty_raw listlengths 3 8 |
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A set contains four numbers. The six pairwise sums of distinct elements of the set, in no particular order, are $189$, $320$, $287$, $234$, $x$, and $y$. Find the greatest possible value of $x+y$. | 791 | Let the four numbers be $a$, $b$, $c$, and $d$, in no particular order. Adding the pairwise sums, we have $3a+3b+3c+3d=1030+x+y$, so $x+y=3(a+b+c+d)-1030$. Since we want to maximize $x+y$, we must maximize $a+b+c+d$.
Of the four sums whose values we know, there must be two sums that add to $a+b+c+d$. To maximize this ... | 5.25 | [
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Find the sum of all positive integers $n$ such that $\sqrt{n^2+85n+2017}$ is an integer. | 195 | Notice that $(n+42)^2= n^2+84n+1764$. Also note that $(n+45)^2= n^2+90n+2025$. Thus, \[(n+42)^2< n^2+85n+2017<(n+45)^2\] where $n^2+85n+2017$ is a perfect square. Hence,\[n^2+85n+2017= (n+43)^2\] or \[n^2+85n+2017= (n+44)^2.\] Solving the two equations yields the two solutions $n= 168, 27$. Therefore, our answer is $\b... | 4.875 | [
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Find the number of integer values of $k$ in the closed interval $[-500,500]$ for which the equation $\log(kx)=2\log(x+2)$ has exactly one real solution. | 501 | [asy] Label f; f.p=fontsize(5); xaxis(-3,3,Ticks(f,1.0)); yaxis(-3,26,Ticks(f,1.0)); real f(real x){return (x+2)^2;} real g(real x){return x*-1;} real h(real x){return x*-2;} real i(real x){return x*-3;} real j(real x){return x*8;} draw(graph(f,-2,3),green); draw(graph(g,-2,2),red); draw(graph(h,-2,1),red); draw(graph(... | 4.8125 | [
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Find the number of positive integers $n$ less than $2017$ such that \[1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!}\] is an integer. | 134 | Note that $1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}$ will have a denominator that divides $5!$. Therefore, for the expression to be an integer, $\frac{n^6}{6!}$ must have a denominator that divides $5!$. Thus, $6\mid n^6$, and $6\mid n$. Let $n=6m$. Substituting gives $1+6m+\frac{6^2m^2}{2!}+\fra... | 6 | [
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A special deck of cards contains $49$ cards, each labeled with a number from $1$ to $7$ and colored with one of seven colors. Each number-color combination appears on exactly one card. Sharon will select a set of eight cards from the deck at random. Given that she gets at least one card of each color and at least one c... | 13 | Without loss of generality, assume that the $8$ numbers on Sharon's cards are $1$, $1$, $2$, $3$, $4$, $5$, $6$, and $7$, in that order, and assume the $8$ colors are red, red, and six different arbitrary colors. There are ${8\choose2}-1$ ways of assigning the two red cards to the $8$ numbers; we subtract $1$ because w... | 6.375 | [
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Rectangle $ABCD$ has side lengths $AB=84$ and $AD=42$. Point $M$ is the midpoint of $\overline{AD}$, point $N$ is the trisection point of $\overline{AB}$ closer to $A$, and point $O$ is the intersection of $\overline{CM}$ and $\overline{DN}$. Point $P$ lies on the quadrilateral $BCON$, and $\overline{BP}$ bisects the a... | 546 | [asy] pair A,B,C,D,M,n,O,P; A=(0,42);B=(84,42);C=(84,0);D=(0,0);M=(0,21);n=(28,42);O=(12,18);P=(32,13); fill(C--D--P--cycle,blue); draw(A--B--C--D--cycle); draw(C--M); draw(D--n); draw(B--P); draw(D--P); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$M$",M,W); label("$N$",n,N); labe... | 6 | [
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Five towns are connected by a system of roads. There is exactly one road connecting each pair of towns. Find the number of ways there are to make all the roads one-way in such a way that it is still possible to get from any town to any other town using the roads (possibly passing through other towns on the way). | 544 | As noted before, you can see that there are 2 ways that the condition cannot be met; it either has all roads leading into one city, or all roads leading out of one city. We then use complementary counting to count these cases, with PIE. Obviously there are $2^5*2^5$ ways to make roads (just draw a pentagon with all of ... | 6.75 | [
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Circle $C_0$ has radius $1$, and the point $A_0$ is a point on the circle. Circle $C_1$ has radius $r<1$ and is internally tangent to $C_0$ at point $A_0$. Point $A_1$ lies on circle $C_1$ so that $A_1$ is located $90^{\circ}$ counterclockwise from $A_0$ on $C_1$. Circle $C_2$ has radius $r^2$ and is internally tangent... | 110 | Let the center of circle $C_i$ be $O_i$. Note that $O_0BO_1$ is a right triangle, with right angle at $B$. Also, $O_1B=\frac{11}{60}O_0B$, or $O_0B = \frac{60}{61}O_0O_1$. It is clear that $O_0O_1=1-r=\frac{49}{60}$, so $O_0B=\frac{60}{61}\times\frac{49}{60}=\frac{49}{61}$. Our answer is $49+61=\boxed{110}$
-william12... | 6.125 | [
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For each integer $n\geq3$, let $f(n)$ be the number of $3$-element subsets of the vertices of the regular $n$-gon that are the vertices of an isosceles triangle (including equilateral triangles). Find the sum of all values of $n$ such that $f(n+1)=f(n)+78$. | 245 | Considering $n \pmod{6}$, we have the following formulas:
Even and a multiple of 3: $\frac{n(n-4)}{2} + \frac{n}{3}$
Even and not a multiple of 3: $\frac{n(n-2)}{2}$
Odd and a multiple of 3: $\frac{n(n-3)}{2} + \frac{n}{3}$
Odd and not a multiple of 3: $\frac{n(n-1)}{2}$
To derive these formulas, we note the follo... | 7 | [
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A $10\times10\times10$ grid of points consists of all points in space of the form $(i,j,k)$, where $i$, $j$, and $k$ are integers between $1$ and $10$, inclusive. Find the number of different lines that contain exactly $8$ of these points. | 168 | Considered the cases $(1, 2, ..., 8), (2, 3, ...,9), (3, 4, ..., 10)$ and reverse. Also, consider the constant subsequences of length 8 $(1, 1, ..., 1), (2, 2, ..., 2), ..., (10, 10, ..., 10)$. Of all the triplets that work they cannot be extended to form another point on the line in the $10 \times 10 \times 10$ grid b... | 6.25 | [
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Tetrahedron $ABCD$ has $AD=BC=28$, $AC=BD=44$, and $AB=CD=52$. For any point $X$ in space, suppose $f(X)=AX+BX+CX+DX$. The least possible value of $f(X)$ can be expressed as $m\sqrt{n}$, where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$. | 682 | Set $a=BC=28$, $b=CA=44$, $c=AB=52$. Let $O$ be the point which minimizes $f(X)$.
$\boxed{\textrm{Claim 1: } O \textrm{ is the gravity center } \ \tfrac {1}{4}(\vec A + \vec B + \vec C + \vec D)}$
$\textrm{Proof:}$ Let $M$ and $N$ denote the midpoints of $AB$ and $CD$. From $\triangle ABD \cong \triangle BAC$ and $\tri... | 7.5 | [
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Let $S$ be the number of ordered pairs of integers $(a,b)$ with $1 \leq a \leq 100$ and $b \geq 0$ such that the polynomial $x^2+ax+b$ can be factored into the product of two (not necessarily distinct) linear factors with integer coefficients. Find the remainder when $S$ is divided by $1000$. | 600 | By Vietas, the sum of the roots is $-a$ and the product is $b$. Therefore, both roots are nonpositive. For each value of $a$ from $1$ to $100$, the number of $b$ values is the number of ways to sum two numbers between $0$ and $a-1$ inclusive to $a$. This is just $1 + 2 + 2 + 3 + 3 +... 50 + 50 + 51 = 2600$. Thus, the a... | 4.75 | [
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The number $n$ can be written in base $14$ as $\underline{a}\text{ }\underline{b}\text{ }\underline{c}$, can be written in base $15$ as $\underline{a}\text{ }\underline{c}\text{ }\underline{b}$, and can be written in base $6$ as $\underline{a}\text{ }\underline{c}\text{ }\underline{a}\text{ }\underline{c}\text{ }$, whe... | 925 | We have these equations: $196a+14b+c=225a+15c+b=222a+37c$. Taking the last two we get $3a+b=22c$. Because $c \neq 0$ otherwise $a \ngtr 0$, and $a \leq 5$, $c=1$.
Then we know $3a+b=22$. Taking the first two equations we see that $29a+14c=13b$. Combining the two gives $a=4, b=10, c=1$. Then we see that $222 \times 4+3... | 6 | [
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Kathy has $5$ red cards and $5$ green cards. She shuffles the $10$ cards and lays out $5$ of the cards in a row in a random order. She will be happy if and only if all the red cards laid out are adjacent and all the green cards laid out are adjacent. For example, card orders RRGGG, GGGGR, or RRRRR will make Kathy happy... | 157 | Assume without loss of generality that the first card laid out is red. Then the arrangements that satisfy Kathy’s requirements are RRRRR, RRRRG, RRRGG, RRGGG, and RGGGG. The probability that Kathy will lay out one of these arrangements is \[\frac49\cdot\frac38\cdot\frac27\cdot\frac16\] \[\frac49\cdot\frac38\cdot\frac27... | 4.25 | [
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In $\triangle ABC, AB = AC = 10$ and $BC = 12$. Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$. Then $AD$ can be expressed in the form $\dfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find ... | 289 | By the Law of Cosines, we have: \[\cos(A) = \frac{10^2+10^2-12^2}{2*10*10} = \frac{7}{25}\] Let $M$ be midpoint of $AE$, then \[\frac{7}{25} = \frac{10-x}{2x} \iff x =\frac{250}{39}\] So, our answer is $250+39=\boxed{289}$. | 5 | [
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For each ordered pair of real numbers $(x,y)$ satisfying \[\log_2(2x+y) = \log_4(x^2+xy+7y^2)\]there is a real number $K$ such that \[\log_3(3x+y) = \log_9(3x^2+4xy+Ky^2).\]Find the product of all possible values of $K$. | 189 | Using the logarithmic property $\log_{a^n}b^n = \log_{a}b$, we note that \[(2x+y)^2 = x^2+xy+7y^2\] That gives \[x^2+xy-2y^2=0\] upon simplification and division by $3$. Factoring $x^2+xy-2y^2=0$ by Simon's Favorite Factoring Trick gives \[(x+2y)(x-y)=0\] Then, \[x=y \text{ or }x=-2y\] From the second equation, \[9x^2+... | 6 | [
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Let $N$ be the number of complex numbers $z$ with the properties that $|z|=1$ and $z^{6!}-z^{5!}$ is a real number. Find the remainder when $N$ is divided by $1000$. | 440 | As mentioned in solution one, for the difference of two complex numbers to be real, their imaginary parts must be equal. We use exponential form of complex numbers. Let $z = e^{i \theta}$. We have two cases to consider. Either $z^{6!} = z^{5!}$, or $z^{6!}$ and $z^{5!}$ are reflections across the imaginary axis. If $z^... | 6.25 | [
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A right hexagonal prism has height $2$. The bases are regular hexagons with side length $1$. Any $3$ of the $12$ vertices determine a triangle. Find the number of these triangles that are isosceles (including equilateral triangles). | 52 | If there are two edges on a single diameter, there would be six diameters. There are four ways to put the third number, and four equilateral triangles. There are $4+ 2 \cdot 2\cdot 6 = 28$ ways. Then, if one length was $\sqrt{3}$ but no side on the diameter, there would be twelve was to put the $\sqrt{3}$ side, and two... | 6.125 | [
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Let $ABCDEF$ be an equiangular hexagon such that $AB=6, BC=8, CD=10$, and $DE=12$. Denote by $d$ the diameter of the largest circle that fits inside the hexagon. Find $d^2$. | 147 | First of all, draw a good diagram! This is always the key to solving any geometry problem. Once you draw it, realize that $EF=2, FA=16$. Why? Because since the hexagon is equiangular, we can put an equilateral triangle around it, with side length $6+8+10=24$. Then, if you drew it to scale, notice that the "widest" this... | 6.125 | [
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Find the number of four-element subsets of $\{1,2,3,4,\dots, 20\}$ with the property that two distinct elements of a subset have a sum of $16$, and two distinct elements of a subset have a sum of $24$. For example, $\{3,5,13,19\}$ and $\{6,10,20,18\}$ are two such subsets. | 210 | Let's say our four elements in our subset are $a,b,c,d$. We have two cases. Note that the order of the elements / the element letters themselves don't matter since they are all on equal grounds at the start.
$\textrm{Case } 1 \textrm{:}$ $a+b = 16$ and $c+d = 24$.
List out possibilities for $a+b$ $(\text{i.e. } 1+15, ... | 6.125 | [
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The wheel shown below consists of two circles and five spokes, with a label at each point where a spoke meets a circle. A bug walks along the wheel, starting at point $A$. At every step of the process, the bug walks from one labeled point to an adjacent labeled point. Along the inner circle the bug only walks in a coun... | 4 | Let an $O$ signal a move that ends in the outer circle and $I$ signal a move that ends in the inner circle. Now notice that for a string of $15$ moves to end at $A$, the difference between $O$'s and $I$'s in the string must be a multiple of $5$.
$15$ $I$'s: Trivially $1$ case.
$5$ $O$'s and $10$ $I$'s: Since the string... | 6.125 | [
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Find the least positive integer $n$ such that when $3^n$ is written in base $143$, its two right-most digits in base $143$ are $01$. | 195 | We have that \[3^n \equiv 1 \pmod{143^2}.\]Now, $3^{110} \equiv 1 \pmod{11^2}$ so by the Fundamental Theorem of Orders, $\text{ord}_{11^2}(3)|110$ and with some bashing, we get that it is $5$. Similarly, we get that $\text{ord}_{13^2}(3)=39$. Now, $\text{lcm}(39,5)=\boxed{195}$ which is our desired solution. | 6.875 | [
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For every subset $T$ of $U = \{ 1,2,3,\ldots,18 \}$, let $s(T)$ be the sum of the elements of $T$, with $s(\emptyset)$ defined to be $0$. If $T$ is chosen at random among all subsets of $U$, the probability that $s(T)$ is divisible by $3$ is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find ... | 683 | Consider the elements of $U$ modulo $3.$
Ignore the $0$'s because we're gonna multiply $\binom{6}{0}+..+\binom{6}{6}=2^6$ at the end. Let $a$ be the $1's$ and $b$ be the $2's$. The key here is that $2 \equiv -1 \pmod{3}$ so the difference between the number of $a$ and $b$ is a multiple of $3$.
1. Counted twice becaus... | 6.75 | [
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Let $\triangle ABC$ have side lengths $AB=30$, $BC=32$, and $AC=34$. Point $X$ lies in the interior of $\overline{BC}$, and points $I_1$ and $I_2$ are the incenters of $\triangle ABX$ and $\triangle ACX$, respectively. Find the minimum possible area of $\triangle AI_1I_2$ as $X$ varies along $\overline{BC}$. | 126 | First note that \[\angle I_1AI_2 = \angle I_1AX + \angle XAI_2 = \frac{\angle BAX}2 + \frac{\angle CAX}2 = \frac{\angle A}2\] is a constant not depending on $X$, so by $[AI_1I_2] = \tfrac12(AI_1)(AI_2)\sin\angle I_1AI_2$ it suffices to minimize $(AI_1)(AI_2)$. Let $a = BC$, $b = AC$, $c = AB$, and $\alpha = \angle AXB$... | 6.75 | [
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Let $SP_1P_2P_3EP_4P_5$ be a heptagon. A frog starts jumping at vertex $S$. From any vertex of the heptagon except $E$, the frog may jump to either of the two adjacent vertices. When it reaches vertex $E$, the frog stops and stays there. Find the number of distinct sequences of jumps of no more than $12$ jumps that end... | 351 | Let $E_n$ denotes the number of sequences with length $n$ that ends at $E$. Define similarly for the other vertices. We seek for a recursive formula for $E_n$. \begin{align*} E_n&=P_{3_{n-1}}+P_{4_{n-1}} \\ &=P_{2_{n-2}}+P_{5_{n-2}} \\ &=P_{1_{n-3}}+P_{3_{n-3}}+S_{n-3}+P_{4_{n-3}} \\ &=(P_{3_{n-3}}+P_{4_{n-3}})+S_{n-3}... | 6 | [
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David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, $A,\text{ }B,\text{ }C$, which can each be inscribed in a circle with radius $1$. Let $\varphi_A$ denote the measure of the acute angle made by the diagonals of quadrilateral $A$, and define $\varphi_... | 59 | Let the sides of the quadrilaterals be $a,b,c,$ and $d$ in some order such that $A$ has $a$ opposite of $c$, $B$ has $a$ opposite of $b$, and $C$ has $a$ opposite of $d$. Then, let the diagonals of $A$ be $e$ and $f$. Similarly to solution $2$, we get that $\tfrac{2}{3}(ac+bd)=\tfrac{3}{5}(ab+cd)=\tfrac{6}{7}(ad+bc)=2K... | 7 | [
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Points $A$, $B$, and $C$ lie in that order along a straight path where the distance from $A$ to $C$ is $1800$ meters. Ina runs twice as fast as Eve, and Paul runs twice as fast as Ina. The three runners start running at the same time with Ina starting at $A$ and running toward $C$, Paul starting at $B$ and running towa... | 800 | Let $x$ be the distance from $A$ to $B$. Then the distance from $B$ to $C$ is $1800-x$. Since Eve is the slowest, we can call her speed $v$, so that Ina's speed is $2v$ and Paul's speed is $4v$.
For Paul and Eve to meet, they must cover a total distance of $1800-x$ which takes them a time of $\frac{1800-x}{4v+v}$. Pau... | 6 | [
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Let $a_{0} = 2$, $a_{1} = 5$, and $a_{2} = 8$, and for $n > 2$ define $a_{n}$ recursively to be the remainder when $4$($a_{n-1}$ $+$ $a_{n-2}$ $+$ $a_{n-3}$) is divided by $11$. Find $a_{2018} \cdot a_{2020} \cdot a_{2022}$. | 112 | When given a sequence problem, one good thing to do is to check if the sequence repeats itself or if there is a pattern.
After computing more values of the sequence, it can be observed that the sequence repeats itself every 10 terms starting at $a_{0}$.
$a_{0} = 2$, $a_{1} = 5$, $a_{2} = 8$, $a_{3} = 5$, $a_{4} = 6$, ... | 4 | [
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Find the sum of all positive integers $b < 1000$ such that the base-$b$ integer $36_{b}$ is a perfect square and the base-$b$ integer $27_{b}$ is a perfect cube. | 371 | The conditions are: \[3b+6 = n^2\] \[2b+7 = m^3\] We can see $n$ is multiple is 3, so let $n=3k$, then $b= 3k^2-2$. Substitute $b$ into second condition and we get $m^3=3(2k^2+1)$. Now we know $m$ is both a multiple of 3 and odd. Also, $m$ must be smaller than 13 for $b$ to be smaller than 1000. So the only two possibl... | 5.875 | [
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In equiangular octagon $CAROLINE$, $CA = RO = LI = NE =$ $\sqrt{2}$ and $AR = OL = IN = EC = 1$. The self-intersecting octagon $CORNELIA$ encloses six non-overlapping triangular regions. Let $K$ be the area enclosed by $CORNELIA$, that is, the total area of the six triangular regions. Then $K =$ $\dfrac{a}{b}$, where $... | 23 | We can draw $CORNELIA$ and introduce some points.
The diagram is essentially a 3x3 grid where each of the 9 squares making up the grid have a side length of 1.
In order to find the area of $CORNELIA$, we need to find 4 times the area of $\bigtriangleup$$ACY$ and 2 times the area of $\bigtriangleup$$YZW$.
Using simil... | 6 | [
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Suppose that $x$, $y$, and $z$ are complex numbers such that $xy = -80 - 320i$, $yz = 60$, and $zx = -96 + 24i$, where $i$ $=$ $\sqrt{-1}$. Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi$. Find $a^2 + b^2$. | 74 | We can turn the expression $x+y+z$ into $\sqrt{x^2+y^2+z^2+2xy+2yz+2xz}$, and this would allow us to plug in the values after some computations.
Based off of the given products, we have
$xy^2z=60(-80-320i)$
$xyz^2=60(-96+24i)$
$x^2yz=(-96+24i)(-80-320i)$.
Dividing by the given products, we have
$y^2=\frac{60(-80-320i... | 6.75 | [
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A real number $a$ is chosen randomly and uniformly from the interval $[-20, 18]$. The probability that the roots of the polynomial
$x^4 + 2ax^3 + (2a - 2)x^2 + (-4a + 3)x - 2$
are all real can be written in the form $\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 37 | The polynomial we are given is rather complicated, so we could use Rational Root Theorem to turn the given polynomial into a degree-2 polynomial. With Rational Root Theorem, $x = 1, -1, 2, -2$ are all possible rational roots. Upon plugging these roots into the polynomial, $x = -2$ and $x = 1$ make the polynomial equal ... | 6.5 | [
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Triangle $ABC$ has side lengths $AB = 9$, $BC =$ $5\sqrt{3}$, and $AC = 12$. Points $A = P_{0}, P_{1}, P_{2}, ... , P_{2450} = B$ are on segment $\overline{AB}$ with $P_{k}$ between $P_{k-1}$ and $P_{k+1}$ for $k = 1, 2, ..., 2449$, and points $A = Q_{0}, Q_{1}, Q_{2}, ... , Q_{2450} = C$ are on segment $\overline{AC}$... | 20 | Let $T_1$ stand for $AP_1Q_1$, and $T_k = AP_kQ_k$. All triangles $T$ are similar by AA. Let the area of $T_1$ be $x$. The next trapezoid will also have an area of $x$, as given. Therefore, $T_k$ has an area of $kx$. The ratio of the areas is equal to the square of the scale factor for any plane figure and its image. T... | 7 | [
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A frog is positioned at the origin of the coordinate plane. From the point $(x, y)$, the frog can jump to any of the points $(x + 1, y)$, $(x + 2, y)$, $(x, y + 1)$, or $(x, y + 2)$. Find the number of distinct sequences of jumps in which the frog begins at $(0, 0)$ and ends at $(4, 4)$. | 556 | Casework Solution: x-distribution: 1-1-1-1 (1 way to order) y-distribution: 1-1-1-1 (1 way to order) $\dbinom{8}{4} = 70$ ways total
x-distribution: 1-1-1-1 (1 way to order) y-distribution: 1-1-2 (3 ways to order) $\dbinom{7}{3} \times 3= 105$ ways total
x-distribution: 1-1-1-1 (1 way to order) y-distribution: 2-2 (1... | 4.5 | [
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Octagon $ABCDEFGH$ with side lengths $AB = CD = EF = GH = 10$ and $BC = DE = FG = HA = 11$ is formed by removing 6-8-10 triangles from the corners of a $23$ $\times$ $27$ rectangle with side $\overline{AH}$ on a short side of the rectangle, as shown. Let $J$ be the midpoint of $\overline{AH}$, and partition the octagon... | 184 | Draw the heptagon whose vertices are the midpoints of octagon $ABCDEFGH$ except $J$. We have a homothety since:
1. $J$ passes through corresponding vertices of the two heptagons.
2. By centroid properties, our ratio between the sidelengths is $\frac{2}{3},$ and their area ratio is hence $\frac{4}{9}.$
Compute the ar... | 6.75 | [
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Find the number of functions $f(x)$ from $\{1, 2, 3, 4, 5\}$ to $\{1, 2, 3, 4, 5\}$ that satisfy $f(f(x)) = f(f(f(x)))$ for all $x$ in $\{1, 2, 3, 4, 5\}$. | 756 | Just to visualize solution 1. If we list all possible $(x,f(x))$, from ${1,2,3,4,5}$ to ${1,2,3,4,5}$ in a specific order, we get $5 \cdot 5 = 25$ different $(x,f(x))$ 's. Namely:
\[(1,1) (1,2) (1,3) (1,4) (1,5)\] \[(2,1) (2,2) (2,3) (2,4) (2,5)\] \[(3,1) (3,2) (3,3) (3,4) (3,5)\] \[(4,1) (4,2) (4,3) (4,4) (4,5)\] \[(5... | 6.125 | [
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6,
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] |
Find the number of permutations of $1, 2, 3, 4, 5, 6$ such that for each $k$ with $1$ $\leq$ $k$ $\leq$ $5$, at least one of the first $k$ terms of the permutation is greater than $k$. | 461 | If $6$ is the first number, then there are no restrictions. There are $5!$, or $120$ ways to place the other $5$ numbers.
If $6$ is the second number, then the first number can be $2, 3, 4,$ or $5$, and there are $4!$ ways to place the other $4$ numbers. $4 \cdot 4! = 96$ ways.
If $6$ is the third number, then we can... | 6.25 | [
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Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$, $BC = 14$, and $AD = 2\sqrt{65}$. Assume that the diagonals of $ABCD$ intersect at point $P$, and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$. Find the area of quadrilateral $ABCD$.
Diagram
Let... | 112 | Either $PA=PC$ or $PD=PB$. Let $PD=PB=s$. Applying Stewart's Theorem on $\triangle ABD$ and $\triangle BCD$, dividing by $2s$ and rearranging, \[\tag{1}CP^2+s^2=148\] \[\tag{2}AP^2+s^2=180\] Applying Stewart on $\triangle CAB$ and $\triangle CAD$, \[\tag{3} 5CP^2=3AP^2\] Substituting equations 1 and 2 into 3 and rearra... | 6.75 | [
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] |
Misha rolls a standard, fair six-sided die until she rolls $1-2-3$ in that order on three consecutive rolls. The probability that she will roll the die an odd number of times is $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 647 | Let $P_n$ be the probability of getting consecutive $1,2,3$ rolls in $n$ rolls and not rolling $1,2,3$ prior to the nth roll.
Let $x = P_3+P_5+...=1-(P_4+P_6+..)$. Following Solution 2, one can see that \[P_{n+1}=P_{n}-\frac{P_{n-2}}{6^3}\] for all positive integers $n \ge 5$. Summing for $n=5,7,...$ gives \[(1-x)-\fr... | 6.125 | [
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The incircle $\omega$ of triangle $ABC$ is tangent to $\overline{BC}$ at $X$. Let $Y \neq X$ be the other intersection of $\overline{AX}$ with $\omega$. Points $P$ and $Q$ lie on $\overline{AB}$ and $\overline{AC}$, respectively, so that $\overline{PQ}$ is tangent to $\omega$ at $Y$. Assume that $AP = 3$, $PB = 4$, $AC... | 227 | Let the incircle of $ABC$ be tangent to $AB$ and $AC$ at $M$ and $N$. By Brianchon's theorem on tangential hexagons $QNCBMP$ and $PYQCXB$, we know that $MN,CP,BQ$ and $XY$ are concurrent at a point $O$. Let $PQ \cap BC = Z$. Then by La Hire's $A$ lies on the polar of $Z$ so $Z$ lies on the polar of $A$. Therefore, $MN$... | 7.125 | [
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] |
Find the number of functions $f$ from $\{0, 1, 2, 3, 4, 5, 6\}$ to the integers such that $f(0) = 0$, $f(6) = 12$, and \[|x - y| \leq |f(x) - f(y)| \leq 3|x - y|\] for all $x$ and $y$ in $\{0, 1, 2, 3, 4, 5, 6\}$. | 185 | Because $f(n)$ and $f(n+1)$ can differ by at most 3 for $n=0,1,2,3,4,5$, $f$ can decrease at most once. This is because if $f$ decreases $2$ times (or more) then we get a contradiction: $12 = f(6) \le 3\cdot 4 + (-1)\cdot 2 = 10$.
If $f$ never decreases, then $f(n+1)-f(n)\in \{1,2,3\}$ for all $n$. Let $a, b$, and $c$... | 7 | [
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Consider the integer \[N = 9 + 99 + 999 + 9999 + \cdots + \underbrace{99\ldots 99}_\text{321 digits}.\]Find the sum of the digits of $N$. | 342 | Observe how adding results in the last term but with a $1$ concatenated in front and also a $1$ subtracted ($09$, $108$, $1107$, $11106$). Then for any index of terms, $n$, the sum is $11...10-n$, where the first term is of length $n+1$. Here, that is $\boxed{342}$.
~BJHHar | 3 | [
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Jenn randomly chooses a number $J$ from $1, 2, 3,\ldots, 19, 20$. Bela then randomly chooses a number $B$ from $1, 2, 3,\ldots, 19, 20$ distinct from $J$. The value of $B - J$ is at least $2$ with a probability that can be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Fin... | 29 | $B-J \ne 0$ because $B \ne J$, so the probability that $B-J < 0$ is $\frac{1}{2}$ by symmetry.
The probability that $B-J = 1$ is $\frac{19}{20 \times 19} = \frac{1}{20}$ because there are 19 pairs: $(B,J) = (2,1), \ldots, (20,19)$.
The probability that $B-J \ge 2$ is $1-\frac{1}{2}-\frac{1}{20} = \frac{9}{20} \implie... | 3.625 | [
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In $\triangle PQR$, $PR=15$, $QR=20$, and $PQ=25$. Points $A$ and $B$ lie on $\overline{PQ}$, points $C$ and $D$ lie on $\overline{QR}$, and points $E$ and $F$ lie on $\overline{PR}$, with $PA=QB=QC=RD=RE=PF=5$. Find the area of hexagon $ABCDEF$.
Diagram
[asy] dot((0,0)); dot((15,0)); dot((15,20)); draw((0,0)--(15,0)--... | 120 | Note that $\triangle{PQR}$ has area $150$ and is a $3$-$4$-$5$ right triangle. Then, by similar triangles, the altitude from $B$ to $QC$ has length $3$ and the altitude from $A$ to $FP$ has length $4$, so $[QBC]+[DRE]+[AFP]=\frac{15}{2}+\frac{25}{2}+\frac{20}{2}=30$, meaning that $[ABCDEF]=\boxed{120}$. -Stormersyle | 5.5 | [
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A soccer team has $22$ available players. A fixed set of $11$ players starts the game, while the other $11$ are available as substitutes. During the game, the coach may make as many as $3$ substitutions, where any one of the $11$ players in the game is replaced by one of the substitutes. No player removed from the game... | 122 | There are $0-3$ substitutions. The number of ways to sub any number of times must be multiplied by the previous number. This is defined recursively. The case for $0$ subs is $1$, and the ways to reorganize after $n$ subs is the product of the number of new subs ($12-n$) and the players that can be ejected ($11$). The f... | 5.25 | [
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A moving particle starts at the point $(4,4)$ and moves until it hits one of the coordinate axes for the first time. When the particle is at the point $(a,b)$, it moves at random to one of the points $(a-1,b)$, $(a,b-1)$, or $(a-1,b-1)$, each with probability $\frac{1}{3}$, independently of its previous moves. The prob... | 252 | One could recursively compute the probabilities of reaching $(0,0)$ as the first axes point from any point $(x,y)$ as \[P(x,y) = \frac{1}{3} P(x-1,y) + \frac{1}{3} P(x,y-1) + \frac{1}{3} P(x-1,y-1)\] for $x,y \geq 1,$ and the base cases are $P(0,0) = 1, P(x,0) = P(y,0) = 0$ for any $x,y$ not equal to zero. We then recu... | 6.375 | [
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] |
In convex quadrilateral $KLMN$ side $\overline{MN}$ is perpendicular to diagonal $\overline{KM}$, side $\overline{KL}$ is perpendicular to diagonal $\overline{LN}$, $MN = 65$, and $KL = 28$. The line through $L$ perpendicular to side $\overline{KN}$ intersects diagonal $\overline{KM}$ at $O$ with $KO = 8$. Find $MO$. | 90 | Let $\angle MKN=\alpha$ and $\angle LNK=\beta$. Note $\angle KLP=\beta$.
Then, $KP=28\sin\beta=8\cos\alpha$. Furthermore, $KN=\frac{65}{\sin\alpha}=\frac{28}{\sin\beta} \Rightarrow 65\sin\beta=28\sin\alpha$.
Dividing the equations gives \[\frac{65}{28}=\frac{28\sin\alpha}{8\cos\alpha}=\frac{7}{2}\tan\alpha\Rightarrow... | 6 | [
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There are positive integers $x$ and $y$ that satisfy the system of equations \begin{align*} \log_{10} x + 2 \log_{10} (\text{gcd}(x,y)) &= 60\\ \log_{10} y + 2 \log_{10} (\text{lcm}(x,y)) &= 570. \end{align*} Let $m$ be the number of (not necessarily distinct) prime factors in the prime factorization of $x$, and let $n... | 880 | Let $x=10^a$ and $y=10^b$ and $a<b$. Then the given equations become $3a=60$ and $3b=570$. Therefore, $x=10^{20}=2^{20}\cdot5^{20}$ and $y=10^{190}=2^{190}\cdot5^{190}$. Our answer is $3(20+20)+2(190+190)=\boxed{880}$. | 6 | [
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Let $x$ be a real number such that $\sin^{10}x+\cos^{10} x = \tfrac{11}{36}$. Then $\sin^{12}x+\cos^{12} x = \tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 67 | We can substitute $y = \sin^2{x}$. Since we know that $\cos^2{x}=1-\sin^2{x}$, we can do some simplification.
This yields $y^5+(1-y)^5=\frac{11}{36}$. From this, we can substitute again to get some cancellation through binomials. If we let $z=\frac{1}{2}-y$, we can simplify the equation to $(\frac{1}{2}+z)^5+(\frac{1}... | 6.125 | [
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Let $\tau(n)$ denote the number of positive integer divisors of $n$. Find the sum of the six least positive integers $n$ that are solutions to $\tau (n) + \tau (n+1) = 7$. | 540 | In order to obtain a sum of $7$, we must have:
either a number with $5$ divisors (a fourth power of a prime) and a number with $2$ divisors (a prime), or
a number with $4$ divisors (a semiprime or a cube of a prime) and a number with $3$ divisors (a square of a prime). (No integer greater than $1$ can have fewer than $... | 6 | [
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For distinct complex numbers $z_1,z_2,\dots,z_{673}$, the polynomial \[(x-z_1)^3(x-z_2)^3 \cdots (x-z_{673})^3\]can be expressed as $x^{2019} + 20x^{2018} + 19x^{2017}+g(x)$, where $g(x)$ is a polynomial with complex coefficients and with degree at most $2016$. The sum $\left| \sum_{1 \le j <k \le 673} z_jz_k \right|$ ... | 352 | This is a quick fake solve using $z_i = 0$ where $3 \le i \le 673$ and only $z_1,z_2 \neq 0$ .
By Vieta's, \[3z_1+3z_2=-20\] and \[3z_1^2+3z_2^2+9z_1z_2 = 19.\] Rearranging gives $z_1 + z_2 = \dfrac{-20}{3}$ and $3(z_1^2 + 2z_1z_2 + z_2^2) + 3z_1z_2 = 19$ giving $3(z_1 + z_2)^2 + 3z_1z_2 = 19$.
Substituting gives $3\... | 8.125 | [
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In $\triangle ABC$, the sides have integer lengths and $AB=AC$. Circle $\omega$ has its center at the incenter of $\triangle ABC$. An excircle of $\triangle ABC$ is a circle in the exterior of $\triangle ABC$ that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that ... | 20 | Notice that the $A$-excircle would have to be very small to fit the property that it is internally tangent to $\omega$ and the other two excircles are both externally tangent, given that circle $\omega$'s centre is at the incenter of $\triangle ABC$. If $BC=2$, we see that $AB=AC$ must be somewhere in the $6$ to $13$ r... | 6.375 | [
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Given $f(z) = z^2-19z$, there are complex numbers $z$ with the property that $z$, $f(z)$, and $f(f(z))$ are the vertices of a right triangle in the complex plane with a right angle at $f(z)$. There are positive integers $m$ and $n$ such that one such value of $z$ is $m+\sqrt{n}+11i$. Find $m+n$. | 230 | Notice that we must have \[\frac{f(f(z))-f(z)}{f(z)-z}=-\frac{f(f(z))-f(z)}{z-f(z)}\in i\mathbb R .\]However, $f(t)-t=t(t-20)$, so \begin{align*} \frac{f(f(z))-f(z)}{f(z)-z}&=\frac{(z^2-19z)(z^2-19z-20)}{z(z-20)}\\ &=\frac{z(z-19)(z-20)(z+1)}{z(z-20)}\\ &=(z-19)(z+1)\\ &=(z-9)^2-100. \end{align*} Then, the real part of... | 6.5 | [
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Triangle $ABC$ has side lengths $AB=4$, $BC=5$, and $CA=6$. Points $D$ and $E$ are on ray $AB$ with $AB<AD<AE$. The point $F \neq C$ is a point of intersection of the circumcircles of $\triangle ACD$ and $\triangle EBC$ satisfying $DF=2$ and $EF=7$. Then $BE$ can be expressed as $\tfrac{a+b\sqrt{c}}{d}$, where $a$, $b$... | 32 | Let $P=AE\cap CF$. Let $CP=5x$ and $BP=5y$; from $\triangle{CBP}\sim\triangle{EFP}$ we have $EP=7x$ and $FP=7y$. From $\triangle{CAP}\sim\triangle{DFP}$ we have $\frac{6}{4+5y}=\frac{2}{7y}$ giving $y=\frac{1}{4}$. So $BP=\frac{5}{4}$ and $FP=\frac{7}{4}$. These similar triangles also gives us $DP=\frac{5}{3}x$ so $DE=... | 6.875 | [
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Find the least odd prime factor of $2019^8+1$. | 97 | We know that $2019^8 \equiv -1 \pmod{p}$ for some prime $p$. We want to find the smallest odd possible value of $p$. By squaring both sides of the congruence, we find $2019^{16} \equiv 1 \pmod{p}$.
Since $2019^{16} \equiv 1 \pmod{p}$, the order of $2019$ modulo $p$ is a positive divisor of $16$.
However, if the order... | 6.125 | [
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Let $\overline{AB}$ be a chord of a circle $\omega$, and let $P$ be a point on the chord $\overline{AB}$. Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$. Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$. Circles $\omega_1$ and $\omega_2$ intersect at... | 65 | Connect $AQ,QB$, since $\angle{AO_1P}=\angle{AOB}=\angle{BO_2P}$, so $\angle{AQP}=\frac{\angle{AO_1P}}{2}=\angle{BQP}=\frac{\angle{BO_2P}}{2}, \angle{AQB}=\angle{AOB}$ then, so $A,O,Q,B$ are concyclic
We let $\angle{AO_1P}=\angle{AOB}=\angle{BO_2P}=2\alpha$, it is clear that $\angle{BQP}=\alpha, \angle{O_1AP}=90^{\cir... | 6.5 | [
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Two different points, $C$ and $D$, lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$, $BC=AD=10$, and $CA=DB=17$. The intersection of these two triangular regions has area $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 59 | Let $a = \angle{CAB}$. By Law of Cosines, \[\cos a = \frac{17^2+9^2-10^2}{2*9*17} = \frac{15}{17}\] \[\sin a = \sqrt{1-\cos^2 a} = \frac{8}{17}\] \[\tan a = \frac{8}{15}\] \[A = \frac{1}{2}* 9*\frac{9}{2}\tan a = \frac{54}{5}\] And $54+5=\boxed{059}.$
- by Mathdummy | 5.625 | [
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Lily pads $1,2,3,\ldots$ lie in a row on a pond. A frog makes a sequence of jumps starting on pad $1$. From any pad $k$ the frog jumps to either pad $k+1$ or pad $k+2$ chosen randomly with probability $\tfrac{1}{2}$ and independently of other jumps. The probability that the frog visits pad $7$ is $\tfrac{p}{q}$, where ... | 107 | Let $P_n$ be the probability the frog visits pad $7$ starting from pad $n$. Then $P_7 = 1$, $P_6 = \frac12$, and $P_n = \frac12(P_{n + 1} + P_{n + 2})$ for all integers $1 \leq n \leq 5$. Working our way down, we find \[P_5 = \frac{3}{4}\] \[P_4 = \frac{5}{8}\] \[P_3 = \frac{11}{16}\] \[P_2 = \frac{21}{32}\] \[P_1 = \f... | 5.375 | [
5,
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Find the number of $7$-tuples of positive integers $(a,b,c,d,e,f,g)$ that satisfy the following systems of equations: \begin{align*} abc&=70,\\ cde&=71,\\ efg&=72. \end{align*} | 96 | We know that any two consecutive numbers are coprime. Using this, we can figure out that $c=1$ and $e=1$. $d$ then has to be $71$. Now we have two equations left. $ab=70$ and $fg=72$. To solve these we just need to figure out all of the factors. Doing the prime factorization of $70$ and $72$, we find that they have $8$... | 5.125 | [
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A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 187 | Note that rolling a 1/4 will not affect whether or not the product is a perfect square. This means that in order for the product to be a perfect square, all non 1/4 numbers rolled must come in pairs, with the only exception being the triplet 2,3, 6. Now we can do casework:
If there are four 1/4's, then there are $2^4=... | 5.875 | [
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Four ambassadors and one advisor for each of them are to be seated at a round table with $12$ chairs numbered in order $1$ to $12$. Each ambassador must sit in an even-numbered chair. Each advisor must sit in a chair adjacent to his or her ambassador. There are $N$ ways for the $8$ people to be seated at the table unde... | 520 | We see that for every 2 adjacent spots on the table, there is exactly one way for an ambassador and his or her partner to sit. There are 2 cases:
There is an ambassador at the 12th chair and his or her partner at the 1st chair
There is no pair that has their chairs numbered as 12 and 1
For the first case, there are $\... | 5.5 | [
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In a Martian civilization, all logarithms whose bases are not specified as assumed to be base $b$, for some fixed $b\ge2$. A Martian student writes down \[3\log(\sqrt{x}\log x)=56\] \[\log_{\log x}(x)=54\] and finds that this system of equations has a single real number solution $x>1$. Find $b$. | 216 | Let $y = \log _{b} x$ Then we have \[3\log _{b} (y\sqrt{x}) = 56\] \[\log _{y} x = 54\] which gives \[y^{54} = x\] Plugging this in gives \[3\log _{b} (y \cdot y^{27}) = 3\log _{b} y^{28} = 56\] which gives \[\log _{b} y = \dfrac{2}{3}\] so \[b^{2/3} = y\] By substitution we have \[b^{36} = x\] which gives \[y = \log _... | 6.375 | [
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Triangle $ABC$ has side lengths $AB=120,BC=220$, and $AC=180$. Lines $\ell_A,\ell_B$, and $\ell_C$ are drawn parallel to $\overline{BC},\overline{AC}$, and $\overline{AB}$, respectively, such that the intersections of $\ell_A,\ell_B$, and $\ell_C$ with the interior of $\triangle ABC$ are segments of lengths $55,45$, an... | 715 | Notation shown on diagram. By similar triangles we have \[k_1 = \frac{EF}{BC} = \frac{AE}{AB} = \frac {AF}{AC} = \frac {1}{4},\] \[k_2 = \frac{F''E''}{AC} = \frac {BF''}{AB} = \frac{1}{4},\] \[k_3 = \frac{E'F'}{AB} = \frac{E'C }{AC} = \frac{1}{8}.\] So, \[\frac{ZE}{BC} = \frac{F''E}{AB} = \frac{AB - AE - BF''}{AB} = 1 ... | 6.625 | [
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The polynomial $f(z)=az^{2018}+bz^{2017}+cz^{2016}$ has real coefficients not exceeding $2019$, and $f\left(\tfrac{1+\sqrt{3}i}{2}\right)=2015+2019\sqrt{3}i$. Find the remainder when $f(1)$ is divided by $1000$. | 53 | We have $\frac{1+\sqrt{3}i}{2} = \omega$ where $\omega = e^{\frac{i\pi}{3}}$ is a primitive 6th root of unity. Then we have
\begin{align*} f(\omega) &= a\omega^{2018} + b\omega^{2017} + c\omega^{2016}\\ &= a\omega^2 + b\omega + c \end{align*}
We wish to find $f(1) = a+b+c$. We first look at the real parts. As $\text{R... | 7.25 | [
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Call a positive integer $n$ $k$-pretty if $n$ has exactly $k$ positive divisors and $n$ is divisible by $k$. For example, $18$ is $6$-pretty. Let $S$ be the sum of positive integers less than $2019$ that are $20$-pretty. Find $\tfrac{S}{20}$. | 472 | For $n$ to have exactly $20$ positive divisors, $n$ can only take on certain prime factorization forms: namely, $p^{19}, p^9q, p^4q^3, p^4qr$. No number that is a multiple of $20$ can be expressed in the first form, and the only integer divisible by $20$ that has the second form is $2^{9}5$, which is greater than $2019... | 6.125 | [
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] |
There is a unique angle $\theta$ between $0^{\circ}$ and $90^{\circ}$ such that for nonnegative integers $n$, the value of $\tan{\left(2^{n}\theta\right)}$ is positive when $n$ is a multiple of $3$, and negative otherwise. The degree measure of $\theta$ is $\tfrac{p}{q}$, where $p$ and $q$ are relatively prime integers... | 547 | Note that if $\tan \theta$ is positive, then $\theta$ is in the first or third quadrant, so $0^{\circ} < \theta < 90^{\circ} \pmod{180^{\circ}}$. Also notice that the only way $\tan{\left(2^{n}\theta\right)}$ can be positive for all $n$ that are multiples of $3$ is when $2^0\theta, 2^3\theta, 2^6\theta$, etc. are all t... | 7 | [
7,
8,
7,
7,
6,
7,
7,
7
] |
Triangle $ABC$ has side lengths $AB=7, BC=8,$ and $CA=9.$ Circle $\omega_1$ passes through $B$ and is tangent to line $AC$ at $A.$ Circle $\omega_2$ passes through $C$ and is tangent to line $AB$ at $A.$ Let $K$ be the intersection of circles $\omega_1$ and $\omega_2$ not equal to $A.$ Then $AK=\tfrac mn,$ where $m$ an... | 11 | By the definition of $K$, it is the spiral center mapping $BA\to AC$, which means that it is the midpoint of the $A$-symmedian chord. In particular, if $M$ is the midpoint of $BC$ and $M'$ is the reflection of $A$ across $K$, we have $\triangle ABM'\sim\triangle AMC$. By Stewart's Theorem, it then follows that \[AK = \... | 5.5 | [
6,
5,
5,
6,
6,
5,
5,
6
] |
For $n \ge 1$ call a finite sequence $(a_1, a_2 \ldots a_n)$ of positive integers progressive if $a_i < a_{i+1}$ and $a_i$ divides $a_{i+1}$ for all $1 \le i \le n-1$. Find the number of progressive sequences such that the sum of the terms in the sequence is equal to $360$. | 47 | If the first term is $x$, then dividing through by $x$, we see that we can find the number of progressive sequences whose sum is $\frac{360}{x} - 1$, and whose first term is not 1. If $a(k)$ denotes the number of progressive sequences whose sum is $k$ and whose first term is not 1, then we can express the answer $N$ as... | 6.625 | [
7,
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7,
6,
8,
7,
6,
6
] |
Regular octagon $A_1A_2A_3A_4A_5A_6A_7A_8$ is inscribed in a circle of area $1.$ Point $P$ lies inside the circle so that the region bounded by $\overline{PA_1},\overline{PA_2},$ and the minor arc $\widehat{A_1A_2}$ of the circle has area $\tfrac{1}{7},$ while the region bounded by $\overline{PA_3},\overline{PA_4},$ an... | 504 | The actual size of the diagram doesn't matter. To make calculation easier, we discard the original area of the circle, $1$, and assume the side length of the octagon is $2$. Let $r$ denote the radius of the circle, $O$ be the center of the circle. Then $r^2= 1^2 + (\sqrt{2}+1)^2= 4+2\sqrt{2}$. Now, we need to find the ... | 7.125 | [
7,
7,
7,
7,
8,
7,
7,
7
] |
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. | 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 solu... | 6.375 | [
6,
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7
] |
In acute triangle $ABC$ points $P$ and $Q$ are the feet of the perpendiculars from $C$ to $\overline{AB}$ and from $B$ to $\overline{AC}$, respectively. Line $PQ$ intersects the circumcircle of $\triangle ABC$ in two distinct points, $X$ and $Y$. Suppose $XP=10$, $PQ=25$, and $QY=15$. The value of $AB\cdot AC$ can be w... | 574 | Let $BC=a$, $AC=b$, and $AB=c$. Let $\cos\angle A=k$. Then $AP=bk$ and $AQ=ck$.
By Power of a Point theorem, \begin{align} AP\cdot BP=XP\cdot YP \quad &\Longrightarrow \quad b^2k^2-bck+400=0\\ AQ\cdot CQ=YQ\cdot XQ \quad &\Longrightarrow \quad c^2k^2-bck+525=0 \end{align} Thus $bck = (bk)^2+400=(ck)^2+525 = u$. Then $... | 7.625 | [
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8
] |
In $\triangle ABC$ with $AB=AC,$ point $D$ lies strictly between $A$ and $C$ on side $\overline{AC},$ and point $E$ lies strictly between $A$ and $B$ on side $\overline{AB}$ such that $AE=ED=DB=BC.$ The degree measure of $\angle ABC$ is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n... | 547 | Let $\angle{BAC}$ be $x$ in degrees. $\angle{ADE}=x$. By Exterior Angle Theorem on triangle $AED$, $\angle{BED}=2x$. By Exterior Angle Theorem on triangle $ADB$, $\angle{BDC}=3x$. This tells us $\angle{BCA}=\angle{ABC}=3x$ and $3x+3x+x=180$. Thus $x=\frac{180}{7}$ and we want $\angle{ABC}=3x=\frac{540}{7}$ to get an an... | 6.375 | [
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There is a unique positive real number $x$ such that the three numbers $\log_8{2x}$, $\log_4{x}$, and $\log_2{x}$, in that order, form a geometric progression with positive common ratio. The number $x$ can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 17 | If we set $x=2^y$, we can obtain three terms of a geometric sequence through logarithm properties. The three terms are \[\frac{y+1}{3}, \frac{y}{2}, y.\] In a three-term geometric sequence, the middle term squared is equal to the product of the other two terms, so we obtain the following: \[\frac{y^2+y}{3} = \frac{y^2}... | 5.5 | [
5,
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6
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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... | 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}... | 5.375 | [
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Let $S$ be the set of positive integers $N$ with the property that the last four digits of $N$ are $2020,$ and when the last four digits are removed, the result is a divisor of $N.$ For example, $42,020$ is in $S$ because $4$ is a divisor of $42,020.$ Find the sum of all the digits of all the numbers in $S.$ For exampl... | 93 | We note that any number in $S$ can be expressed as $a(10,000) + 2,020$ for some integer $a$. The problem requires that $a$ divides this number, and since we know $a$ divides $a(10,000)$, we need that $a$ divides 2020. Each number contributes the sum of the digits of $a$, as well as $2 + 0 + 2 +0 = 4$. Since $2020$ can ... | 6 | [
6,
6,
6,
6,
6,
6,
6,
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] |
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. | 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... | 5.5 | [
5,
5,
6,
5,
5,
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6,
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] |
A flat board has a circular hole with radius $1$ and a circular hole with radius $2$ such that the distance between the centers of the two holes is $7.$ Two spheres with equal radii sit in the two holes such that the spheres are tangent to each other. The square of the radius of the spheres is $\tfrac{m}{n},$ where $m$... | 173 | [asy] size(10cm); pair A, B, C, D, O, P, H, L, X, Y; A = (-1, 0); B = (1, 0); H = (0, 0); C = (5, 0); D = (9, 0); L = (7, 0); O = (0, sqrt(160/13 - 1)); P = (7, sqrt(160/13 - 4)); X = (0, sqrt(160/13 - 4)); Y = (O + P) / 2; draw(A -- O -- B -- cycle); draw(C -- P -- D -- cycle); draw(B -- C); draw(O -- P); draw(P -- X,... | 6.25 | [
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] |
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... | 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 indi... | 5.375 | [
6,
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5,
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6,
5
] |
A bug walks all day and sleeps all night. On the first day, it starts at point $O$, faces east, and walks a distance of $5$ units due east. Each night the bug rotates $60^\circ$ counterclockwise. Each day it walks in this new direction half as far as it walked the previous day. The bug gets arbitrarily close to the poi... | 103 | The ant goes in the opposite direction every $3$ moves, going $(1/2)^3=1/8$ the distance backwards. Using geometric series, he travels $1-1/8+1/64-1/512...=(7/8)(1+1/64+1/4096...)=(7/8)(64/63)=8/9$ the distance of the first three moves over infinity moves. Now, we use coordinates meaning $(5+5/4-5/8, 0+5\sqrt3/4+5\sqrt... | 6.625 | [
7,
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] |
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 ... | 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 m... | 6.25 | [
6,
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] |
Let $m$ and $n$ be positive integers satisfying the conditions
$\quad\bullet\ \gcd(m+n,210)=1,$
$\quad\bullet\ m^m$ is a multiple of $n^n,$ and
$\quad\bullet\ m$ is not a multiple of $n.$
Find the least possible value of $m+n.$ | 407 | Taking inspiration from $4^4 \mid 10^{10}$ we are inspired to take $n$ to be $p^2$, the lowest prime not dividing $210$, or $11 \implies n = 121$. Now, there are $242$ factors of $11$, so $11^{242} \mid m^m$, and then $m = 11k$ for $k \geq 22$. Now, $\gcd(m+n, 210) = \gcd(11+k,210) = 1$. Noting $k = 26$ is the minimal ... | 6.5 | [
6,
7,
7,
6,
6,
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] |
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 $g(f(2))=g(f(4))=0.$ | 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... | 6.25 | [
6,
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6,
6,
7,
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] |
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 $n.$ | 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}... | 7.25 | [
7,
7,
8,
7,
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] |
Point $D$ lies on side $\overline{BC}$ of $\triangle ABC$ so that $\overline{AD}$ bisects $\angle BAC.$ The perpendicular bisector of $\overline{AD}$ intersects the bisectors of $\angle ABC$ and $\angle ACB$ in points $E$ and $F,$ respectively. Given that $AB=4,BC=5,$ and $CA=6,$ the area of $\triangle AEF$ can be writ... | 36 | Points are defined as shown. It is pretty easy to show that $\triangle AFE \sim \triangle AGH$ by spiral similarity at $A$ by some short angle chasing. Now, note that $AD$ is the altitude of $\triangle AFE$, as the altitude of $AGH$. We need to compare these altitudes in order to compare their areas. Note that Stewart'... | 6.625 | [
7,
7,
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5,
6,
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] |
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 $(a+b)^2.$ | 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)... | 6.5 | [
6,
6,
7,
7,
6,
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7,
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] |
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 i... | 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 ... | 8.25 | [
8,
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8,
8,
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8,
9
] |
Find the number of ordered pairs of positive integers $(m,n)$ such that ${m^2n = 20 ^{20}}$. | 231 | In this problem, we want to find the number of ordered pairs $(m, n)$ such that $m^2n = 20^{20}$. Let $x = m^2$. Therefore, we want two numbers, $x$ and $n$, such that their product is $20^{20}$ and $x$ is a perfect square. Note that there is exactly one valid $n$ for a unique $x$, which is $\tfrac{20^{20}}{x}$. This r... | 4.125 | [
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] |
Let $P$ be a point chosen uniformly at random in the interior of the unit square with vertices at $(0,0), (1,0), (1,1)$, and $(0,1)$. The probability that the slope of the line determined by $P$ and the point $\left(\frac58, \frac38 \right)$ is greater than or equal to $\frac12$ can be written as $\frac{m}{n}$, where $... | 171 | The areas bounded by the unit square and alternately bounded by the lines through $\left(\frac{5}{8},\frac{3}{8}\right)$ that are vertical or have a slope of $1/2$ show where $P$ can be placed to satisfy the condition. One of the areas is a trapezoid with bases $1/16$ and $3/8$ and height $5/8$. The other area is a tra... | 6.125 | [
6,
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] |
The value of $x$ that satisfies $\log_{2^x} 3^{20} = \log_{2^{x+3}} 3^{2020}$ can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 103 | Recall the identity $\log_{a^n} b^{m} = \frac{m}{n}\log_{a} b$ (which is easily proven using exponents or change of base). Then this problem turns into \[\frac{20}{x}\log_{2} 3 = \frac{2020}{x+3}\log_{2} 3\] Divide $\log_{2} 3$ from both sides. And we are left with $\frac{20}{x}=\frac{2020}{x+3}$.Solving this simple eq... | 3.875 | [
3,
4,
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4,
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] |
Triangles $\triangle ABC$ and $\triangle A'B'C'$ lie in the coordinate plane with vertices $A(0,0)$, $B(0,12)$, $C(16,0)$, $A'(24,18)$, $B'(36,18)$, $C'(24,2)$. A rotation of $m$ degrees clockwise around the point $(x,y)$ where $0<m<180$, will transform $\triangle ABC$ to $\triangle A'B'C'$. Find $m+x+y$. | 108 | After sketching, it is clear a $90^{\circ}$ rotation is done about $(x,y)$. Looking between $A$ and $A'$, $x+y=18$ and $x-y=24$. Solving gives $(x,y)\implies(21,-3)$. Thus $90+21-3=\boxed{108}$. ~mn28407 | 5.375 | [
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For each positive integer $n$, let $f(n)$ be the sum of the digits in the base-four representation of $n$ and let $g(n)$ be the sum of the digits in the base-eight representation of $f(n)$. For example, $f(2020) = f(133210_{\text{4}}) = 10 = 12_{\text{8}}$, and $g(2020) = \text{the digit sum of }12_{\text{8}} = 3$. Let... | 151 | Let's work backwards. The minimum base-sixteen representation of $g(n)$ that cannot be expressed using only the digits $0$ through $9$ is $A_{16}$, which is equal to $10$ in base 10. Thus, the sum of the digits of the base-eight representation of the sum of the digits of $f(n)$ is $10$. The minimum value for which this... | 7.125 | [
7,
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7,
7,
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] |
Define a sequence recursively by $t_1 = 20$, $t_2 = 21$, and\[t_n = \frac{5t_{n-1}+1}{25t_{n-2}}\]for all $n \ge 3$. Then $t_{2020}$ can be expressed as $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. | 626 | Let $t_n=\frac{s_n}{5}$. Then, we have $s_n=\frac{s_{n-1}+1}{s_{n-2}}$ where $s_1 = 100$ and $s_2 = 105$. By substitution, we find $s_3 = \frac{53}{50}$, $s_4=\frac{103}{105\cdot50}$, $s_5=\frac{101}{105}$, $s_6=100$, and $s_7=105$. So $s_n$ has a period of $5$. Thus $s_{2020}=s_5=\frac{101}{105}$. So, $\frac{101}{105\... | 6.125 | [
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] |
Two congruent right circular cones each with base radius $3$ and height $8$ have the axes of symmetry that intersect at right angles at a point in the interior of the cones a distance $3$ from the base of each cone. A sphere with radius $r$ lies within both cones. The maximum possible value of $r^2$ is $\frac{m}{n}$, w... | 298 | Consider the cross section of the cones and sphere by a plane that contains the two axes of symmetry of the cones as shown below. The sphere with maximum radius will be tangent to the sides of each of the cones. The center of that sphere must be on the axis of symmetry of each of the cones and thus must be at the inter... | 7.125 | [
7,
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8
] |
Define a sequence recursively by $f_1(x)=|x-1|$ and $f_n(x)=f_{n-1}(|x-n|)$ for integers $n>1$. Find the least value of $n$ such that the sum of the zeros of $f_n$ exceeds $500,000$. | 101 | First it will be shown by induction that the zeros of $f_n$ are the integers $a, {a+2,} {a+4,} \dots, {a + n(n-1)}$, where $a = n - \frac{n(n-1)}2.$
This is certainly true for $n=1$. Suppose that it is true for $n = m-1 \ge 1$, and note that the zeros of $f_m$ are the solutions of $|x - m| = k$, where $k$ is a nonnega... | 6.5 | [
6,
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6,
7,
7,
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7,
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] |
While watching a show, Ayako, Billy, Carlos, Dahlia, Ehuang, and Frank sat in that order in a row of six chairs. During the break, they went to the kitchen for a snack. When they came back, they sat on those six chairs in such a way that if two of them sat next to each other before the break, then they did not sit next... | 90 | We proceed with casework based on the person who sits first after the break.
$\textbf{Case 1:}$ A is first. Then the first three people in the row can be ACE, ACF, ADB, ADF, AEB, AEC, AFB, AFC, or AFD, which yield 2, 1, 2, 2, 1, 2, 0, 1, and 1 possible configurations, respectively, implying 2 + 1 + 2 + 2 + 1 + 2 + 0 + ... | 6.375 | [
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] |
Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3+\cdots +n^3$ is divided by $n+5$, the remainder is $17$. | 239 | The formula for the sum of cubes is as follows: \[1^3+2^3+3^3+...+n^3=(1+2+3+...+n)^2=\left(\frac{n(n+1)}{2}\right)^2\] So let's apply this to this problem.
Let $m=n+5$. Then we have \[1^3+2^3+3^3+\cdots+(m-5)^3\equiv 17 \mod m\] \[\left(\frac{(m-5)(m-4)}{2}\right)^2\equiv 17 \mod m\] \[\frac{400}{4}\equiv 17 \mod m\]... | 5.625 | [
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] |
Let $P(x) = x^2 - 3x - 7$, and let $Q(x)$ and $R(x)$ be two quadratic polynomials also with the coefficient of $x^2$ equal to $1$. David computes each of the three sums $P + Q$, $P + R$, and $Q + R$ and is surprised to find that each pair of these sums has a common root, and these three common roots are distinct. If $Q... | 71 | Let $Q(x) = x^2 + ax + 2$ and $R(x) = x^2 + bx + c$. We can write the following: \[P + Q = 2x^2 + (a - 3)x - 5\] \[P + R = 2x^2 + (b - 3)x + (c - 7)\] \[Q + R = 2x^2 + (a + b)x + (c + 2)\] Let the common root of $P+Q,P+R$ be $r$; $P+R,Q+R$ be $s$; and $P+Q,Q+R$ be $t$. We then have that the roots of $P+Q$ are $r,t$, th... | 6.125 | [
6,
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] |
Let $m$ and $n$ be odd integers greater than $1.$ An $m\times n$ rectangle is made up of unit squares where the squares in the top row are numbered left to right with the integers $1$ through $n$, those in the second row are numbered left to right with the integers $n + 1$ through $2n$, and so on. Square $200$ is in th... | 248 | Let us take some cases. Since $m$ and $n$ are odds, and $200$ is in the top row and $2000$ in the bottom, $m$ has to be $3$, $5$, $7$, or $9$. Also, taking a look at the diagram, the slope of the line connecting those centers has to have an absolute value of $< 1$. Therefore, $m < 1800 \mod n < 1800-m$.
If $m=3$, $n$ ... | 6.75 | [
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Convex pentagon $ABCDE$ has side lengths $AB=5$, $BC=CD=DE=6$, and $EA=7$. Moreover, the pentagon has an inscribed circle (a circle tangent to each side of the pentagon). Find the area of $ABCDE$. | 60 | This pentagon is very close to a regular pentagon with side lengths $6$. The area of a regular pentagon with side lengths $s$ is $\frac{5s^2}{4\sqrt{5-2\sqrt{5}}}$. $5-2\sqrt{5}$ is slightly greater than $\frac{1}{2}$ given that $2\sqrt{5}$ is slightly less than $\frac{9}{2}$. $4\sqrt{5-2\sqrt{5}}$ is then slightly gre... | 5.125 | [
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5
] |
For a real number $x$ let $\lfloor x\rfloor$ be the greatest integer less than or equal to $x$, and define $\{x\} = x - \lfloor x \rfloor$ to be the fractional part of $x$. For example, $\{3\} = 0$ and $\{4.56\} = 0.56$. Define $f(x)=x\{x\}$, and let $N$ be the number of real-valued solutions to the equation $f(f(f(x))... | 10 | To solve $f(f(f(x)))=17$, we need to solve $f(x) = y$ where $f(f(y))=17$, and to solve that we need to solve $f(y) = z$ where $f(z) = 17$.
It is clear to see for some integer $a \geq 17$ there is exactly one value of $z$ in the interval $[a, a+1)$ where $f(z) = 17$. To understand this, imagine the graph of $f(z)$ on t... | 7.75 | [
8,
7,
8,
8,
8,
8,
8,
7
] |
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