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 disk with radius $1$ is externally tangent to a disk with radius $5$. Let $A$ be the point where the disks are tangent, $C$ be the center of the smaller disk, and $E$ be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to roll along the outside of the larger disk until t... | 58 | First, we determine how far the small circle goes. For the small circle to rotate completely around the circumference, it must rotate $5$ times (the circumference of the small circle is $2\pi$ while the larger one has a circumference of $10\pi$) plus the extra rotation the circle gets for rotating around the circle, fo... | 7.125 | [
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A token starts at the point $(0,0)$ of an $xy$-coordinate grid and then makes a sequence of six moves. Each move is 1 unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. The probability the token ends at a poi... | 391 | Perform the coordinate transformation $(x, y)\rightarrow (x+y, x-y)$. Then we can see that a movement up, right, left, or down in the old coordinates adds the vectors $\langle 1, -1 \rangle$, $\langle 1, 1 \rangle$, $\langle -1, -1 \rangle$, $\langle -1, 1 \rangle$ respectively. Moreover, the transformation takes the e... | 6 | [
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Let $A=\{1,2,3,4\}$, and $f$ and $g$ be randomly chosen (not necessarily distinct) functions from $A$ to $A$. The probability that the range of $f$ and the range of $g$ are disjoint is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m$. | 453 | As before, there are 4 functions with a range of size 1, 84 with a range of size 2, and 144 with a range of size 3. If the range of $f$ has size $k$, the codomain of $g$ is restricted to a set of size $4 - k$. Any function from $A$ into this codomain will do, so there are $(4 - k)^4$ possibilities for $g$ given a funct... | 5.875 | [
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On square $ABCD$, points $E,F,G$, and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34$. Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P$, and the areas of the quadrilaterals $AEPH, BFPE, CGP... | 850 | Continue in the same way as solution 1 to get that $POK$ has area $3a$, and $OK = \frac{d}{10}$. You can then find $PK$ has length $\frac 32$.
Then, if we drop a perpendicular from $H$ to $BC$ at $L$, We get $\triangle HLF \sim \triangle OPK$.
Thus, $LF = \frac{15\cdot 34}{d}$, and we know $HL = d$, and $HF = 34$. Th... | 6.625 | [
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Let $m$ be the largest real solution to the equation
$\frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}=x^2-11x-4$
There are positive integers $a$, $b$, and $c$ such that $m=a+\sqrt{b+\sqrt{c}}$. Find $a+b+c$. | 263 | The first step is to notice that the 4 on the right hand side can simplify the terms on the left hand side. If we distribute 1 to $\frac{3}{x-3}$, then the fraction becomes of the form $\frac{x}{x - 3}$. A similar cancellation happens with the other four terms. If we assume $x = 0$ is not the highest solution (which is... | 6.875 | [
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In $\triangle ABC$, $AB = 3$, $BC = 4$, and $CA = 5$. Circle $\omega$ intersects $\overline{AB}$ at $E$ and $B$, $\overline{BC}$ at $B$ and $D$, and $\overline{AC}$ at $F$ and $G$. Given that $EF=DF$ and $\frac{DG}{EG} = \frac{3}{4}$, length $DE=\frac{a\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive inte... | 41 | Call $DE=x$ and as a result $DF=EF=\frac{x\sqrt{2}}{2}, EG=\frac{4x}{5}, GD=\frac{3x}{5}$. Since $EFGD$ is cyclic we just need to get $DG$ and using LoS(for more detail see the $2$nd paragraph of Solution $2$) we get $AG=\frac{5}{2}$ and using a similar argument(use LoS again) and subtracting you get $FG=\frac{5}{14}$ ... | 6.25 | [
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Abe can paint the room in 15 hours, Bea can paint 50 percent faster than Abe, and Coe can paint twice as fast as Abe. Abe begins to paint the room and works alone for the first hour and a half. Then Bea joins Abe, and they work together until half the room is painted. Then Coe joins Abe and Bea, and they work together ... | 334 | From the given information, we can see that Abe can paint $\frac{1}{15}$ of the room in an hour, Bea can paint $\frac{1}{15}\times\frac{3}{2} = \frac{1}{10}$ of the room in an hour, and Coe can paint the room in $\frac{1}{15}\times 2 = \frac{2}{15}$ of the room in an hour. After $90$ minutes, Abe has painted $\frac{1}{... | 6.125 | [
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Arnold is studying the prevalence of three health risk factors, denoted by A, B, and C, within a population of men. For each of the three factors, the probability that a randomly selected man in the population has only this risk factor (and none of the others) is 0.1. For any two of the three factors, the probability t... | 76 | We first assume a population of $100$ to facilitate solving. Then we simply organize the statistics given into a Venn diagram.
[asy] pair A,B,C,D,E,F,G; A=(0,55); B=(60,55); C=(60,0); D=(0,0); draw(A--B--C--D--A); E=(30,35); F=(20,20); G=(40,20); draw(circle(E,15)); draw(circle(F,15)); draw(circle(G,15)); draw("$A$",(3... | 6.25 | [
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A rectangle has sides of length $a$ and 36. A hinge is installed at each vertex of the rectangle, and at the midpoint of each side of length 36. The sides of length $a$ can be pressed toward each other keeping those two sides parallel so the rectangle becomes a convex hexagon as shown. When the figure is a hexagon with... | 720 | Alternatively, use basic geometry. First, scale everything down by dividing everything by 6. Let $a/6=p$. Then, the dimensions of the central rectangle in the hexagon is p x 4, and the original rectangle is 6 x p. By Pythagorean theorem and splitting the end triangles of the hexagon into two right triangles, the altitu... | 5.375 | [
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The repeating decimals $0.abab\overline{ab}$ and $0.abcabc\overline{abc}$ satisfy
\[0.abab\overline{ab}+0.abcabc\overline{abc}=\frac{33}{37},\]
where $a$, $b$, and $c$ are (not necessarily distinct) digits. Find the three digit number $abc$. | 447 | Note that $\frac{33}{37}=\frac{891}{999} = 0.\overline{891}$. Also note that the period of $0.abab\overline{ab}+0.abcabc\overline{abc}$ is at most $6$. Therefore, we only need to worry about the sum $0.ababab+ 0.abcabc$. Adding the two, we get \[\begin{array}{ccccccc}&a&b&a&b&a&b\\ +&a&b&c&a&b&c\\ \hline &8&9&1&8&9&1\e... | 5.875 | [
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Real numbers $r$ and $s$ are roots of $p(x)=x^3+ax+b$, and $r+4$ and $s-3$ are roots of $q(x)=x^3+ax+b+240$. Find the sum of all possible values of $|b|$.
Hint
\[\color{red}\boxed{\boxed{\color{blue}\textbf{Use Vieta's Formulae!}}}\] | 420 | Because the coefficient of $x^2$ in both $p(x)$ and $q(x)$ is 0, the remaining root of $p(x)$ is $-(r+s)$, and the remaining root of $q(x)$ is $-(r+s+1)$. The coefficients of $x$ in $p(x)$ and $q(x)$ are both equal to $a$, and equating the two coefficients gives \[rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2\]from which $s = \tfr... | 6.5 | [
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Charles has two six-sided die. One of the die is fair, and the other die is biased so that it comes up six with probability $\frac{2}{3}$ and each of the other five sides has probability $\frac{1}{15}$. Charles chooses one of the two dice at random and rolls it three times. Given that the first two rolls are both sixes... | 167 | The probability that he rolls a six twice when using the fair die is $\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}$. The probability that he rolls a six twice using the biased die is $\frac{2}{3}\times \frac{2}{3}=\frac{4}{9}=\frac{16}{36}$. Given that Charles rolled two sixes, we can see that it is $16$ times more likel... | 6.125 | [
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Let $f(x)=(x^2+3x+2)^{\cos(\pi x)}$. Find the sum of all positive integers $n$ for which \[\left |\sum_{k=1}^n\log_{10}f(k)\right|=1.\] | 21 | Note that $\cos(\pi x)$ is $-1$ when $x$ is odd and $1$ when $x$ is even. Also note that $x^2+3x+2=(x+1)(x+2)$ for all $x$. Therefore \[\log_{10}f(x)=\log_{10}(x+1)+\log_{10}(x+2)\ \ \ \text{if }x \text{ is even}\] \[\log_{10}f(x)=-\log_{10}(x+1)-\log_{10}(x+2)\ \ \ \text{if }x \text{ is odd}\] Because of this, $\sum_{... | 6.25 | [
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Circle $C$ with radius 2 has diameter $\overline{AB}$. Circle D is internally tangent to circle $C$ at $A$. Circle $E$ is internally tangent to circle $C$, externally tangent to circle $D$, and tangent to $\overline{AB}$. The radius of circle $D$ is three times the radius of circle $E$, and can be written in the form $... | 254 | We use the notation of Solution 1 for triangle $\triangle DEC$ \[\sin \angle EDC = \frac {EF}{DE} = \frac {1}{4} \implies \cos \angle EDC = \frac {\sqrt{15}}{4}.\] We use Cosine Law for $\triangle DEC$ and get: \[(4r)^2 +(2 – 3r)^2 – 2 \cdot 4r \cdot (2 – 3r) \cdot \frac {\sqrt{15}}{4} = (2 – r)^2\]. \[(24 + 6 \sqrt{15... | 7 | [
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Ten chairs are arranged in a circle. Find the number of subsets of this set of chairs that contain at least three adjacent chairs. | 581 | Starting with small cases, we see that four chairs give $4 + 1 = 5$, five chairs give $5 + 5 + 1 = 11$, and six chairs give $6 + 6 + 6 + 6 + 1 = 25.$ Thus, n chairs should give $n 2^{n-4} + 1$, as confirmed above. This claim can be verified by the principle of inclusion-exclusion: there are $n 2^{n-3}$ ways to arrange ... | 6.125 | [
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Let $z$ be a complex number with $|z|=2014$. Let $P$ be the polygon in the complex plane whose vertices are $z$ and every $w$ such that $\frac{1}{z+w}=\frac{1}{z}+\frac{1}{w}$. Then the area enclosed by $P$ can be written in the form $n\sqrt{3}$, where $n$ is an integer. Find the remainder when $n$ is divided by $1000$... | 147 | Notice that \[\frac1{w+z} = \frac{w+z}{wz} \implies 0 = w^2 + wz + z^2 = \frac{w^3-z^3}{w-z}.\] Hence, $w=ze^{2\pi i/3},ze^{4\pi i/3}$, and $P$ is an equilateral triangle with circumradius $2014$. Then, \[[P]=\frac{3}{2}\cdot 2014^2\cdot\sin\frac{\pi}3=3\cdot 1007^2\sqrt3,\] and the answer is $3\cdot 1007^2\equiv 3\cdo... | 7.125 | [
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In $\triangle RED$, $\measuredangle DRE=75^{\circ}$ and $\measuredangle RED=45^{\circ}$. $RD=1$. Let $M$ be the midpoint of segment $\overline{RD}$. Point $C$ lies on side $\overline{ED}$ such that $\overline{RC}\perp\overline{EM}$. Extend segment $\overline{DE}$ through $E$ to point $A$ such that $CA=AR$. Then $AE=\fr... | 56 | Let $P$ be the foot of the perpendicular from $A$ to $\overline{CR}$, so $\overline{AP}\parallel\overline{EM}$. Since triangle $ARC$ is isosceles, $P$ is the midpoint of $\overline{CR}$, and $\overline{PM}\parallel\overline{CD}$. Thus, $APME$ is a parallelogram and $AE = PM = \frac{CD}{2}$. We can then use coordinates.... | 7.125 | [
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Suppose that the angles of $\triangle ABC$ satisfy $\cos(3A)+\cos(3B)+\cos(3C)=1.$ Two sides of the triangle have lengths 10 and 13. There is a positive integer $m$ so that the maximum possible length for the remaining side of $\triangle ABC$ is $\sqrt{m}.$ Find $m.$ | 399 | \[\cos3A+\cos3B=1-\cos(3C)=1+\cos(3A+3B)\] \[2\cos\frac{3}{2}(A+B)\cos\frac{3}{2}(A-B)=2\cos^2\frac{3}{2}(A+B)\] If $\cos\frac{3}{2}(A+B) = 0$, then $\frac{3}{2}(A+B)=90$, $A+B=60$, so $C=120$; otherwise, \[2\cos\frac{3}{2}(A-B)=2cos\frac{3}{2}(A+B)\] \[\sin\frac{3}{2}A\sin\frac{3}{2}B=0\] so either $\sin\frac{3}{2}A=0... | 6.75 | [
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Ten adults enter a room, remove their shoes, and toss their shoes into a pile. Later, a child randomly pairs each left shoe with a right shoe without regard to which shoes belong together. The probability that for every positive integer $k<5$, no collection of $k$ pairs made by the child contains the shoes from exactly... | 28 | Label the left shoes be $L_1,\dots, L_{10}$ and the right shoes $R_1,\dots, R_{10}$. Notice that there are $10!$ possible pairings.
Let a pairing be "bad" if it violates the stated condition. We would like a better condition to determine if a given pairing is bad.
Note that, in order to have a bad pairing, there must... | 6.875 | [
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In $\triangle{ABC}, AB=10, \angle{A}=30^\circ$ , and $\angle{C=45^\circ}$. Let $H, D,$ and $M$ be points on the line $BC$ such that $AH\perp{BC}$, $\angle{BAD}=\angle{CAD}$, and $BM=CM$. Point $N$ is the midpoint of the segment $HM$, and point $P$ is on ray $AD$ such that $PN\perp{BC}$. Then $AP^2=\dfrac{m}{n}$, where ... | 77 | Break our diagram into 2 special right triangle by dropping an altitude from $B$ to $AC$ we then get that \[AC=5+5\sqrt{3}, BC=5\sqrt{2}.\] Since $\triangle{HCA}$ is a 45-45-90,
\[HC=\frac{5\sqrt2+5\sqrt6}{2}\] $MC=\frac{BM}{2},$ \[HM=\frac{5\sqrt6}{2}\] \[HN=\frac{5\sqrt6}{4}\] We know that $\triangle{AHD}\simeq \tria... | 6.375 | [
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For any integer $k\geq 1$, let $p(k)$ be the smallest prime which does not divide $k.$ Define the integer function $X(k)$ to be the product of all primes less than $p(k)$ if $p(k)>2$, and $X(k)=1$ if $p(k)=2.$ Let $\{x_n\}$ be the sequence defined by $x_0=1$, and $x_{n+1}X(x_n)=x_np(x_n)$ for $n\geq 0.$ Find the smalle... | 149 | We go through the terms and look for a pattern. We find that
$x_0 = 1$ $x_8 = 7$
$x_1 = 2$ $x_9 = 14$
$x_2 = 3$ $x_{10} = 21$
$x_3 = 6$ $x_{11} = 42$
$x_4 = 5$ $x_{12} = 35$
$x_5 = 10$ $x_{13} = 70$
$x_6 = 15$ $x_{14} = 105$
$x_7 = 30$ $x_{15} = 210$
Commit to the bash. Eventually, you will receive that $x_{149} = 209... | 6.875 | [
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The expressions $A$ = $1 \times 2 + 3 \times 4 + 5 \times 6 + \cdots + 37 \times 38 + 39$ and $B$ = $1 + 2 \times 3 + 4 \times 5 + \cdots + 36 \times 37 + 38 \times 39$ are obtained by writing multiplication and addition operators in an alternating pattern between successive integers. Find the positive difference betwe... | 722 | We have \[|A-B|=|1+3(4-2)+5(6-4)+ \cdots + 37(38-36)-39(1-38)|\]\[\implies |2(1+3+5+7+ \cdots +37)-1-39(37)|\]\[\implies |361(2)-1-39(37)|=|722-1-1443|=|-722|\implies \boxed{722}\] | 3.125 | [
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The nine delegates to the Economic Cooperation Conference include $2$ officials from Mexico, $3$ officials from Canada, and $4$ officials from the United States. During the opening session, three of the delegates fall asleep. Assuming that the three sleepers were determined randomly, the probability that exactly two of... | 139 | Like in the solution above, there are $84$ ways to pick $3$ delegates. We can use casework to find the probability that there aren't exactly $2$ sleepers from a county, then subtract from $1$.
If no country has at least $2$ delegates sleeping, then every country must have $1$ delegate sleeping. There are $2*3*4=24$ way... | 4.875 | [
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There is a prime number $p$ such that $16p+1$ is the cube of a positive integer. Find $p$.
~ pi_is_3.14 | 307 | If $16p+1 =k$, we have $k \equiv 1 \mod 16$, so $k^3 \equiv 1 \mod 16$. If $k=1$ we have $p=0$, which is not prime. If $k=17$ we have $16p+1=4913$, or $p=\boxed{307}$ | 3.875 | [
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Point $B$ lies on line segment $\overline{AC}$ with $AB=16$ and $BC=4$. Points $D$ and $E$ lie on the same side of line $AC$ forming equilateral triangles $\triangle ABD$ and $\triangle BCE$. Let $M$ be the midpoint of $\overline{AE}$, and $N$ be the midpoint of $\overline{CD}$. The area of $\triangle BMN$ is $x$. Find... | 507 | Let $A$ be the origin, so $B=(16,0)$ and $C=(20,0).$ Using equilateral triangle properties tells us that $D=(8,8\sqrt3)$ and $E=(18,2\sqrt3)$ as well. Therefore, $M=(9,\sqrt3)$ and $N=(14,4\sqrt3).$ Applying the Shoelace Theorem to triangle $BMN$ gives
\[x=\dfrac 1 2 |16\sqrt3+36\sqrt3+0-(0+14\sqrt3+64\sqrt3)| =13\sqrt... | 5.75 | [
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In a drawer Sandy has $5$ pairs of socks, each pair a different color. On Monday Sandy selects two individual socks at random from the $10$ socks in the drawer. On Tuesday Sandy selects $2$ of the remaining $8$ socks at random and on Wednesday two of the remaining $6$ socks at random. The probability that Wednesday is ... | 341 | The key is to count backwards. First, choose the pair which you pick on Wednesday in $5$ ways. Then there are four pairs of socks for you to pick a pair of on Tuesday, and you don't want to pick a pair. Since there are $4$ pairs, the number of ways to do this is $\dbinom{8}{2}-4$. Then, there are two pairs and two nonm... | 4.875 | [
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Point $A,B,C,D,$ and $E$ are equally spaced on a minor arc of a circle. Points $E,F,G,H,I$ and $A$ are equally spaced on a minor arc of a second circle with center $C$ as shown in the figure below. The angle $\angle ABD$ exceeds $\angle AHG$ by $12^\circ$. Find the degree measure of $\angle BAG$.
[asy] pair A,B,C,D,E,F... | 58 | Let $O$ be the center of the circle with $ABCDE$ on it.
Let $x$ be the degree measurement of $\overarc{ED}=\overarc{DC}=\overarc{CB}=\overarc{BA}$ in circle $O$
and $y$ be the degree measurement of $\overarc{EF}=\overarc{FG}=\overarc{GH}=\overarc{HI}=\overarc{IA}$ in circle $C$.
$\angle ECA$ is, therefore, $5y$ by wa... | 6.125 | [
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In the diagram below, $ABCD$ is a square. Point $E$ is the midpoint of $\overline{AD}$. Points $F$ and $G$ lie on $\overline{CE}$, and $H$ and $J$ lie on $\overline{AB}$ and $\overline{BC}$, respectively, so that $FGHJ$ is a square. Points $K$ and $L$ lie on $\overline{GH}$, and $M$ and $N$ lie on $\overline{AD}$ and $... | 539 | This is a relatively quick solution but a fakesolve. We see that with a ruler, $KL = \frac{3}{2}$ cm and $HG = \frac{7}{2}$ cm. Thus if $KL$ corresponds with an area of $99$, then $HG$ ($FGHJ$'s area) would correspond with $99*(\frac{7}{3})^2 = \boxed{539}$ - aops5234 | 5.625 | [
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For positive integer $n$, let $s(n)$ denote the sum of the digits of $n$. Find the smallest positive integer satisfying $s(n) = s(n+864) = 20$. | 695 | First of all, notice that the smallest $n$ with $s(n) = 20$ is $299$. Also, if $s(n + 864) = 20$, $s(n - 136) = 19$ (because subtracting $1000$ from the number removes the $1$ in the thousands place). After checking $s(n - 136)$ for various $n$ with $s(n) = 20$, we see that we need to have a carry when subtracting $136... | 5 | [
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Let $S$ be the set of all ordered triple of integers $(a_1,a_2,a_3)$ with $1 \le a_1,a_2,a_3 \le 10$. Each ordered triple in $S$ generates a sequence according to the rule $a_n=a_{n-1}\cdot | a_{n-2}-a_{n-3} |$ for all $n\ge 4$. Find the number of such sequences for which $a_n=0$ for some $n$. | 494 | Let $a_1=x, a_2=y, a_3=z$. First note that if any absolute value equals 0, then $a_n=0$. Also note that if at any position, $a_n=a_{n-1}$, then $a_{n+2}=0$. Then, if any absolute value equals 1, then $a_n=0$. Therefore, if either $|y-x|$ or $|z-y|$ is less than or equal to 1, then that ordered triple meets the criteria... | 6 | [
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Let $f(x)$ be a third-degree polynomial with real coefficients satisfying \[|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.\] Find $|f(0)|$. | 72 | Because a cubic must come in a "wave form" with two points of inflection, we can see that $f(1)=f(5)=f(6)$, and $f(2)=f(3)=f(7)$. By symmetry, $f(4)=0$. Now, WLOG let $f(1)=12$, and $f(2)=f(3)=-12$. Then, we can use finite differences to get that the third (constant) difference is $12$, and therefore $f(0)=12+(24+(24+1... | 6.375 | [
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Triangle $ABC$ has positive integer side lengths with $AB=AC$. Let $I$ be the intersection of the bisectors of $\angle B$ and $\angle C$. Suppose $BI=8$. Find the smallest possible perimeter of $\triangle ABC$. | 108 | Let $M$ be midpoint $BC, BM = x, AB = y, \angle IBM = \alpha.$
$BI$ is the bisector of $\angle ABM$ in $\triangle ABM.$ $BI = \frac {2 xy \cos \alpha}{x+y} = 8, \cos \alpha = \frac {x}{8} \implies \frac {x^2 y}{x+y} = 32.$ \[y = \frac {32 x} {x^2 - 32}.\] $BC = 2x$ is integer, $5.5^2 < 32 \implies x \ge 6.$ $BM < BI \i... | 5.375 | [
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Consider all 1000-element subsets of the set $\{1, 2, 3, ... , 2015\}$. From each such subset choose the least element. The arithmetic mean of all of these least elements is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$.
Hint
Use the Hockey Stick Identity in the form
\[\binom{a}... | 431 | Let $p$ be the size of the large set and $q$ be the size of the subset (i.e. in this problem, $p = 2015$ and $q = 1000$). We can easily find the answers for smaller values of $p$ and $q$:
For $p = 2$ and $q = 2$, the answer is $1$.
For $p = 3$ and $q = 2$, the answer is $\frac43$.
For $p = 4$ and $q = 2$, the answer... | 6.375 | [
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With all angles measured in degrees, the product $\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n$, where $m$ and $n$ are integers greater than 1. Find $m+n$. | 91 | Let $p=\sin1\sin3\sin5...\sin89$
\[p=\sqrt{\sin1\sin3\sin5...\sin177\sin179}\]
\[=\sqrt{\frac{\sin1\sin2\sin3\sin4...\sin177\sin178\sin179}{\sin2\sin4\sin6\sin8...\sin176\sin178}}\]
\[=\sqrt{\frac{\sin1\sin2\sin3\sin4...\sin177\sin178\sin179}{(2\sin1\cos1)\cdot(2\sin2\cos2)\cdot(2\sin3\cos3)\cdot....\cdot(2\sin89\cos89... | 7.5 | [
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For each integer $n \ge 2$, let $A(n)$ be the area of the region in the coordinate plane defined by the inequalities $1\le x \le n$ and $0\le y \le x \left\lfloor \sqrt x \right\rfloor$, where $\left\lfloor \sqrt x \right\rfloor$ is the greatest integer not exceeding $\sqrt x$. Find the number of values of $n$ with $2\... | 483 | By considering the graph of this function, it is shown that the graph is composed of trapezoids ranging from $a^2$ to $(a+1)^2$ with the top made of diagonal line $y=ax$. The width of each trapezoid is $3, 5, 7$, etc. Whenever $a$ is odd, the value of $A(n)$ increases by an integer value, plus $\frac{1}{2}$. Whenever $... | 7.125 | [
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A block of wood has the shape of a right circular cylinder with radius $6$ and height $8$, and its entire surface has been painted blue. Points $A$ and $B$ are chosen on the edge of one of the circular faces of the cylinder so that $\overarc{AB}$ on that face measures $120^\text{o}$. The block is then sliced in half al... | 53 | Label the points where the plane intersects the top face of the cylinder as $C$ and $D$, and the center of the cylinder as $O$, such that $C,O,$ and $A$ are collinear. Let $T$ be the center of the bottom face, and $M$ the midpoint of $\overline{AB}$. Then $OT=4$, $TM=3$ (because of the 120 degree angle), and so $OM=5$.... | 6.375 | [
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Let $N$ be the least positive integer that is both $22$ percent less than one integer and $16$ percent greater than another integer. Find the remainder when $N$ is divided by $1000$. | 131 | Continuing from Solution 1, we have $N=\frac{39}{50}k$ and $N=\frac{29}{25}m$. It follows that $k=\frac{50}{39}N$ and $m=\frac{25}{29}N$. Both $m$ and $k$ have to be integers, so, in order for that to be true, $N$ has to cancel the denominators of both $\frac{50}{39}$ and $\frac{25}{29}$. In other words, $N$ is a multi... | 5 | [
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In a new school $40$ percent of the students are freshmen, $30$ percent are sophomores, $20$ percent are juniors, and $10$ percent are seniors. All freshmen are required to take Latin, and $80$ percent of the sophomores, $50$ percent of the juniors, and $20$ percent of the seniors elect to take Latin. The probability t... | 25 | We see that $40\% \cdot 100\% + 30\% \cdot 80\% + 20\% \cdot 50\% + 10\% \cdot 20\% = 76\%$ of students are learning Latin. In addition, $30\% \cdot 80\% = 24\%$ of students are sophomores learning Latin. Thus, our desired probability is $\dfrac{24}{76}=\dfrac{6}{19}$ and our answer is $6+19=\boxed{025}$. | 3.875 | [
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Let $m$ be the least positive integer divisible by $17$ whose digits sum to $17$. Find $m$. | 476 | The three-digit integers divisible by $17$, and their digit sum: \[\begin{array}{c|c} m & s(m)\\ \hline 102 & 3 \\ 119 & 11\\ 136 & 10\\ 153 & 9\\ 170 & 8\\ 187 & 16\\ 204 & 6\\ 221 & 5\\ 238 & 13\\ 255 & 12\\ 272 & 11\\ 289 & 19\\ 306 & 9\\ 323 & 8\\ 340 & 7\\ 357 & 15\\ 374 & 14\\ 391 & 13\\ 408 & 12\\ 425 & 11\\ 442... | 3.625 | [
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In an isosceles trapezoid, the parallel bases have lengths $\log 3$ and $\log 192$, and the altitude to these bases has length $\log 16$. The perimeter of the trapezoid can be written in the form $\log 2^p 3^q$, where $p$ and $q$ are positive integers. Find $p + q$. | 18 | Call the trapezoid $ABCD$ with $AB$ as the smaller base and $CD$ as the longer. Let the point where an altitude intersects the larger base be $E$, where $E$ is closer to $D$.
Subtract the two bases and divide to find that $ED$ is $\log 8$. The altitude can be expressed as $\frac{4}{3} \log 8$. Therefore, the two legs ... | 6 | [
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] |
Two unit squares are selected at random without replacement from an $n \times n$ grid of unit squares. Find the least positive integer $n$ such that the probability that the two selected unit squares are horizontally or vertically adjacent is less than $\frac{1}{2015}$. | 90 | There are 3 cases in this problem. Number one, the center squares. Two, the edges, and three, the corners. The probability of getting one center square and an adjacent square is $\frac{(n-2)(n-2)}{n^2}$ multiplied by $\frac{4}{n^2 -1}$. Add that to the probability of an edge and an adjacent square( $\frac{4n-8}{n^2}$ m... | 5.125 | [
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Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form $P(x) = 2x^3-2ax^2+(a^2-81)x-c$ for some positive integers $a$ and $c$. Can you tell me the values of $a$ and $c$?"
After some calculations, Jon says, "There is more than one such polynomial."
Steve sa... | 440 | Since each of the roots is positive, the local maximum of the function must occur at a positive value of $x$. Taking $\frac{d}{dx}$ of the polynomial yields $6x^2-4ax+a^2-81$, which is equal to $0$ at the local maximum. Since this is a quadratic in $a$, we can find an expression for $a$ in terms of $x$. The quadratic f... | 6.625 | [
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Triangle $ABC$ has side lengths $AB = 12$, $BC = 25$, and $CA = 17$. Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$, vertex $Q$ on $\overline{AC}$, and vertices $R$ and $S$ on $\overline{BC}$. In terms of the side length $PQ = \omega$, the area of $PQRS$ can be expressed as the quadratic polynomial \[Area(PQRS) = \... | 161 | Proceed as in solution 1. When $\omega$ is equal to zero, $\alpha - \beta\omega=\alpha$ is equal to the altitude. This means that $25\beta$ is equal to $\frac{36}{5}$, so $\beta = \frac{36}{125}$, yielding $\boxed{161}$. | 6.25 | [
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Let $a$ and $b$ be positive integers satisfying $\frac{ab+1}{a+b} < \frac{3}{2}$. The maximum possible value of $\frac{a^3b^3+1}{a^3+b^3}$ is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. | 36 | Notice that for $\frac{a^3b^3+1}{a^3+b^3}$ to be maximized, $\frac{ab+1}{a+b}$ has to be maximized. We simplify as above to $2ab + 2 < 3a + 3b$, which is $(a-\frac{3}{2})(b-\frac{3}{2}) < \frac{5}{4}$. To maximize, $a$ has to be as close to $b$ as possible, making $a$ close to $\frac{3+\sqrt{5}}{2}$. Because $a$ and $b... | 6.625 | [
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A cylindrical barrel with radius $4$ feet and height $10$ feet is full of water. A solid cube with side length $8$ feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is $v$ cubic feet. Find $v^2$.
[asy] import three; import solids; size(5cm); currentprojection=o... | 384 | We can use the same method as in Solution 2 to find the side length of the equilateral triangle, which is $4\sqrt3$. From here, its area is \[\dfrac{\bigl(4\sqrt3\bigr)^2\sqrt3}4=12\sqrt3.\] The leg of the isosceles right triangle is $\dfrac{4\sqrt3}{\sqrt2}=2\sqrt6$, and the horizontal distance from the vertex to the ... | 6.5 | [
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Call a permutation $a_1, a_2, \ldots, a_n$ of the integers $1, 2, \ldots, n$ quasi-increasing if $a_k \leq a_{k+1} + 2$ for each $1 \leq k \leq n-1$. For example, 53421 and 14253 are quasi-increasing permutations of the integers $1, 2, 3, 4, 5$, but 45123 is not. Find the number of quasi-increasing permutations of the ... | 486 | The simple recurrence can be found.
When inserting an integer $n$ into a string with $n - 1$ integers, we notice that the integer $n$ has 3 spots where it can go: before $n - 1$, before $n - 2$, and at the very end.
EXAMPLE: Putting 4 into the string 123: 4 can go before the 2: 1423, Before the 3: 1243, And at the ve... | 6.125 | [
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The circumcircle of acute $\triangle ABC$ has center $O$. The line passing through point $O$ perpendicular to $\overline{OB}$ intersects lines $AB$ and $BC$ and $P$ and $Q$, respectively. Also $AB=5$, $BC=4$, $BQ=4.5$, and $BP=\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
Diagram
[... | 23 | Let $r=BO$. Drawing perpendiculars, $BM=MC=2$ and $BN=NA=2.5$. From there, \[OM=\sqrt{r^2-4}\] Thus, \[OQ=\frac{\sqrt{4r^2+9}}{2}\] Using $\triangle{BOQ}$, we get $r=3$. Now let's find $NP$. After some calculations with $\triangle{BON}$ ~ $\triangle{OPN}$, ${NP=11/10}$. Therefore, \[BP=\frac{5}{2}+\frac{11}{10}=18/5\] ... | 6 | [
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There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical. | 548 | Let $a_{n}$ be the number of ways to form $n$-letter strings made up of As and Bs such that no more than $3$ adjacent letters are identical.
Note that, at the end of each $n$-letter string, there are $3$ possibilities for the last letter chain: it must be either $1$, $2$, or $3$ letters long. Removing this last chain ... | 5.75 | [
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Define the sequence $a_1, a_2, a_3, \ldots$ by $a_n = \sum\limits_{k=1}^n \sin{k}$, where $k$ represents radian measure. Find the index of the 100th term for which $a_n < 0$. | 628 | If $n = 1$, $a_n = \sin(1) > 0$. Then if $n$ satisfies $a_n < 0$, $n \ge 2$, and \[a_n = \sum_{k=1}^n \sin(k) = \cfrac{1}{\sin{1}} \sum_{k=1}^n\sin(1)\sin(k) = \cfrac{1}{2\sin{1}} \sum_{k=1}^n\cos(k - 1) - \cos(k + 1) = \cfrac{1}{2\sin(1)} [\cos(0) + \cos(1) - \cos(n) - \cos(n + 1)].\] Since $2\sin 1$ is positive, it d... | 6.75 | [
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Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$. Evaluate $2x^3+(xy)^3+2y^3$. | 89 | $x^4y^4(x+y)=810; x^3y^3(x^3+y^3)=945, \frac{x^2-xy+y^2}{xy}=\frac{7}{6}, \frac{x^2+y^2}{xy}=\frac{13}{6}$
Let $x^2+y^2=13k; xy=6k$, then we can see $(x+y)^2-12k=13k, x+y=5\sqrt{k}$, now, we see $x^4y^4\cdot (x+y)=1296k^4\cdot 5\sqrt{k}, k=\frac{1}{\sqrt[3]{4}}$.
The rest is easy, $2(x^3+y^3)+x^3y^3=216k^3+2[(x+y)^3-... | 6.75 | [
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Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$, respectively, and are externally tangent at point $A$. Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and in... | 129 | Let $P$ and $Q$ be the centers of circles $\mathcal{P}$ and $\mathcal{Q}$ , respectively.
Let $M$ be midpoint $BC, \beta = \angle ACB.$
Upper diagram shows that
$\sin 2\beta = \frac {4}{5}$ and $AC = 2 AB.$ Therefore $\cos 2\beta = \frac {3}{5}.$
Let $CH\perp l, BH'\perp l.$ Lower diagram shows that
$\angle CAE = \a... | 7 | [
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For $-1<r<1$, let $S(r)$ denote the sum of the geometric series \[12+12r+12r^2+12r^3+\cdots .\] Let $a$ between $-1$ and $1$ satisfy $S(a)S(-a)=2016$. Find $S(a)+S(-a)$. | 336 | The sum of an infinite geometric series is $\frac{a}{1-r}\rightarrow \frac{12}{1\mp a}$. The product $S(a)S(-a)=\frac{144}{1-a^2}=2016$. $\frac{12}{1-a}+\frac{12}{1+a}=\frac{24}{1-a^2}$, so the answer is $\frac{2016}{6}=\boxed{336}$.
~ pi_is_3.14 | 5.375 | [
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Two dice appear to be normal dice with their faces numbered from $1$ to $6$, but each die is weighted so that the probability of rolling the number $k$ is directly proportional to $k$. The probability of rolling a $7$ with this pair of dice is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Fin... | 71 | It is easier to think of the dice as $21$ sided dice with $6$ sixes, $5$ fives, etc. Then there are $21^2=441$ possible rolls. There are $2\cdot(1\cdot 6+2\cdot 5+3\cdot 4)=56$ rolls that will result in a seven. The odds are therefore $\frac{56}{441}=\frac{8}{63}$. The answer is $8+63=\boxed{071}$
See also 2006 AMC 12... | 4.625 | [
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A regular icosahedron is a $20$-faced solid where each face is an equilateral triangle and five triangles meet at every vertex. The regular icosahedron shown below has one vertex at the top, one vertex at the bottom, an upper pentagon of five vertices all adjacent to the top vertex and all in the same horizontal plane,... | 810 | Go to 2020 AMC 10A #19, and connect all of the centers of the faces on the dodecahedron to get the icosahedron. The answer is $\boxed{810}$. | 6.25 | [
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A right prism with height $h$ has bases that are regular hexagons with sides of length $12$. A vertex $A$ of the prism and its three adjacent vertices are the vertices of a triangular pyramid. The dihedral angle (the angle between the two planes) formed by the face of the pyramid that lies in a base of the prism and th... | 108 | Let $B$ and $C$ be the vertices adjacent to $A$ on the same base as $A$, and let $D$ be the last vertex of the triangular pyramid. Notice that we can already find some lengths. We have $AB=AC=12$ (given) and $BC=BD=\sqrt{144+h^2}$ by the Pythagorean Theorem. Let $M$ be the midpoint of $BC$. Then, we have $AM=6$ (30-60-... | 5.875 | [
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Anh read a book. On the first day she read $n$ pages in $t$ minutes, where $n$ and $t$ are positive integers. On the second day Anh read $n + 1$ pages in $t + 1$ minutes. Each day thereafter Anh read one more page than she read on the previous day, and it took her one more minute than on the previous day until she comp... | 53 | We could see that both $374$ and $319$ are divisible by $11$ in the outset, and that $34$ and $29$, the quotients, are relatively prime. Both are the $average$ number of minutes across the $11$ days, so we need to subtract $\left \lfloor{\frac{11}{2}}\right \rfloor=5$ from each to get $(n,t)=(29,24)$ and $29+24=\boxed{... | 5.5 | [
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In $\triangle ABC$ let $I$ be the center of the inscribed circle, and let the bisector of $\angle ACB$ intersect $AB$ at $L$. The line through $C$ and $L$ intersects the circumscribed circle of $\triangle ABC$ at the two points $C$ and $D$. If $LI=2$ and $LD=3$, then $IC=\tfrac{m}{n}$, where $m$ and $n$ are relatively ... | 13 | Let $AB=c,BC=a,CA=b$, and $x=\tfrac{a+b}{c}$. Then, notice that $\tfrac{CI}{IL}=\tfrac{a+b}{c}=x$, so $CI=IL\cdot{}x=2x$. Also, by the incenter-excenter lemma, $AD=BD=ID=IL+LD=5$. Therefore, by Ptolemy's Theorem on cyclic quadrilateral $ABCD$, $5a+5b=c(2x+5)$, so $5\left(\tfrac{a+b}{c}\right)=2x+5$, so $5x=2x+5$. Solvi... | 6 | [
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For integers $a$ and $b$ consider the complex number \[\frac{\sqrt{ab+2016}}{ab+100}-\left({\frac{\sqrt{|a+b|}}{ab+100}}\right)i\]
Find the number of ordered pairs of integers $(a,b)$ such that this complex number is a real number. | 103 | Similar to Solution 1, but concise:
First, we set the imaginary expression to $0$, so that $|a+b|=0$ or $a=-b$, of which there are $44\cdot 2+1=89$ possibilities. But $(a,b)\ne(\pm 10,\mp 10)$ because the denominator would be $0$. So this gives $89-2=87$ solutions.
Then we try to cancel the imaginary part with the sq... | 7 | [
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For a permutation $p = (a_1,a_2,\ldots,a_9)$ of the digits $1,2,\ldots,9$, let $s(p)$ denote the sum of the three $3$-digit numbers $a_1a_2a_3$, $a_4a_5a_6$, and $a_7a_8a_9$. Let $m$ be the minimum value of $s(p)$ subject to the condition that the units digit of $s(p)$ is $0$. Let $n$ denote the number of permutations ... | 162 | To minimize $s(p)$, the numbers $1$, $2$, and $3$ (which sum to $6$) must be in the hundreds places. For the units digit of $s(p)$ to be $0$, the numbers in the ones places must have a sum of either $10$ or $20$. However, since the tens digit contributes more to the final sum $s(p)$ than the ones digit, and we are look... | 6 | [
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Triangle $ABC$ has $AB=40,AC=31,$ and $\sin{A}=\frac{1}{5}$. This triangle is inscribed in rectangle $AQRS$ with $B$ on $\overline{QR}$ and $C$ on $\overline{RS}$. Find the maximum possible area of $AQRS$. | 744 | Note that if angle $BAC$ is obtuse, it would be impossible for the triangle to inscribed in a rectangle. This can easily be shown by drawing triangle ABC, where $A$ is obtuse. Therefore, angle A is acute. Let angle $CAS=n$ and angle $BAQ=m$. Then, $\overline{AS}=31\cos(n)$ and $\overline{AQ}=40\cos(m)$. Then the area o... | 6.125 | [
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A strictly increasing sequence of positive integers $a_1$, $a_2$, $a_3$, $\cdots$ has the property that for every positive integer $k$, the subsequence $a_{2k-1}$, $a_{2k}$, $a_{2k+1}$ is geometric and the subsequence $a_{2k}$, $a_{2k+1}$, $a_{2k+2}$ is arithmetic. Suppose that $a_{13} = 2016$. Find $a_1$. | 504 | We first create a similar sequence where $a_1=1$ and $a_2=2$. Continuing the sequence,
\[1, 2,4,6,9,12,16,20,25,30,36,42,49,\cdots\]
Here we can see a pattern; every second term (starting from the first) is a square, and every second term (starting from the third) is the end of a geometric sequence. This can be proven... | 6.5 | [
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Let $P(x)$ be a nonzero polynomial such that $(x-1)P(x+1)=(x+2)P(x)$ for every real $x$, and $\left(P(2)\right)^2 = P(3)$. Then $P(\tfrac72)=\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 109 | Substituting $x=2$ into the given equation, we find that $P(3)=4P(2)=P(2)^2$. Therefore, either $P(2)=0$ or $P(2)=4$. Now for integers $n\ge 2$, we know that \[P(n+1)=\frac{n+2}{n-1}P(n).\] Applying this repeatedly, we find that \[P(n+1)=\frac{(n+2)!/3!}{(n-1)!}P(2).\] If $P(2)=0$, this shows that $P(x)$ has infinitely... | 6.875 | [
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Find the least positive integer $m$ such that $m^2 - m + 11$ is a product of at least four not necessarily distinct primes. | 132 | Suppose $p=11$; then $m^2-m+11=11qrs$. Reducing modulo 11, we get $m\equiv 1,0 \pmod{11}$ so $k(11k\pm 1)+1 = qrs$.
Suppose $q=11$. Then we must have $11k^2\pm k + 1 = 11rs$, which leads to $k\equiv \mp 1 \pmod{11}$, i.e., $k\in \{1,10,12,21,23,\ldots\}$.
$k=1$ leads to $rs=1$ (impossible)! Then $k=10$ leads to $rs=10... | 6 | [
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Freddy the frog is jumping around the coordinate plane searching for a river, which lies on the horizontal line $y = 24$. A fence is located at the horizontal line $y = 0$. On each jump Freddy randomly chooses a direction parallel to one of the coordinate axes and moves one unit in that direction. When he is at a point... | 273 | Clearly Freddy's $x$-coordinate is irrelevant, so we let $E(y)$ be the expected value of the number of jumps it will take him to reach the river from a given $y$-coordinate. Observe that $E(24)=0$, and \[E(y)=1+\frac{E(y+1)+E(y-1)+2E(y)}{4}\] for all $y$ such that $1\le y\le 23$. Also note that $E(0)=1+\frac{2E(0)+E(1)... | 6.5 | [
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Centered at each lattice point in the coordinate plane are a circle radius $\frac{1}{10}$ and a square with sides of length $\frac{1}{5}$ whose sides are parallel to the coordinate axes. The line segment from $(0,0)$ to $(1001, 429)$ intersects $m$ of the squares and $n$ of the circles. Find $m + n$. | 574 | This is mostly a clarification to Solution 1, but let's take the diagram for the origin to $(7,3)$. We have the origin circle and square intersected, then two squares, then the circle and square at $(7,3)$. If we take the circle and square at the origin out of the diagram, we will be able to repeat the resulting segmen... | 6.375 | [
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Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$. Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$, respectively, with line $AB$ closer to point $X$ than to $Y$. Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again at... | 270 | $AB^2 = 4 AM^2 =2x(2x+ 2 XY) =(XP - XY) (XP + XY) = XP^2 - XY^2 = XC \cdot XD - XY^2 = 67 \cdot 37 - 47^2 = \boxed{270}.$
vladimir.shelomovskii@gmail.com, vvsss
~MathProblemSolvingSkills.com | 6.375 | [
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Initially Alex, Betty, and Charlie had a total of $444$ peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats $5$ of his peanuts, Betty eats $9$ of her peanuts, and Charlie eats $25$ of his peanuts. Now the three number... | 108 | Let $b$ be the finish number of Betty's peanuts. Then \[3b = 444-(5 + 9 + 25) = 405 = 3 \cdot 135 \implies b = 135, b+ 9 = 144.\]
Let $k > 1$ be the common ratio. Then \[\frac{144}{k} + 144 + k \cdot 144 = 444 \implies \frac{144}{k} + k \cdot 144 = 300\implies \frac{12}{k} + k \cdot 12 = 25\implies k = \frac{4}{3} \im... | 5.75 | [
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There is a $40\%$ chance of rain on Saturday and a $30\%$ chance of rain on Sunday. However, it is twice as likely to rain on Sunday if it rains on Saturday than if it does not rain on Saturday. The probability that it rains at least one day this weekend is $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive... | 107 | Let $x$ be the probability that it rains on Sunday given that it doesn't rain on Saturday. We then have $\dfrac{3}{5}x+\dfrac{2}{5}2x = \dfrac{3}{10} \implies \dfrac{7}{5}x=\dfrac{3}{10}$ $\implies x=\dfrac{3}{14}$. Therefore, the probability that it doesn't rain on either day is $\left(1-\dfrac{3}{14}\right)\left(\dfr... | 5.125 | [
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6
] |
Let $x,y,$ and $z$ be real numbers satisfying the system \begin{align*} \log_2(xyz-3+\log_5 x)&=5,\\ \log_3(xyz-3+\log_5 y)&=4,\\ \log_4(xyz-3+\log_5 z)&=4.\\ \end{align*} Find the value of $|\log_5 x|+|\log_5 y|+|\log_5 z|$. | 265 | First, we get rid of logs by taking powers: $xyz-3+\log_5 x=2^{5}=32$, $xyz-3+\log_5 y=3^{4}=81$, and $(xyz-3+\log_5 z)=4^{4}=256$. Adding all the equations up and using the $\log {xy}=\log {x}+\log{y}$ property, we have $3xyz+\log_5{xyz} = 378$, so we have $xyz=125$. Solving for $x,y,z$ by substituting $125$ for $xyz$... | 6.75 | [
7,
6,
7,
7,
6,
7,
7,
7
] |
An $a \times b \times c$ rectangular box is built from $a \cdot b \cdot c$ unit cubes. Each unit cube is colored red, green, or yellow. Each of the $a$ layers of size $1 \times b \times c$ parallel to the $(b \times c)$ faces of the box contains exactly $9$ red cubes, exactly $12$ green cubes, and some yellow cubes. Ea... | 180 | The total number of green cubes is given by $12a=20b\Longrightarrow a=\frac{5}{3}b$.
Let $r$ be the number of red cubes on each one of the $b$ layers then the total number of red cubes is $9a=br$. Substitute $a=\frac{5}{3}b$ gives $r=15$.
Repeating the procedure on the number of yellow cubes $y$ on each of the $a$ la... | 5.25 | [
5,
5,
5,
6,
6,
5,
4,
6
] |
Triangle $ABC_0$ has a right angle at $C_0$. Its side lengths are pairwise relatively prime positive integers, and its perimeter is $p$. Let $C_1$ be the foot of the altitude to $\overline{AB}$, and for $n \geq 2$, let $C_n$ be the foot of the altitude to $\overline{C_{n-2}B}$ in $\triangle C_{n-2}C_{n-1}B$. The sum $\... | 182 | For this problem, first notice that its an infinite geometric series of $6(a+b+c)=\frac{ab}{c-b}$ if $c$ is the hypotenuse. WLOG $a<b$, we can generalize a pythagorean triple of $x^2-y^2, 2xy, x^2+y^2$. Let $b=2xy$, then this generalization gives $6(a+b+c)(c-b)=ab$ \[(x^2-y^2)2xy=6(2x^2+2xy)(x-y)^2\] \[(x+y)xy=6(x^2+xy... | 7.375 | [
7,
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7,
7,
8,
8
] |
For polynomial $P(x)=1-\dfrac{1}{3}x+\dfrac{1}{6}x^{2}$, define $Q(x)=P(x)P(x^{3})P(x^{5})P(x^{7})P(x^{9})=\sum_{i=0}^{50} a_ix^{i}$. Then $\sum_{i=0}^{50} |a_i|=\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 275 | Multiply $P(x)P(x^3)$ and notice that the odd degree terms have a negative coefficient. Observing that this is probably true for all polynomials like this (including $P(x)P(x^3)P(x^5)P(x^7)P(x^9)$), we plug in $-1$ to get $\frac{243}{32} \implies \boxed{275}$. | 5.875 | [
6,
6,
6,
6,
5,
6,
6,
6
] |
Squares $ABCD$ and $EFGH$ have a common center and $\overline{AB} || \overline{EF}$. The area of $ABCD$ is 2016, and the area of $EFGH$ is a smaller positive integer. Square $IJKL$ is constructed so that each of its vertices lies on a side of $ABCD$ and each vertex of $EFGH$ lies on a side of $IJKL$. Find the differenc... | 840 | Letting $AI=a$ and $IB=b$, we have \[IJ^{2}=a^{2}+b^{2} \geq 1008\] by AM-GM inequality. Also, since $EFGH||ABCD$, the angles that each square cuts another are equal, so all the triangles are formed by a vertex of a larger square and $2$ adjacent vertices of a smaller square are similar. Therefore, the areas form a geo... | 6 | [
6,
6,
7,
6,
6,
5,
6,
6
] |
Find the number of sets ${a,b,c}$ of three distinct positive integers with the property that the product of $a,b,$ and $c$ is equal to the product of $11,21,31,41,51,61$. | 728 | Note that the prime factorization of the product is $3^{2}\cdot 7 \cdot 11 \cdot 17 \cdot 31 \cdot 41 \cdot 61$. Ignoring overcounting, by stars and bars there are $6$ ways to choose how to distribute the factors of $3$, and $3$ ways to distribute the factors of the other primes, so we have $3^{6} \cdot 6$ ways. Howeve... | 5.875 | [
5,
7,
6,
6,
6,
5,
5,
7
] |
The sequences of positive integers $1,a_2, a_3,...$ and $1,b_2, b_3,...$ are an increasing arithmetic sequence and an increasing geometric sequence, respectively. Let $c_n=a_n+b_n$. There is an integer $k$ such that $c_{k-1}=100$ and $c_{k+1}=1000$. Find $c_k$. | 262 | Since all the terms of the sequences are integers, and 100 isn't very big, we should just try out the possibilities for $b_2$. When we get to $b_2=9$ and $a_2=91$, we have $a_4=271$ and $b_4=729$, which works, therefore, the answer is $b_3+a_3=81+181=\boxed{262}$. | 5.375 | [
6,
4,
5,
5,
5,
6,
7,
5
] |
Triangle $ABC$ is inscribed in circle $\omega$. Points $P$ and $Q$ are on side $\overline{AB}$ with $AP<AQ$. Rays $CP$ and $CQ$ meet $\omega$ again at $S$ and $T$ (other than $C$), respectively. If $AP=4,PQ=3,QB=6,BT=5,$ and $AS=7$, then $ST=\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $... | 43 | By Ptolemy's Theorem applied to quadrilateral $ASTB$, we find \[AS\cdot BT+AB\cdot ST=AT\cdot BS.\] Projecting through $C$ we have \[\frac{AQ \cdot PB}{PQ \cdot AB} = (A,Q; P,B)\stackrel{C}{=}(A,T; S,B)=\frac{AT \cdot BS}{ST \cdot AB}.\] Therefore \[AT \cdot BS = \frac {AQ \cdot PB}{PQ} \times ST \implies\] \[\left(\fr... | 6.625 | [
6,
7,
6,
7,
6,
7,
7,
7
] |
For positive integers $N$ and $k$, define $N$ to be $k$-nice if there exists a positive integer $a$ such that $a^{k}$ has exactly $N$ positive divisors. Find the number of positive integers less than $1000$ that are neither $7$-nice nor $8$-nice. | 749 | All integers $a$ will have factorization $2^a3^b5^c7^d...$. Therefore, the number of factors in $a^7$ is $(7a+1)(7b+1)...$, and for $a^8$ is $(8a+1)(8b+1)...$. The most salient step afterwards is to realize that all numbers $N$ not $1 \pmod{7}$ and also not $1 \pmod{8}$ satisfy the criterion. The cycle repeats every $5... | 6.375 | [
7,
6,
6,
6,
6,
6,
7,
7
] |
The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] dr... | 732 | This is equivalent to a node coloring of a cycle with 6 nodes. After repeatedly using deletion-contraction, the solution comes out to be $\boxed{732}$ | 5.5 | [
5,
6,
6,
5,
6,
4,
5,
7
] |
Beatrix is going to place six rooks on a $6 \times 6$ chessboard where both the rows and columns are labeled $1$ to $6$; the rooks are placed so that no two rooks are in the same row or the same column. The $value$ of a square is the sum of its row number and column number. The $score$ of an arrangement of rooks is the... | 371 | So we first count the number of permutations with score $\ge 2$. This is obviously $6!=720$. Then, the number of permutations with score $\ge 3$ can also be computed: in the first column, there are five ways to place a rook- anywhere but the place with score $1$. In the next column, there are $5$ ways to place a rook- ... | 6.25 | [
7,
6,
7,
6,
6,
6,
6,
6
] |
Equilateral $\triangle ABC$ has side length $600$. Points $P$ and $Q$ lie outside the plane of $\triangle ABC$ and are on opposite sides of the plane. Furthermore, $PA=PB=PC$, and $QA=QB=QC$, and the planes of $\triangle PAB$ and $\triangle QAB$ form a $120^{\circ}$ dihedral angle (the angle between the two planes). Th... | 450 | Draw a good diagram. Draw $CH$ as an altitude of the triangle. Scale everything down by a factor of $100\sqrt{3}$, so that $AB=2\sqrt{3}$. Finally, call the center of the triangle U. Draw a cross-section of the triangle via line $CH$, which of course includes $P, Q$. From there, we can call $OU=h$. There are two crucia... | 7.75 | [
8,
7,
8,
7,
8,
8,
8,
8
] |
For $1 \leq i \leq 215$ let $a_i = \dfrac{1}{2^{i}}$ and $a_{216} = \dfrac{1}{2^{215}}$. Let $x_1, x_2, ..., x_{216}$ be positive real numbers such that $\sum_{i=1}^{216} x_i=1$ and $\sum_{1 \leq i < j \leq 216} x_ix_j = \dfrac{107}{215} + \sum_{i=1}^{216} \dfrac{a_i x_i^{2}}{2(1-a_i)}$. The maximum possible value of $... | 863 | Note that \begin{align*}\sum_{1 \leq i < j \leq 216} x_ix_j &= \frac12\left(\left(\sum_{i=1}^{216} x_i\right)^2-\sum_{i=1}^{216} x_i^2\right)\\&=\frac12\left(1-\sum x_i^2\right).\end{align*} Substituting this into the second equation and collecting $x_i^2$ terms, we find \[\sum_{i=1}^{216}\frac{x_i^2}{1-a_i}=\frac{1}{2... | 7.25 | [
8,
8,
7,
8,
6,
7,
6,
8
] |
Fifteen distinct points are designated on $\triangle ABC$: the 3 vertices $A$, $B$, and $C$; $3$ other points on side $\overline{AB}$; $4$ other points on side $\overline{BC}$; and $5$ other points on side $\overline{CA}$. Find the number of triangles with positive area whose vertices are among these $15$ points. | 390 | Every triangle is uniquely determined by 3 points. There are $\binom{15}{3}=455$ ways to choose 3 points, but that counts the degenerate triangles formed by choosing three points on a line. There are $\binom{5}{3}$ invalid cases on segment $AB$, $\binom{6}{3}$ invalid cases on segment $BC$, and $\binom{7}{3}$ invalid c... | 4.75 | [
4,
5,
5,
5,
4,
5,
5,
5
] |
When each of $702$, $787$, and $855$ is divided by the positive integer $m$, the remainder is always the positive integer $r$. When each of $412$, $722$, and $815$ is divided by the positive integer $n$, the remainder is always the positive integer $s \neq r$. Find $m+n+r+s$. | 62 | We know that $702 = am + r, 787 = bm + r,$ and $855 = cm+r$ where $a-c$ are integers.
Subtracting the first two, the first and third, and the last two, we get $85 = (b-a)m, 153=(c-a)m,$ and $68=(c-b)m.$
We know that $b-a, c-a$ and $c-b$ must be integers, so all the numbers are divisible by $m.$
Factorizing the numbe... | 5.375 | [
5,
5,
6,
6,
5,
5,
5,
6
] |
For a positive integer $n$, let $d_n$ be the units digit of $1 + 2 + \dots + n$. Find the remainder when \[\sum_{n=1}^{2017} d_n\]is divided by $1000$. | 69 | We see that $d_n$ appears in cycles of $20$ and the cycles are \[1,3,6,0,5,1,8,6,5,5,6,8,1,5,0,6,3,1,0,0,\] adding a total of $70$ each cycle. Since $\left\lfloor\frac{2017}{20}\right\rfloor=100$, we know that by $2017$, there have been $100$ cycles and $7000$ has been added. This can be discarded as we're just looking... | 4.875 | [
5,
5,
5,
4,
5,
4,
5,
6
] |
A pyramid has a triangular base with side lengths $20$, $20$, and $24$. The three edges of the pyramid from the three corners of the base to the fourth vertex of the pyramid all have length $25$. The volume of the pyramid is $m\sqrt{n}$, where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of... | 803 | Notation is shown on diagram. \[AM = MB = c = 12, AC = BC = b = 20,\] \[DA = DB = DC = a = 25.\] \[CM = x + y = \sqrt{b^2-c^2} = 16,\] \[x^2 - y^2 = CD^2 – DM^2 = CD^2 – (BD^2 – BM^2) = c^2 = 144,\] \[x – y = \frac{x^2 – y^2}{x+y} = \frac {c^2} {16} = 9,\] \[x = \frac {16 + 9}{2} = \frac {a}{2},\] \[h = \sqrt{a^2 -\fra... | 6.375 | [
6,
7,
6,
7,
6,
6,
6,
7
] |
A rational number written in base eight is $\underline{ab} . \underline{cd}$, where all digits are nonzero. The same number in base twelve is $\underline{bb} . \underline{ba}$. Find the base-ten number $\underline{abc}$. | 321 | The parts before and after the decimal points must be equal. Therefore $8a + b = 12b + b$ and $c/8 + d/64 = b/12 + a/144$. Simplifying the first equation gives $a = (3/2)b$. Plugging this into the second equation gives $3b/32 = c/8 + d/64$. Multiplying both sides by 64 gives $6b = 8c + d$. $a$ and $b$ are both digits b... | 4.75 | [
5,
4,
4,
5,
6,
4,
5,
5
] |
A circle is circumscribed around an isosceles triangle whose two congruent angles have degree measure $x$. Two points are chosen independently and uniformly at random on the circle, and a chord is drawn between them. The probability that the chord intersects the triangle is $\frac{14}{25}$. Find the difference between ... | 48 | Because we know that we have an isosceles triangle with angles of $x$ (and we know that x is an inscribed angle), that means that the arc that is intercepted by this angle is $2x$. We form this same conclusion for the other angle $x$, and $180-2x$. Therefore we get $3$ arcs, namely, $2x$, $2x$, and $360-4x$. To have th... | 7.125 | [
7,
7,
7,
7,
6,
8,
8,
7
] |
For nonnegative integers $a$ and $b$ with $a + b \leq 6$, let $T(a, b) = \binom{6}{a} \binom{6}{b} \binom{6}{a + b}$. Let $S$ denote the sum of all $T(a, b)$, where $a$ and $b$ are nonnegative integers with $a + b \leq 6$. Find the remainder when $S$ is divided by $1000$.
Major Note
Most solutions use committee formin... | 564 | Let $c=6-(a+b)$, and note that $\binom{6}{a + b}=\binom{6}{c}$. The problem thus asks for the sum $\binom{6}{a} \binom{6}{b} \binom{6}{c}$ over all $a,b,c$ such that $a+b+c=6$. Consider an array of 18 dots, with 3 columns of 6 dots each. The desired expression counts the total number of ways to select 6 dots by conside... | 5.75 | [
5,
6,
6,
5,
6,
6,
6,
6
] |
Two real numbers $a$ and $b$ are chosen independently and uniformly at random from the interval $(0, 75)$. Let $O$ and $P$ be two points on the plane with $OP = 200$. Let $Q$ and $R$ be on the same side of line $OP$ such that the degree measures of $\angle POQ$ and $\angle POR$ are $a$ and $b$ respectively, and $\angle... | 41 | Noting that $\angle OQP$ and $\angle ORP$ are right angles, we realize that we can draw a semicircle with diameter $\overline{OP}$ and points $Q$ and $R$ on the semicircle. Since the radius of the semicircle is $100$, if $\overline{QR} \leq 100$, then $\overarc{QR}$ must be less than or equal to $60^{\circ}$.
This sim... | 6.25 | [
6,
6,
6,
7,
6,
7,
6,
6
] |
Let $a_{10} = 10$, and for each positive integer $n >10$ let $a_n = 100a_{n - 1} + n$. Find the least positive $n > 10$ such that $a_n$ is a multiple of $99$. | 45 | We just notice that $100 \equiv 1 \pmod{99}$, so we are just trying to find $10 + 11 + 12 + \cdots + n$ modulo $99$, or $\dfrac{n(n+1)}{2} - 45$ modulo $99$. Also, the sum to $44$ is divisible by $99$, and is the first one that is. Thus, if we sum to $45$ the $45$ is cut off and thus is just a sum to $44$.
Without che... | 4.75 | [
5,
5,
4,
4,
5,
5,
5,
5
] |
Let $z_1=18+83i,~z_2=18+39i,$ and $z_3=78+99i,$ where $i=\sqrt{-1}.$ Let $z$ be the unique complex number with the properties that $\frac{z_3-z_1}{z_2-z_1}~\cdot~\frac{z-z_2}{z-z_3}$ is a real number and the imaginary part of $z$ is the greatest possible. Find the real part of $z$. | 56 | We will just bash. Let $z=a+bi$ where $a,b\in\mathbb{R}$. We see that $\frac{z_3-z_1}{z_2-z_1}=\frac{-4+15i}{11}$ after doing some calculations. We also see that $\frac{[(a-18)+(b-39)i][(a-78)-(b-99)i]}{\text{some real stuff}}.$ We note that $[(a-18)+(b-39)i][(a-78)-(b-99)i]$ is a multiple of $-4-15i$ because the numer... | 6.875 | [
6,
7,
7,
7,
7,
7,
7,
7
] |
Consider arrangements of the $9$ numbers $1, 2, 3, \dots, 9$ in a $3 \times 3$ array. For each such arrangement, let $a_1$, $a_2$, and $a_3$ be the medians of the numbers in rows $1$, $2$, and $3$ respectively, and let $m$ be the median of $\{a_1, a_2, a_3\}$. Let $Q$ be the number of arrangements for which $m = 5$. Fi... | 360 | We see that if one of the medians is 5, then there are two remaining numbers greater than 5 and two less than 5, so it follows that $m=5$. There are 3 ways to choose which row to have 5 in, $4\cdot 4=16$ ways to choose the other two numbers in that row, $3!=6$ ways to arrange the numbers in that row, and $6!=720$ ways ... | 5.875 | [
5,
6,
6,
5,
7,
7,
6,
5
] |
Call a set $S$ product-free if there do not exist $a, b, c \in S$ (not necessarily distinct) such that $a b = c$. For example, the empty set and the set $\{16, 20\}$ are product-free, whereas the sets $\{4, 16\}$ and $\{2, 8, 16\}$ are not product-free. Find the number of product-free subsets of the set $\{1, 2, 3, 4,.... | 252 | Let $X$ be a product-free subset, and note that 1 is not in $x$. We consider four cases:
1.) both 2 and 3 are not in $X$. Then there are $2^7=128$ possible subsets for this case.
2.) 2 is in $X$, but 3 is not. Then 4 in not in $X$, so there are $2^6=64$ subsets; however, there is a $\frac{1}{4}$ chance that 5 and 10 ... | 6.625 | [
7,
7,
7,
6,
7,
6,
7,
6
] |
For every $m \geq 2$, let $Q(m)$ be the least positive integer with the following property: For every $n \geq Q(m)$, there is always a perfect cube $k^3$ in the range $n < k^3 \leq m \cdot n$. Find the remainder when \[\sum_{m = 2}^{2017} Q(m)\]is divided by 1000. | 59 | We claim that $Q(m) = 1$ when $m \ge 8$.
When $m \ge 8$, for every $n \ge Q(m) = 1$, we need to prove there exists an integer $k$, such that $n < k^3 \le m*n$.
That because $\sqrt[3]{m*n} - \sqrt[3]{n} \ge 2\sqrt[3]{n} - \sqrt[3]{n} = \sqrt[3]{n} \ge 1$, so k exists between $\sqrt[3]{m*n}$ and $\sqrt[3]{n}$
$\sqrt[3]... | 6.25 | [
7,
6,
6,
7,
6,
6,
7,
5
] |
Let $a > 1$ and $x > 1$ satisfy $\log_a(\log_a(\log_a 2) + \log_a 24 - 128) = 128$ and $\log_a(\log_a x) = 256$. Find the remainder when $x$ is divided by $1000$. | 896 | The first condition implies
\[a^{128} = \log_a\log_a 2 + \log_a 24 - 128\]
\[128+a^{128} = \log_a\log_a 2^{24}\]
\[a^{a^{128}a^{a^{128}}} = 2^{24}\]
\[\left(a^{a^{128}}\right)^{\left(a^{a^{128}}\right)} = 2^{24} = 8^8\]
So $a^{a^{128}} = 8$.
Putting each side to the power of $128$:
\[\left(a^{128}\right)^{\left(a^{12... | 7.75 | [
7,
7,
8,
8,
8,
8,
8,
8
] |
The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths $2\sqrt{3},~5,$ and $\sqrt{37},$ as shown, is $\frac{m\sqrt{p}}{n},$ where $m,~n,$ and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any pri... | 145 | Let $AB=2\sqrt{3}, BC=5$, $D$ lies on $BC$, $F$ lies on $AB$ and $E$ lies on $AC$
Set $D$ as the origin, $BD=a,BF=b$, $F$ can be expressed as $-a+bi$ in argand plane, the distance of $CD$ is $5-a$
We know that $(-a+bi)\cdot cis(-\frac{\pi}{3})=(-\frac{a+\sqrt{3}b}{2}+\frac{\sqrt{3}a+b}{2}i)$. We know that the slope o... | 7 | [
7,
8,
7,
6,
7,
7,
7,
7
] |
Find the number of subsets of $\{1, 2, 3, 4, 5, 6, 7, 8\}$ that are subsets of neither $\{1, 2, 3, 4, 5\}$ nor $\{4, 5, 6, 7, 8\}$. | 196 | Note that by Principle of Inclusion and Exclusion, the total number of subsets must be $2^8-2^5-2^5+2^2$ as denoted by above. Thus our answer is $64(3)+4 = \boxed{196}$ | 4.125 | [
4,
4,
5,
4,
4,
4,
4,
4
] |
The teams $T_1$, $T_2$, $T_3$, and $T_4$ are in the playoffs. In the semifinal matches, $T_1$ plays $T_4$, and $T_2$ plays $T_3$. The winners of those two matches will play each other in the final match to determine the champion. When $T_i$ plays $T_j$, the probability that $T_i$ wins is $\frac{i}{i+j}$, and the outcom... | 781 | There are two scenarios in which $T_4$ wins. The first scenario is where $T_4$ beats $T_1$, $T_3$ beats $T_2$, and $T_4$ beats $T_3$, and the second scenario is where $T_4$ beats $T_1$, $T_2$ beats $T_3$, and $T_4$ beats $T_2$. Consider the first scenario. The probability $T_4$ beats $T_1$ is $\frac{4}{4+1}$, the proba... | 6 | [
7,
6,
6,
5,
6,
6,
7,
5
] |
A triangle has vertices $A(0,0)$, $B(12,0)$, and $C(8,10)$. The probability that a randomly chosen point inside the triangle is closer to vertex $B$ than to either vertex $A$ or vertex $C$ can be written as $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. | 409 | Calculate the area of the triangle using the Shoelace Theorem on $(0,0), (12,0), (8,10)$ \[\frac{1}{2}|(0+120+0)-(0+0+0)|=60\]
Get the four points $(6,0), (6,\frac{17}{5}), (10,5)$, and $(12,0)$ by any method from the above solutions. Then use the Shoelace Theorem to find the area of the region we want: \[\frac{1}{2}|... | 5.875 | [
6,
5,
6,
6,
6,
6,
6,
6
] |
Find the number of positive integers less than or equal to $2017$ whose base-three representation contains no digit equal to $0$. | 222 | Note that $2017=2202201_{3}$, and $2187=3^7=10000000_{3}$. There can be a $1,2,...,7$ digit number less than $2187$, and each digit can either be $1$ or $2$. So $2^1$ one digit numbers and so on up to $2^7$ $7$ digit.
Now we have to subtract out numbers from $2018$ to $2187$
Then either the number must begin $221...$... | 5 | [
5,
4,
4,
5,
6,
5,
6,
5
] |
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