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 |
|---|---|---|---|---|
A circle of radius 1 is randomly placed in a 15-by-36 rectangle $ABCD$ so that the circle lies completely within the rectangle. Given that the probability that the circle will not touch diagonal $AC$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$ | 817 | Let the bisector of $\angle CAD$ be $AE$, with $E$ on $CD$. By the angle bisector theorem, $DE = 36/5$. Since $\triangle AOR \sim \triangle AED$ ($O$ is the center of the circle), we find that $AR = 5$ since $OR = 1$. Also $AT = 35$ so $RT = OQ = 30$.
We can apply the same principle again to find that $PT = 27/2$, and... | 5.875 | [
6,
7,
5,
6,
6,
6,
6,
5
] |
A solid in the shape of a right circular cone is 4 inches tall and its base has a 3-inch radius. The entire surface of the cone, including its base, is painted. A plane parallel to the base of the cone divides the cone into two solids, a smaller cone-shaped solid $C$ and a frustum-shaped solid $F,$ in such a way that t... | 512 | Our original solid $V$ has surface area $A_v = \pi r^2 + \pi r \ell$, where $\ell$ is the slant height of the cone. Using the Pythagorean Theorem or Pythagorean Triple knowledge, we obtain $\ell = 5$ and lateral area $A_\ell = 15\pi$. The area of the base is $A_B = 3^2\pi = 9\pi$.
$V$ and $C$ are similar cones, because... | 6.5 | [
7,
6,
6,
7,
6,
7,
6,
7
] |
Let $S$ be the set of ordered pairs $(x, y)$ such that $0 < x \le 1, 0<y\le 1,$ and $\left[\log_2{\left(\frac 1x\right)}\right]$ and $\left[\log_5{\left(\frac 1y\right)}\right]$ are both even. Given that the area of the graph of $S$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$ The not... | 14 | $\left\lfloor\log_2\left(\frac{1}{x}\right)\right\rfloor$ is even when
\[x \in \left(\frac{1}{2},1\right) \cup \left(\frac{1}{8},\frac{1}{4}\right) \cup \left(\frac{1}{32},\frac{1}{16}\right) \cup \cdots\]
Likewise: $\left\lfloor\log_5\left(\frac{1}{y}\right)\right\rfloor$ is even when
\[y \in \left(\frac{1}{5},1\righ... | 6 | [
6,
6,
6,
6,
6,
6,
6,
6
] |
The polynomial $P(x)=(1+x+x^2+\cdots+x^{17})^2-x^{17}$ has $34$ complex roots of the form $z_k = r_k[\cos(2\pi a_k)+i\sin(2\pi a_k)], k=1, 2, 3,\ldots, 34,$ with $0 < a_1 \le a_2 \le a_3 \le \cdots \le a_{34} < 1$ and $r_k>0.$ Given that $a_1 + a_2 + a_3 + a_4 + a_5 = m/n,$ where $m$ and $n$ are relatively prime positi... | 482 | We see that the expression for the polynomial $P$ is very difficult to work with directly, but there is one obvious transformation to make: sum the geometric series:
\begin{align*} P(x) &= \left(\frac{x^{18} - 1}{x - 1}\right)^2 - x^{17} = \frac{x^{36} - 2x^{18} + 1}{x^2 - 2x + 1} - x^{17}\\ &= \frac{x^{36} - x^{19} - ... | 7 | [
7,
7,
8,
7,
7,
7,
7,
6
] |
A unicorn is tethered by a $20$-foot silver rope to the base of a magician's cylindrical tower whose radius is $8$ feet. The rope is attached to the tower at ground level and to the unicorn at a height of $4$ feet. The unicorn has pulled the rope taut, the end of the rope is $4$ feet from the nearest point on the tower... | 813 | Note that by Power of a Point, the point the unicorn is at has power $4 \cdot 20 = 80$ which implies that the tangent from that point to the tower is of length $\sqrt{80}=4\sqrt{5},$ however this is length of the rope projected into 2-D. If we let $\theta$ be the angle between the horizontal and the rope, we have that ... | 7.25 | [
7,
7,
8,
8,
7,
7,
6,
8
] |
For all positive integers $x$, let \[f(x)=\begin{cases}1 & \text{if }x = 1\\ \frac x{10} & \text{if }x\text{ is divisible by 10}\\ x+1 & \text{otherwise}\end{cases}\] and define a sequence as follows: $x_1=x$ and $x_{n+1}=f(x_n)$ for all positive integers $n$. Let $d(x)$ be the smallest $n$ such that $x_n=1$. (For exam... | 511 | We backcount the number of ways. Namely, we start at $x_{20} = 1$, which can only be reached if $x_{19} = 10$, and then we perform $18$ operations that either consist of $A: (-1)$ or $B: (\times 10)$. We represent these operations in a string format, starting with the operation that sends $f(x_{18}) = x_{19}$ and so fo... | 7 | [
7,
7,
7,
7,
7,
7,
8,
6
] |
A chord of a circle is perpendicular to a radius at the midpoint of the radius. The ratio of the area of the larger of the two regions into which the chord divides the circle to the smaller can be expressed in the form $\frac{a\pi+b\sqrt{c}}{d\pi-e\sqrt{f}},$ where $a, b, c, d, e,$ and $f$ are positive integers, $a$ an... | 592 | Let $r$ be the length of the radius of the circle. A right triangle is formed by half of the chord, half of the radius (since the chord bisects it), and the radius. Thus, it is a $30^\circ$ - $60^\circ$ - $90^\circ$ triangle, and the area of two such triangles is $2 \cdot \frac{1}{2} \cdot \frac{r}{2} \cdot \frac{r\sqr... | 6.875 | [
7,
7,
7,
6,
7,
7,
7,
7
] |
A jar has $10$ red candies and $10$ blue candies. Terry picks two candies at random, then Mary picks two of the remaining candies at random. Given that the probability that they get the same color combination, irrespective of order, is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$ | 441 | The probability that Terry picks two red candies is $\frac{10 \cdot 9}{20 \cdot 19} = \frac{9}{38}$, and the probability that Mary picks two red candies after Terry chooses two red candies is $\frac{7\cdot8}{18\cdot17} = \frac{28}{153}$. So the probability that they both pick two red candies is $\frac{9}{38} \cdot \fra... | 5.75 | [
6,
6,
6,
6,
6,
5,
6,
5
] |
A solid rectangular block is formed by gluing together $N$ congruent 1-cm cubes face to face. When the block is viewed so that three of its faces are visible, exactly $231$ of the 1-cm cubes cannot be seen. Find the smallest possible value of $N.$ | 384 | The $231$ cubes which are not visible must lie below exactly one layer of cubes. Thus, they form a rectangular solid which is one unit shorter in each dimension. If the original block has dimensions $l \times m \times n$, we must have $(l - 1)\times(m-1) \times(n - 1) = 231$. The prime factorization of $231 = 3\cdot7\c... | 5.5 | [
6,
6,
5,
5,
5,
6,
6,
5
] |
How many positive integers less than 10,000 have at most two different digits? | 927 | We use casework on the number of digits for this problem.
If the number has a single digit, namely the number $n \in [1,9],$ we can clearly all such $n$ work.
If the number has two digits, or the number $n \in [10,99]$ we can clearly see all such $n$ work.
If the number $n$ has three digits, there are a total of $90... | 4.125 | [
4,
4,
5,
4,
4,
4,
4,
4
] |
In order to complete a large job, $1000$ workers were hired, just enough to complete the job on schedule. All the workers stayed on the job while the first quarter of the work was done, so the first quarter of the work was completed on schedule. Then $100$ workers were laid off, so the second quarter of the work was co... | 766 | Suppose $1000$ workers can complete one quarter of the job in one day. After the first day, there were $900$ workers remaining so the second quarter was completed in $\frac{10}{9}$ days. Now there are only $800$ workers remaining so the third quarter can be completed in $\frac{10}{8}$ days. It has been $1+\frac{10}{9}+... | 6.25 | [
6,
6,
6,
7,
6,
6,
7,
6
] |
Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third m... | 408 | Let $A,B,C$ be the fraction of bananas taken by the first, second, and third monkeys respectively. Then we have the system of equations \[\frac{3}{4}A+\frac{3}{8}B+\frac{11}{24}C=\frac{1}{2}\] \[\frac{1}{8}A+\frac{1}{4}B+\frac{11}{24}C=\frac{1}{3}\] \[\frac{1}{8}A+\frac{3}{8}B+\frac{2}{24}C=\frac{1}{6}.\] Solve this yo... | 6 | [
6,
7,
6,
6,
5,
6,
6,
6
] |
$ABCD$ is a rectangular sheet of paper that has been folded so that corner $B$ is matched with point $B'$ on edge $AD.$ The crease is $EF,$ where $E$ is on $AB$ and $F$ is on $CD.$ The dimensions $AE=8, BE=17,$ and $CF=3$ are given. The perimeter of rectangle $ABCD$ is $m/n,$ where $m$ and $n$ are relatively prime posi... | 293 | Use the prepared diagram for this solution.
Call the intersection of DF and B'C' G. AB'E is an 8-15-17 right triangle, and so are B'DG and C'FG. Since C'F is 3, then using the properties of similar triangles GF is 51/8. DF is 22, so DG is 125/8. Finally, DB can to calculated to be 25/3. Add all the sides together to g... | 5.625 | [
5,
6,
5,
6,
6,
5,
5,
7
] |
How many positive integer divisors of $2004^{2004}$ are divisible by exactly 2004 positive integers? | 54 | Clearly we need to find a group of numbers that multiply to 2004. We can list them all out since we know that 2004 is only $167 * 2^2 * 3$.
167, 2, 2, 3
4, 3, 167
12, 167
4, 501
2, 1002
2, 3, 334
2, 2, 501*
6, 2, 167
3, 668
6, 334
2004*
To begin, the first multiple doesn't work because there are only 3 prim... | 6 | [
5,
6,
6,
6,
6,
6,
6,
7
] |
A sequence of positive integers with $a_1=1$ and $a_9+a_{10}=646$ is formed so that the first three terms are in geometric progression, the second, third, and fourth terms are in arithmetic progression, and, in general, for all $n\ge1,$ the terms $a_{2n-1}, a_{2n}, a_{2n+1}$ are in geometric progression, and the terms ... | 973 | Let $x = a_2$. It is apparent that the sequence grows relatively fast, so we start trying positive integers to see what $x$ can be. Finding that $x = 5$ works, after bashing out the rest of the terms we find that $a_{16} = 957$ and $a_{17} = 1089$, hence our answer is $957 + 16 = \boxed{973}$. | 6.875 | [
6,
7,
7,
7,
7,
7,
7,
7
] |
Let $S$ be the set of integers between $1$ and $2^{40}$ whose binary expansions have exactly two $1$'s. If a number is chosen at random from $S,$ the probability that it is divisible by $9$ is $p/q,$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$ | 913 | A positive integer $n$ has exactly two 1s in its binary representation exactly when $n = 2^j + 2^k$ for $j \neq k$ nonnegative integers. Thus, the set $S$ is equal to the set $\{n \in \mathbb{Z} \mid n = 2^j + 2^k \,\mathrm{ and }\, 0 \leq j < k \leq 39\}$. (The second condition ensures simultaneously that $j \neq k$ a... | 6.75 | [
7,
7,
7,
7,
7,
6,
7,
6
] |
A right circular cone has a base with radius $600$ and height $200\sqrt{7}.$ A fly starts at a point on the surface of the cone whose distance from the vertex of the cone is $125$, and crawls along the surface of the cone to a point on the exact opposite side of the cone whose distance from the vertex is $375\sqrt{2}.$... | 625 | The easiest way is to unwrap the cone into a circular sector. Center the sector at the origin with one radius on the positive $x$-axis and the angle $\theta$ going counterclockwise. The circumference of the base is $C=1200\pi$. The sector's radius (cone's sweep) is $R=\sqrt{r^2+h^2}=\sqrt{600^2+(200\sqrt{7})^2}=\sqrt{3... | 7.375 | [
8,
7,
7,
7,
7,
7,
8,
8
] |
Let $ABCD$ be an isosceles trapezoid, whose dimensions are $AB = 6, BC=5=DA,$and $CD=4.$ Draw circles of radius 3 centered at $A$ and $B,$ and circles of radius 2 centered at $C$ and $D.$ A circle contained within the trapezoid is tangent to all four of these circles. Its radius is $\frac{-k+m\sqrt{n}}p,$ where $k, m, ... | 134 | Let the radius of the center circle be $r$ and its center be denoted as $O$.
[asy] pointpen = black; pathpen = black+linewidth(0.7); pen d = linewidth(0.7) + linetype("4 4"); pen f = fontsize(8); real r = (-60 + 48 * 3^.5)/23; pair A=(0,0), B=(6,0), D=(1, 24^.5), C=(5,D.y), O = (3,(r^2 + 6*r)^.5); D(MP("A",A)--MP("B",B... | 6.75 | [
7,
7,
7,
6,
7,
7,
6,
7
] |
Let $ABCDE$ be a convex pentagon with $AB \parallel CE, BC \parallel AD, AC \parallel DE, \angle ABC=120^\circ, AB=3, BC=5,$ and $DE = 15.$ Given that the ratio between the area of triangle $ABC$ and the area of triangle $EBD$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$ | 484 | Let the intersection of $\overline{AD}$ and $\overline{CE}$ be $F$. Since $AB \parallel CE, BC \parallel AD,$ it follows that $ABCF$ is a parallelogram, and so $\triangle ABC \cong \triangle CFA$. Also, as $AC \parallel DE$, it follows that $\triangle ABC \sim \triangle EFD$.
[asy] pointpen = black; pathpen = black+lin... | 6.75 | [
7,
7,
7,
6,
6,
7,
7,
7
] |
Consider a string of $n$ $7$'s, $7777\cdots77,$ into which $+$ signs are inserted to produce an arithmetic expression. For example, $7+77+777+7+7=875$ could be obtained from eight $7$'s in this way. For how many values of $n$ is it possible to insert $+$ signs so that the resulting expression has value $7000$? | 108 | To simplify, replace all the $7$’s with $1$’s. Because the sum is congruent to $n \pmod 9$ and \[1000 \equiv 1 \pmod 9 \implies n \equiv 1 \pmod 9\] Also, $n \leq 1000$. There are \[\left \lfloor \frac{1000}{9} \right \rfloor + 1 = 112 \textrm{ positive integers that satisfy both conditions i.e. } \{1, 10, 19, 28, 37, ... | 7.25 | [
7,
7,
7,
7,
7,
8,
8,
7
] |
A long thin strip of paper is $1024$ units in length, $1$ unit in width, and is divided into $1024$ unit squares. The paper is folded in half repeatedly. For the first fold, the right end of the paper is folded over to coincide with and lie on top of the left end. The result is a $512$ by $1$ strip of double thickness.... | 593 | We can keep track of the position of the square labeled 942 in each step. We use an $(x,y)$ coordinate system, so originally the 942 square is in the position $(942,1)$. In general, suppose that we've folded the strip into an array $r=2^k$ squares wide and $c=1024/r=2^{10-k}$ squares tall (so we've made $10-k$ folds). ... | 6.875 | [
7,
8,
7,
7,
6,
7,
6,
7
] |
Six congruent circles form a ring with each circle externally tangent to two circles adjacent to it. All circles are internally tangent to a circle $C$ with radius 30. Let $K$ be the area of the region inside circle $C$ and outside of the six circles in the ring. Find $\lfloor K \rfloor$ (the floor function). | 942 | Define the radii of the six congruent circles as $r$. If we draw all of the radii to the points of external tangency, we get a regular hexagon. If we connect the vertices of the hexagon to the center of the circle $C$, we form several equilateral triangles. The length of each side of the triangle is $2r$. Notice that t... | 5.875 | [
6,
6,
6,
6,
6,
5,
6,
6
] |
For each positive integer $k$, let $S_k$ denote the increasing arithmetic sequence of integers whose first term is $1$ and whose common difference is $k$. For example, $S_3$ is the sequence $1,4,7,10,\ldots.$ For how many values of $k$ does $S_k$ contain the term $2005$? | 12 | Suppose that the $n$th term of the sequence $S_k$ is $2005$. Then $1+(n-1)k=2005$ so $k(n-1)=2004=2^2\cdot 3\cdot 167$. The ordered pairs $(k,n-1)$ of positive integers that satisfy the last equation are $(1,2004)$,$(2,1002)$, $(3,668)$, $(4,501)$, $(6,334)$, $(12,167)$, $(167,12)$,$(334,6)$, $(501,4)$, $(668,3)$, $(10... | 3.5 | [
4,
3,
4,
4,
3,
3,
3,
4
] |
How many positive integers have exactly three proper divisors (positive integral divisors excluding itself), each of which is less than 50? | 109 | Suppose $n$ is such an integer. Because $n$ has $3$ proper divisors, it must have $4$ divisors,, so $n$ must be in the form $n=p\cdot q$ or $n=p^3$ for distinct prime numbers $p$ and $q$.
In the first case, the three proper divisors of $n$ are $1$, $p$ and $q$. Thus, we need to pick two prime numbers less than $50$. T... | 5.375 | [
6,
5,
5,
5,
5,
6,
5,
6
] |
The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no ... | 294 | Let there be $m$ members and $n$ members for the square and $c$ for the number of columns of the other formation. We have $n^2 +5 = c(c+7) \implies n^2+5 = \left(c+\frac{7}{2}\right)^2 -\frac{49}{4} \implies n^2 - \left(c+\frac{7}{2}\right)^2 = -\frac{69}{4} \implies \left(n-c-\frac{7}{2}\right)\left(n + c +\frac{7}{2}... | 4.25 | [
4,
5,
4,
5,
4,
4,
4,
4
] |
Robert has 4 indistinguishable gold coins and 4 indistinguishable silver coins. Each coin has an engraving of one face on one side, but not on the other. He wants to stack the eight coins on a table into a single stack so that no two adjacent coins are face to face. Find the number of possible distinguishable arrangeme... | 630 | There are two separate parts to this problem: one is the color (gold vs silver), and the other is the orientation.
There are ${8\choose4} = 70$ ways to position the gold coins in the stack of 8 coins, which determines the positions of the silver coins.
Create a string of letters H and T to denote the orientation of t... | 4.125 | [
4,
4,
5,
4,
4,
4,
4,
4
] |
Let $P$ be the product of the nonreal roots of $x^4-4x^3+6x^2-4x=2005.$ Find $\lfloor P\rfloor.$ | 45 | If we don't see the fourth power, we can always factor the LHS to try to create a quadratic substitution. Checking, we find that $x=0$ and $x=2$ are both roots. Synthetic division gives $(x^2-2x)(x^2-2x+2)=2005$. We now have our quadratic substitution of $y=x^2-2x+1=(x-1)^2$, giving us $(y-1)(y+1)=2005$. From here we p... | 5.5 | [
6,
6,
5,
5,
5,
6,
6,
5
] |
In quadrilateral $ABCD,\ BC=8,\ CD=12,\ AD=10,$ and $m\angle A= m\angle B = 60^\circ.$ Given that $AB = p + \sqrt{q},$ where $p$ and $q$ are positive integers, find $p+q.$ | 150 | Draw the perpendiculars from $C$ and $D$ to $AB$, labeling the intersection points as $E$ and $F$. This forms 2 $30-60-90$ right triangles, so $AE = 5$ and $BF = 4$. Also, if we draw the horizontal line extending from $C$ to a point $G$ on the line $DE$, we find another right triangle $\triangle DGC$. $DG = DE - CF = 5... | 5.625 | [
6,
5,
6,
5,
6,
5,
6,
6
] |
The equation $2^{333x-2} + 2^{111x+2} = 2^{222x+1} + 1$ has three real roots. Given that their sum is $m/n$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$ | 113 | Let $y = 2^{111x}$. Then our equation reads $\frac{1}{4}y^3 + 4y = 2y^2 + 1$ or $y^3 - 8y^2 + 16y - 4 = 0$. Thus, if this equation has roots $r_1, r_2$ and $r_3$, by Vieta's formulas we have $r_1\cdot r_2\cdot r_3 = 4$. Let the corresponding values of $x$ be $x_1, x_2$ and $x_3$. Then the previous statement says that $... | 6 | [
6,
7,
6,
6,
5,
6,
6,
6
] |
Twenty seven unit cubes are painted orange on a set of four faces so that two non-painted faces share an edge. The 27 cubes are randomly arranged to form a $3\times 3 \times 3$ cube. Given the probability of the entire surface area of the larger cube is orange is $\frac{p^a}{q^br^c},$ where $p,q,$ and $r$ are distinct ... | 74 | We can consider the orientation of each of the individual cubes independent=== The unit cube at the center of our large cube has no exterior faces, so all of its orientations work.
For the six unit cubes and the centers of the faces of the large cube, we need that they show an orange face. This happens in $\frac{4}{6}... | 7.625 | [
8,
8,
8,
8,
7,
8,
8,
6
] |
Triangle $ABC$ lies in the cartesian plane and has an area of $70$. The coordinates of $B$ and $C$ are $(12,19)$ and $(23,20),$ respectively, and the coordinates of $A$ are $(p,q).$ The line containing the median to side $BC$ has slope $-5.$ Find the largest possible value of $p+q.$
[asy]defaultpen(fontsize(8)); size(1... | 47 | Using the equation of the median from above, we can write the coordinates of $A$ as $(p,\ -5p + 107)$. The equation of $\overline{BC}$ is $\frac{20 - 19}{23 - 12} = \frac{y - 19}{x - 12}$, so $x - 12 = 11y - 209$. In general form, the line is $x - 11y + 197 = 0$. Use the equation for the distance between a line and poi... | 6 | [
6,
6,
6,
6,
6,
6,
6,
6
] |
A semicircle with diameter $d$ is contained in a square whose sides have length 8. Given the maximum value of $d$ is $m - \sqrt{n},$ find $m+n.$ | 544 | It is easy after getting the image, after drawing labeling the lengths of those segments, assume the radius is $x$, we can see $x=\sqrt{2}(8-x)$ and we get $2x=32-\sqrt{512}$ and we have the answer $\boxed{544}$ ~bluesoul | 5.75 | [
5,
7,
6,
6,
6,
5,
6,
5
] |
For positive integers $n,$ let $\tau (n)$ denote the number of positive integer divisors of $n,$ including 1 and $n.$ For example, $\tau (1)=1$ and $\tau(6) =4.$ Define $S(n)$ by $S(n)=\tau(1)+ \tau(2) + \cdots + \tau(n).$ Let $a$ denote the number of positive integers $n \leq 2005$ with $S(n)$ odd, and let $b$ denote ... | 25 | Let $\Delta n$ denote the sum $1+2+3+ \dots +n-1+n$. We can easily see from the fact "It is well-known that $\tau(n)$ is odd if and only if $n$ is a perfect square.", that
$a = (2^2-1^2) + (4^2-3^2) \dots (44^2 - 43^2) = (2+1)(2-1)+(4+3)(4-3) \dots (44+43)(44-43) = 1+2+3...44 = \Delta 44$.
$b = 3^2-2^2+5^2-4^2...2006-4... | 6.375 | [
6,
7,
6,
7,
6,
7,
6,
6
] |
A particle moves in the Cartesian plane according to the following rules:
From any lattice point $(a,b),$ the particle may only move to $(a+1,b), (a,b+1),$ or $(a+1,b+1).$
There are no right angle turns in the particle's path.
How many different paths can the particle take from $(0,0)$ to $(5,5)$? | 83 | The length of the path (the number of times the particle moves) can range from $l = 5$ to $9$; notice that $d = 10-l$ gives the number of diagonals. Let $R$ represent a move to the right, $U$ represent a move upwards, and $D$ to be a move that is diagonal. Casework upon the number of diagonal moves:
Case $d = 1$: It is... | 6.25 | [
7,
6,
6,
6,
7,
7,
6,
5
] |
Consider the points $A(0,12), B(10,9), C(8,0),$ and $D(-4,7).$ There is a unique square $S$ such that each of the four points is on a different side of $S.$ Let $K$ be the area of $S.$ Find the remainder when $10K$ is divided by $1000$. | 936 | Let $(a,b)$ denote a normal vector of the side containing $A$. Note that $\overline{AC}, \overline{BD}$ intersect and hence must be opposite vertices of the square. The lines containing the sides of the square have the form $ax+by=12b$, $ax+by=8a$, $bx-ay=10b-9a$, and $bx-ay=-4b-7a$. The lines form a square, so the dis... | 6.625 | [
6,
7,
7,
7,
6,
7,
7,
6
] |
Triangle $ABC$ has $BC=20.$ The incircle of the triangle evenly trisects the median $AD.$ If the area of the triangle is $m \sqrt{n}$ where $m$ and $n$ are integers and $n$ is not divisible by the square of a prime, find $m+n.$ | 38 | WLOG let E be be between C & D (as in solution 1). Assume $AD = 3m$. We use power of a point to get that $AG = DE = \sqrt{2}m$ and $AB = AG + GB = AG + BE = 10+2\sqrt{2} m$
Since now we have $AC = 10$, $BC = 20, AB = 10+2\sqrt{2} m$ in triangle $\triangle ABC$ and cevian $AD = 3m$. Now, we can apply Stewart's Theorem.... | 6.5 | [
7,
7,
7,
6,
6,
7,
6,
6
] |
A game uses a deck of $n$ different cards, where $n$ is an integer and $n \geq 6.$ The number of possible sets of 6 cards that can be drawn from the deck is 6 times the number of possible sets of 3 cards that can be drawn. Find $n.$
~ pi_is_3.14 | 13 | The number of ways to draw six cards from $n$ is given by the binomial coefficient ${n \choose 6} = \frac{n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot(n-4)\cdot(n-5)}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$.
The number of ways to choose three cards from $n$ is ${n\choose 3} = \frac{n\cdot(n-1)\cdot(n-2)}{3\cdot2\cdot1}$.
We are gi... | 4.125 | [
4,
5,
4,
4,
4,
4,
4,
4
] |
A hotel packed breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls and once wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each ... | 79 | Call the three different types of rolls as A, B, and C. We need to arrange 3As, 3Bs, and 3Cs in a string such that A, B, and C appear in the first three, second three, and the third three like ABCABCABC or BCABACCAB. This can occur in $\left(\frac{3!}{1!1!1!}\right)^3 = 6^3 = 216$ different manners. The total number of... | 4.875 | [
5,
6,
4,
5,
5,
4,
5,
5
] |
An infinite geometric series has sum 2005. A new series, obtained by squaring each term of the original series, has 10 times the sum of the original series. The common ratio of the original series is $\frac mn$ where $m$ and $n$ are relatively prime integers. Find $m+n.$ | 802 | Let's call the first term of the original geometric series $a$ and the common ratio $r$, so $2005 = a + ar + ar^2 + \ldots$. Using the sum formula for infinite geometric series, we have $\;\;\frac a{1 -r} = 2005$. Then we form a new series, $a^2 + a^2 r^2 + a^2 r^4 + \ldots$. We know this series has sum $20050 = \frac{... | 6 | [
5,
6,
6,
6,
7,
6,
6,
6
] |
Find the number of positive integers that are divisors of at least one of $10^{10},15^7,18^{11}.$ | 435 | $10^{10} = 2^{10}\cdot 5^{10}$ so $10^{10}$ has $11\cdot11 = 121$ divisors.
$15^7 = 3^7\cdot5^7$ so $15^7$ has $8\cdot8 = 64$ divisors.
$18^{11} = 2^{11}\cdot3^{22}$ so $18^{11}$ has $12\cdot23 = 276$ divisors.
Now, we use the Principle of Inclusion-Exclusion. We have $121 + 64 + 276$ total potential divisors so far, ... | 5.625 | [
5,
6,
5,
6,
5,
6,
6,
6
] |
Determine the number of ordered pairs $(a,b)$ of integers such that $\log_a b + 6\log_b a=5, 2 \leq a \leq 2005,$ and $2 \leq b \leq 2005.$ | 54 | Let $k=\log_a b$. Then our equation becomes $k+\frac{6}{k}=5$. Multiplying through by $k$ and solving the quadratic gives us $k=2$ or $k=3$. Hence $a^2=b$ or $a^3=b$.
For the first case $a^2=b$, $a$ can range from 2 to 44, a total of 43 values. For the second case $a^3=b$, $a$ can range from 2 to 12, a total of 11 val... | 4.5 | [
5,
5,
4,
4,
5,
4,
4,
5
] |
The cards in a stack of $2n$ cards are numbered consecutively from 1 through $2n$ from top to bottom. The top $n$ cards are removed, kept in order, and form pile $A.$ The remaining cards form pile $B.$ The cards are then restacked by taking cards alternately from the tops of pile $B$ and $A,$ respectively. In this proc... | 392 | If you index the final stack $1,2,\dots,2n$, you notice that pile A resides only in the odd indices and has maintained its original order aside from flipping over. The same has happened to pile B except replace odd with even. Thus, if 131 is still at index 131, an odd number, then 131 must be from pile A. The numbers i... | 5.875 | [
5,
6,
6,
6,
6,
5,
7,
6
] |
Let $x=\frac{4}{(\sqrt{5}+1)(\sqrt[4]{5}+1)(\sqrt[8]{5}+1)(\sqrt[16]{5}+1)}.$ Find $(x+1)^{48}$. | 125 | Like Solution $2$, let $z=\sqrt[16]{5}$ Then, the expression becomes
$x=\frac{4}{(z+1)(z^2+1)(z^4+1)(z^8+1)}$ Now, multiplying by the conjugate of each binomial in the denominator, we obtain...
$x=\frac{4(z-1)(z^2-1)(z^4-1)(z^8-1)}{(z^2-1)(z^4-1)(z^8-1)(z^{16}-1)}\implies x=\frac{4(z-1)}{z^{16}-1}$ Plugging back in,
$x... | 6.75 | [
7,
6,
7,
7,
6,
7,
7,
7
] |
Circles $C_1$ and $C_2$ are externally tangent, and they are both internally tangent to circle $C_3.$ The radii of $C_1$ and $C_2$ are 4 and 10, respectively, and the centers of the three circles are all collinear. A chord of $C_3$ is also a common external tangent of $C_1$ and $C_2.$ Given that the length of the chord... | 405 | Call our desired length $x$. Note for any $X$ on $\overline{AB}$ and $Y$ on $\overline{O_1O_2}$ such that $\overline{XY}\perp\overline{AB}$ that the function $f$ such that $f(\overline{O_1Y})=\overline{XY}$ is linear. Since $(0,4)$ and $(14,10)$, we can quickly interpolate that $f(10)=\overline{O_3T}=\frac{58}{7}$. The... | 7.375 | [
8,
8,
7,
8,
7,
7,
7,
7
] |
For how many positive integers $n$ less than or equal to $1000$ is $(\sin t + i \cos t)^n = \sin nt + i \cos nt$ true for all real $t$? | 250 | Let $t=0$. Then, we have $i^n=i$ which means $n\equiv 1\pmod{4}$. Thus, the answer is $\boxed{250}$. | 4.375 | [
4,
4,
5,
3,
5,
5,
5,
4
] |
Given that $O$ is a regular octahedron, that $C$ is the cube whose vertices are the centers of the faces of $O,$ and that the ratio of the volume of $O$ to that of $C$ is $\frac mn,$ where $m$ and $n$ are relatively prime integers, find $m+n.$ | 11 | Let the octahedron have vertices $(\pm 3, 0, 0), (0, \pm 3, 0), (0, 0, \pm 3)$. Then the vertices of the cube lie at the centroids of the faces, which have coordinates $(\pm 1, \pm 1, \pm 1)$. The cube has volume 8. The region of the octahedron lying in each octant is a tetrahedron with three edges mutually perpendicul... | 6.5 | [
6,
6,
7,
7,
7,
6,
7,
6
] |
Let $m$ be a positive integer, and let $a_0, a_1,\ldots,a_m$ be a sequence of reals such that $a_0 = 37, a_1 = 72, a_m = 0,$ and $a_{k+1} = a_{k-1} - \frac 3{a_k}$ for $k = 1,2,\ldots, m-1.$ Find $m.$ | 889 | For $0 < k < m$, we have
$a_{k}a_{k+1} = a_{k-1}a_{k} - 3$.
Thus the product $a_{k}a_{k+1}$ is a monovariant: it decreases by 3 each time $k$ increases by 1. For $k = 0$ we have $a_{k}a_{k+1} = 37\cdot 72$, so when $k = \frac{37 \cdot 72}{3} = 888$, $a_{k}a_{k+1}$ will be zero for the first time, which implies that $m... | 6 | [
6,
6,
6,
6,
6,
6,
6,
6
] |
Square $ABCD$ has center $O,\ AB=900,\ E$ and $F$ are on $AB$ with $AE<BF$ and $E$ between $A$ and $F, m\angle EOF =45^\circ,$ and $EF=400.$ Given that $BF=p+q\sqrt{r},$ where $p,q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime, find $p+q+r.$ | 307 | Draw AO, OB, and extend OB to D. Let $\angle{FOB} = \alpha.$ Then, after angle chasing, we find that \[\angle{AEB} = 90 + \alpha\]. Using this, we draw a line perpendicular to $AB$ at $E$ to meet $BD$ at $M$. Since $\angle{MEO} = \alpha$ and $\angle{EMO} = 45$, we have that \[\triangle{EMO} \sim \triangle{OBF}\] Let $F... | 6.875 | [
7,
7,
7,
6,
7,
7,
7,
7
] |
Let $P(x)$ be a polynomial with integer coefficients that satisfies $P(17)=10$ and $P(24)=17.$ Given that $P(n)=n+3$ has two distinct integer solutions $n_1$ and $n_2,$ find the product $n_1\cdot n_2.$ | 418 | We define $Q(x)=P(x)-x+7$, noting that it has roots at $17$ and $24$. Hence $P(x)-x+7=A(x-17)(x-24)$. In particular, this means that $P(x)-x-3=A(x-17)(x-24)-10$. Therefore, $x=n_1,n_2$ satisfy $A(x-17)(x-24)=10$, where $A$, $(x-17)$, and $(x-24)$ are integers. This cannot occur if $x\le 17$ or $x\ge 24$ because the pro... | 6 | [
6,
6,
6,
6,
6,
6,
6,
6
] |
In triangle $ABC, AB=13, BC=15,$ and $CA = 14.$ Point $D$ is on $\overline{BC}$ with $CD=6.$ Point $E$ is on $\overline{BC}$ such that $\angle BAE\cong \angle CAD.$ Given that $BE=\frac pq$ where $p$ and $q$ are relatively prime positive integers, find $q.$ | 463 | Let $ED = x$, such that $BE = 9-x$. Since $\overline{AE}$ and $\overline{AD}$ are isogonal, we get $\frac{9-x}{6+x} \cdot \frac{9}{6} = \frac{13^2}{14^2} \Rightarrow 588(9 - x) = 338(6 + x)$, and we can solve to get $x = \frac{1632}{463}$(and $BE = \frac{1146}{463}$). Hence, our answer is $\boxed{463}$. - Spacesam | 5.75 | [
6,
6,
6,
6,
5,
5,
6,
6
] |
Let $w_1$ and $w_2$ denote the circles $x^2+y^2+10x-24y-87=0$ and $x^2 +y^2-10x-24y+153=0,$ respectively. Let $m$ be the smallest positive value of $a$ for which the line $y=ax$ contains the center of a circle that is externally tangent to $w_2$ and internally tangent to $w_1.$ Given that $m^2=\frac pq,$ where $p$ and ... | 169 | We use the same reflection as in Solution 2. As $OF_1'=OF_2=13$, we know that $\triangle OF_1'F_2$ is isosceles. Hence $\angle F_2F_1'O=\angle F_1'F_2O$. But by symmetry, we also know that $\angle OF_1T=\angle F_2F_1'O$. Hence $\angle OF_1T=\angle F_1'F_2O$. In particular, as $\angle OF_1T=\angle OF_2T$, this implies t... | 6.75 | [
7,
7,
7,
6,
7,
7,
6,
7
] |
In quadrilateral $ABCD$, $\angle B$ is a right angle, diagonal $\overline{AC}$ is perpendicular to $\overline{CD}$, $AB=18$, $BC=21$, and $CD=14$. Find the perimeter of $ABCD$. | 84 | From the problem statement, we construct the following diagram:
[asy] pointpen = black; pathpen = black + linewidth(0.65); pair C=(0,0), D=(0,-14),A=(-(961-196)^.5,0),B=IP(circle(C,21),circle(A,18)); D(MP("A",A,W)--MP("B",B,N)--MP("C",C,E)--MP("D",D,E)--A--C); D(rightanglemark(A,C,D,40)); D(rightanglemark(A,B,C,40)); [... | 4 | [
4,
4,
4,
3,
4,
5,
4,
4
] |
Let set $\mathcal{A}$ be a 90-element subset of $\{1,2,3,\ldots,100\},$ and let $S$ be the sum of the elements of $\mathcal{A}.$ Find the number of possible values of $S.$ | 901 | The smallest $S$ is $1+2+ \ldots +90 = 91 \cdot 45 = 4095$. The largest $S$ is $11+12+ \ldots +100=111\cdot 45=4995$. All numbers between $4095$ and $4995$ are possible values of S, so the number of possible values of S is $4995-4095+1=901$.
Alternatively, for ease of calculation, let set $\mathcal{B}$ be a 10-element... | 3.25 | [
3,
3,
3,
3,
3,
4,
3,
4
] |
Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is $\frac{1}{29}$ of the original integer. | 725 | Suppose the original number is $N = \overline{a_na_{n-1}\ldots a_1a_0},$ where the $a_i$ are digits and the first digit, $a_n,$ is nonzero. Then the number we create is $N_0 = \overline{a_{n-1}\ldots a_1a_0},$ so \[N = 29N_0.\] But $N$ is $N_0$ with the digit $a_n$ added to the left, so $N = N_0 + a_n \cdot 10^n.$ Thus... | 5.125 | [
5,
4,
5,
5,
5,
5,
6,
6
] |
Let $N$ be the number of consecutive $0$'s at the right end of the decimal representation of the product $1!2!3!4!\cdots99!100!.$ Find the remainder when $N$ is divided by $1000$. | 124 | A number in decimal notation ends in a zero for each power of ten which divides it. Thus, we need to count both the number of 5s and the number of 2s dividing into our given expression. Since there are clearly more 2s than 5s, it is sufficient to count the number of 5s.
One way to do this is as follows: $96$ of the nu... | 7.25 | [
7,
7,
7,
7,
8,
8,
7,
7
] |
The number $\sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}$ can be written as $a\sqrt{2}+b\sqrt{3}+c\sqrt{5},$ where $a, b,$ and $c$ are positive integers. Find $abc$. | 936 | We begin by equating the two expressions:
\[a\sqrt{2}+b\sqrt{3}+c\sqrt{5} = \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}\]
Squaring both sides yields:
\[2ab\sqrt{6} + 2ac\sqrt{10} + 2bc\sqrt{15} + 2a^2 + 3b^2 + 5c^2 = 104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006\]
Since $a$, $b$, and $c$ are integers, we can match... | 6.625 | [
6,
7,
6,
6,
7,
7,
7,
7
] |
Let $\mathcal{S}$ be the set of real numbers that can be represented as repeating decimals of the form $0.\overline{abc}$ where $a, b, c$ are distinct digits. Find the sum of the elements of $\mathcal{S}.$ | 360 | By symmetry, the average over all $720=(10)(9)(8)$ numbers is $\frac{1}{2}$. Then, their sum is $\frac{1}{2}(720)=\boxed{360}$. | 3.625 | [
4,
4,
3,
4,
4,
3,
4,
3
] |
An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region $C$ to the area of shaded region $B$ is 11/5. Find the ratio of shaded region $D$ to the area of shaded region $A.$
[asy] defaultpen(linewidth(0.7)+fontsize(10)); for(int i=0; i<4; i=i+1) { fill((2*i,0)--(2*i+1,... | 408 | Let the distances from the apex to the parallel lines be $x$ and $y$ and the distance between the intersections be $a,b.$ We know the area ratio means $\frac{(x+4a)(y+4b)-(x+3a)(y+3b)}{(x+2a)(y+2b)-(x+a)(y+b)} =\frac{5}{11}$ which simplifying yields $ab = 3ay+3bx.$ The ratio we seek is $\frac{(x+6a)(y+6b)-(x+5a)(y+5b)}... | 6.875 | [
7,
7,
7,
7,
7,
7,
6,
7
] |
Hexagon $ABCDEF$ is divided into five rhombuses, $\mathcal{P, Q, R, S,}$ and $\mathcal{T,}$ as shown. Rhombuses $\mathcal{P, Q, R,}$ and $\mathcal{S}$ are congruent, and each has area $\sqrt{2006}.$ Let $K$ be the area of rhombus $\mathcal{T}$. Given that $K$ is a positive integer, find the number of possible values fo... | 89 | Let $x$ denote the common side length of the rhombi. Let $y$ denote one of the smaller interior angles of rhombus $\mathcal{P}$. Then $x^2\sin(y)=\sqrt{2006}$. We also see that $K=x^2\sin(2y) \Longrightarrow K=2x^2\sin y \cdot \cos y \Longrightarrow K = 2\sqrt{2006}\cdot \cos y$. Thus $K$ can be any positive integer in... | 6.375 | [
7,
6,
7,
6,
7,
6,
6,
6
] |
The sequence $a_1, a_2, \ldots$ is geometric with $a_1=a$ and common ratio $r,$ where $a$ and $r$ are positive integers. Given that $\log_8 a_1+\log_8 a_2+\cdots+\log_8 a_{12} = 2006,$ find the number of possible ordered pairs $(a,r).$ | 46 | Using the above method, we can derive that $a^{2}r^{11} = 2^{1003}$. Now, think about what happens when r is an even power of 2. Then $a^{2}$ must be an odd power of 2 in order to satisfy the equation which is clearly not possible. Thus the only restriction r has is that it must be an odd power of 2, so $2^{1}$, $2^{3}... | 5.875 | [
6,
6,
6,
6,
6,
5,
6,
6
] |
Eight circles of diameter 1 are packed in the first quadrant of the coordinate plane as shown. Let region $\mathcal{R}$ be the union of the eight circular regions. Line $l,$ with slope 3, divides $\mathcal{R}$ into two regions of equal area. Line $l$'s equation can be expressed in the form $ax=by+c,$ where $a, b,$ and ... | 65 | The line passing through the tangency point of the bottom left circle and the one to its right and through the tangency of the top circle in the middle column and the one beneath it is the line we are looking for: a line passing through the tangency of two circles cuts congruent areas, so our line cuts through the four... | 6.25 | [
6,
5,
6,
7,
6,
6,
7,
7
] |
A collection of 8 cubes consists of one cube with edge-length $k$ for each integer $k, 1 \le k \le 8.$ A tower is to be built using all 8 cubes according to the rules:
Any cube may be the bottom cube in the tower.
The cube immediately on top of a cube with edge-length $k$ must have edge-length at most $k+2.$
Let $T$ b... | 458 | We proceed recursively. Suppose we can build $T_m$ towers using blocks of size $1, 2, \ldots, m$. How many towers can we build using blocks of size $1, 2, \ldots, m, m + 1$? If we remove the block of size $m + 1$ from such a tower (keeping all other blocks in order), we get a valid tower using blocks $1, 2, \ldots, m$.... | 5.875 | [
6,
6,
6,
6,
8,
6,
5,
4
] |
Find the sum of the values of $x$ such that $\cos^3 3x+ \cos^3 5x = 8 \cos^3 4x \cos^3 x$, where $x$ is measured in degrees and $100< x< 200.$ | 906 | Observe that $2\cos 4x\cos x = \cos 5x + \cos 3x$ by the sum-to-product formulas. Defining $a = \cos 3x$ and $b = \cos 5x$, we have $a^3 + b^3 = (a+b)^3 \rightarrow ab(a+b) = 0$. But $a+b = 2\cos 4x\cos x$, so we require $\cos x = 0$, $\cos 3x = 0$, $\cos 4x = 0$, or $\cos 5x = 0$.
Hence we see by careful analysis of ... | 6.625 | [
6,
7,
7,
7,
6,
7,
6,
7
] |
For each even positive integer $x$, let $g(x)$ denote the greatest power of 2 that divides $x.$ For example, $g(20)=4$ and $g(16)=16.$ For each positive integer $n,$ let $S_n=\sum_{k=1}^{2^{n-1}}g(2k).$ Find the greatest integer $n$ less than 1000 such that $S_n$ is a perfect square. | 899 | First note that $g(k)=1$ if $k$ is odd and $2g(k/2)$ if $k$ is even. so $S_n=\sum_{k=1}^{2^{n-1}}g(2k). = \sum_{k=1}^{2^{n-1}}2g(k) = 2\sum_{k=1}^{2^{n-1}}g(k) = 2\sum_{k=1}^{2^{n-2}} g(2k-1)+g(2k).$ $2k-1$ must be odd so this reduces to $2\sum_{k=1}^{2^{n-2}}1+g(2k) = 2(2^{n-2}+\sum_{k=1}^{2^n-2}g(2k)).$ Thus $S_n=2(2... | 7.125 | [
7,
7,
7,
7,
7,
7,
8,
7
] |
A tripod has three legs each of length $5$ feet. When the tripod is set up, the angle between any pair of legs is equal to the angle between any other pair, and the top of the tripod is $4$ feet from the ground. In setting up the tripod, the lower 1 foot of one leg breaks off. Let $h$ be the height in feet of the top o... | 183 | [asy] size(200); import three; pointpen=black;pathpen=black+linewidth(0.65);pen ddash = dashed+linewidth(0.65); currentprojection = perspective(1,-10,3.3); triple O=(0,0,0),T=(0,0,5),C=(0,3,0),A=(-3*3^.5/2,-3/2,0),B=(3*3^.5/2,-3/2,0); triple M=(B+C)/2,S=(4*A+T)/5; draw(T--S--B--T--C--B--S--C);draw(B--A--C--A--S,ddash);... | 6.75 | [
7,
7,
6,
7,
7,
7,
7,
6
] |
Given that a sequence satisfies $x_0=0$ and $|x_k|=|x_{k-1}+3|$ for all integers $k\ge 1,$ find the minimum possible value of $|x_1+x_2+\cdots+x_{2006}|.$ | 27 | Playing around with a couple numbers, we see that we can generate the sequence $0, 3, -6, 3, -6, \cdots$, and we can also generate the sequence $3, 6, 9, 12, \cdots$ after each $-6$ value. Thus, we will apply this to try and find some bounds. We can test if the first $1000$ pairs of numbers each sum up to $-3$, and the... | 6.375 | [
6,
6,
6,
6,
7,
7,
7,
6
] |
In convex hexagon $ABCDEF$, all six sides are congruent, $\angle A$ and $\angle D$ are right angles, and $\angle B, \angle C, \angle E,$ and $\angle F$ are congruent. The area of the hexagonal region is $2116(\sqrt{2}+1).$ Find $AB$. | 46 | Let the side length be called $x$, so $x=AB=BC=CD=DE=EF=AF$.
The diagonal $BF=\sqrt{AB^2+AF^2}=\sqrt{x^2+x^2}=x\sqrt{2}$. Then the areas of the triangles AFB and CDE in total are $\frac{x^2}{2}\cdot 2$, and the area of the rectangle BCEF equals $x\cdot x\sqrt{2}=x^2\sqrt{2}$
Then we have to solve the equation
$2116(\... | 5.75 | [
6,
6,
5,
6,
6,
5,
5,
7
] |
The lengths of the sides of a triangle with positive area are $\log_{10} 12$, $\log_{10} 75$, and $\log_{10} n$, where $n$ is a positive integer. Find the number of possible values for $n$. | 893 | By the Triangle Inequality and applying the well-known logarithmic property $\log_{c} a + \log_{c} b = \log_{c} ab$, we have that
$\log_{10} 12 + \log_{10} n > \log_{10} 75$
$\log_{10} 12n > \log_{10} 75$
$12n > 75$
$n > \frac{75}{12} = \frac{25}{4} = 6.25$
Also,
$\log_{10} 12 + \log_{10} 75 > \log_{10} n$
$\log_{10} ... | 4.25 | [
4,
4,
5,
5,
4,
4,
4,
4
] |
Let $P$ be the product of the first $100$ positive odd integers. Find the largest integer $k$ such that $P$ is divisible by $3^k .$ | 49 | We are obviously searching for multiples of three set S of odd numbers 1-199. Starting with 3, every number $\equiv 2 \pmod{3}$ in set S will be divisible by 3. In other words, every number $\equiv 3 \pmod{6}$. This is because the LCM must be divisible by 3, and 2, because the set is comprised of only odd numbers. Usin... | 5.875 | [
6,
5,
6,
6,
6,
6,
6,
6
] |
Let $(a_1,a_2,a_3,\ldots,a_{12})$ be a permutation of $(1,2,3,\ldots,12)$ for which
$a_1>a_2>a_3>a_4>a_5>a_6 \mathrm{\ and \ } a_6<a_7<a_8<a_9<a_{10}<a_{11}<a_{12}.$
An example of such a permutation is $(6,5,4,3,2,1,7,8,9,10,11,12).$ Find the number of such permutations. | 462 | Clearly, $a_6=1$. Now, consider selecting $5$ of the remaining $11$ values. Sort these values in descending order, and sort the other $6$ values in ascending order. Now, let the $5$ selected values be $a_1$ through $a_5$, and let the remaining $6$ be $a_7$ through ${a_{12}}$. It is now clear that there is a bijection b... | 5.625 | [
6,
6,
5,
6,
6,
5,
6,
5
] |
When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face $F$ is greater than $1/6$, the probability of obtaining the face opposite is less than $1/6$, the probability of obtaining any one of the other four faces is $1/6$, and the sum of the numbers on oppos... | 29 | We have that the cube probabilities to land on its faces are $\frac{1}{6}$, $\frac{1}{6}$, $\frac{1}{6}$, $\frac{1}{6}$ ,$\frac{1}{6}+x$ ,$\frac{1}{6}-x$ we also know that the sum could be 7 only when the faces in each of the two tosses are opposite hence the probability to get a 7 is: \[4 \cdot \left(\frac{1}{6} \righ... | 6.375 | [
7,
5,
6,
6,
7,
7,
7,
6
] |
Square $ABCD$ has sides of length 1. Points $E$ and $F$ are on $\overline{BC}$ and $\overline{CD},$ respectively, so that $\triangle AEF$ is equilateral. A square with vertex $B$ has sides that are parallel to those of $ABCD$ and a vertex on $\overline{AE}.$ The length of a side of this smaller square is $\frac{a-\sqrt... | 12 | Why not solve in terms of the side $x$ only (single-variable beauty)? By similar triangles we obtain that $BE=\frac{x}{1-x}$, therefore $CE=\frac{1-2x}{1-x}$. Then $AE=\sqrt{2}*\frac{1-2x}{1-x}$. Using Pythagorean Theorem on $\triangle{ABE}$ yields $\frac{x^2}{(1-x)^2} + 1 = 2 * \frac{(1-2x)^2}{(1-x)^2}$. This means $6... | 6.125 | [
6,
6,
5,
6,
6,
6,
7,
7
] |
Find the number of ordered pairs of positive integers $(a,b)$ such that $a+b=1000$ and neither $a$ nor $b$ has a zero digit. | 738 | There are $\left\lfloor\frac{999}{10}\right\rfloor = 99$ numbers up to 1000 that have 0 as their units digit. All of the other excluded possibilities are when $a$ or $b$ have a 0 in the tens digit, and since the equation is symmetric, we will just count when $a$ has a 0 in the tens digit and multiply by 2 (notice that ... | 3.875 | [
4,
4,
3,
4,
4,
4,
4,
4
] |
There is an unlimited supply of congruent equilateral triangles made of colored paper. Each triangle is a solid color with the same color on both sides of the paper. A large equilateral triangle is constructed from four of these paper triangles. Two large triangles are considered distinguishable if it is not possible t... | 336 | If two of our big equilateral triangles have the same color for their center triangle and the same multiset of colors for their outer three triangles, we can carry one onto the other by a combination of rotation and reflection. Thus, to make two triangles distinct, they must differ either in their center triangle or in... | 5.625 | [
5,
6,
6,
5,
6,
6,
6,
5
] |
Circles $\mathcal{C}_1, \mathcal{C}_2,$ and $\mathcal{C}_3$ have their centers at (0,0), (12,0), and (24,0), and have radii 1, 2, and 4, respectively. Line $t_1$ is a common internal tangent to $\mathcal{C}_1$ and $\mathcal{C}_2$ and has a positive slope, and line $t_2$ is a common internal tangent to $\mathcal{C}_2$ a... | 27 | Call the centers $O_1, O_2, O_3$, the points of tangency $r_1, r_2, s_1, s_2$ (with $r$ on $t_1$ and $s$ on $t_2$, and $s_2$ on $\mathcal{C}_2$), and the intersection of each common internal tangent to the X-axis $r, s$. $\triangle O_1r_1r \sim \triangle O_2r_2r$ since both triangles have a right angle and have vertica... | 6.875 | [
6,
7,
7,
8,
7,
6,
7,
7
] |
Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumulate... | 831 | After the first game, there are $10$ games we care about-- those involving $A$ or $B$. There are $3$ cases of these $10$ games: $A$ wins more than $B$, $B$ wins more than $A$, or $A$ and $B$ win the same number of games. Also, there are $2^{10} = 1024$ total outcomes. By symmetry, the first and second cases are equally... | 6.125 | [
6,
6,
5,
6,
6,
8,
6,
6
] |
A sequence is defined as follows $a_1=a_2=a_3=1,$ and, for all positive integers $n, a_{n+3}=a_{n+2}+a_{n+1}+a_n.$ Given that $a_{28}=6090307, a_{29}=11201821,$ and $a_{30}=20603361,$ find the remainder when $\sum^{28}_{k=1} a_k$ is divided by 1000. | 834 | Since the problem only asks for the first 28 terms and we only need to calculate mod 1000, we simply bash the first 28 terms:
$a_{1}\equiv 1 \pmod {1000} \newline a_{2}\equiv 1 \pmod {1000} \newline a_{3}\equiv 1 \pmod {1000} \newline a_{4}\equiv 3 \pmod {1000} \newline a_{5}\equiv 5 \pmod {1000} \newline \cdots \newli... | 4.25 | [
4,
4,
4,
5,
4,
5,
4,
4
] |
Equilateral $\triangle ABC$ is inscribed in a circle of radius $2$. Extend $\overline{AB}$ through $B$ to point $D$ so that $AD=13,$ and extend $\overline{AC}$ through $C$ to point $E$ so that $AE = 11.$ Through $D,$ draw a line $l_1$ parallel to $\overline{AE},$ and through $E,$ draw a line $l_2$ parallel to $\overlin... | 865 | Note that $AB=2\sqrt3$, $DF=11$, and $EF=13$. If we take a homothety of the parallelogram with respect to $A$, such that $F$ maps to $G$, we see that $\frac{[ABG]}{[ACG]}=\frac{11}{13}$. Since $\angle AGB=\angle AGC=60^{\circ}$, from the sine area formula we have $\frac{BG}{CG}=\frac{11}{13}$. Let $BG=11k$ and $CG=13k$... | 6.875 | [
7,
7,
6,
7,
7,
7,
7,
7
] |
How many integers $N$ less than $1000$ can be written as the sum of $j$ consecutive positive odd integers from exactly 5 values of $j\ge 1$? | 15 | Let the largest odd number below the sequence be the $q$th positive odd number, and the largest odd number in the sequence be the $p$th positive odd number. Therefore, the sum is $p^2-q^2=(p+q)(p-q)$ by sum of consecutive odd numbers. Note that $p+q$ and $p-q$ have the same parity, and $q$ can equal $0$. We then perfor... | 6.625 | [
7,
7,
7,
6,
6,
6,
7,
7
] |
Let $S_n$ be the sum of the reciprocals of the non-zero digits of the integers from $1$ to $10^n$ inclusive. Find the smallest positive integer $n$ for which $S_n$ is an integer. | 63 | Let $K = \sum_{i=1}^{9}{\frac{1}{i}}$. Examining the terms in $S_1$, we see that $S_1 = K + 1$ since each digit $n$ appears once and 1 appears an extra time. Now consider writing out $S_2$. Each term of $K$ will appear 10 times in the units place and 10 times in the tens place (plus one extra 1 will appear), so $S_2 = ... | 6.5 | [
7,
6,
6,
7,
6,
7,
7,
6
] |
Given that $x, y,$ and $z$ are real numbers that satisfy: \begin{align*} x &= \sqrt{y^2-\frac{1}{16}}+\sqrt{z^2-\frac{1}{16}}, \\ y &= \sqrt{z^2-\frac{1}{25}}+\sqrt{x^2-\frac{1}{25}}, \\ z &= \sqrt{x^2 - \frac 1{36}}+\sqrt{y^2-\frac 1{36}}, \end{align*} and that $x+y+z = \frac{m}{\sqrt{n}},$ where $m$ and $n$ are posit... | 9 | Note that none of $x,y,z$ can be zero.
Each of the equations is in the form \[a=\sqrt{b^2-d^2}+\sqrt{c^2-d^2}\]
Isolate a radical and square the equation to get \[b^2-d^2=a^2-2a\sqrt{c^2-d^2}+c^2-d^2\]
Now cancel, and again isolate the radical, and square the equation to get \[a^4+b^4+c^4+2a^2c^2-2a^2b^2-2b^2c^2=4a^... | 7 | [
7,
7,
7,
8,
7,
7,
7,
6
] |
How many positive perfect squares less than $10^6$ are multiples of $24$? | 83 | The prime factorization of $24$ is $2^3\cdot3$. Thus, each square must have at least $3$ factors of $2$ and $1$ factor of $3$ and its square root must have $2$ factors of $2$ and $1$ factor of $3$. This means that each square is in the form $(12c)^2$, where $12 c$ is a positive integer less than $\sqrt{10^6}$. There ar... | 3.625 | [
3,
3,
4,
4,
4,
4,
4,
3
] |
A 100 foot long moving walkway moves at a constant rate of 6 feet per second. Al steps onto the start of the walkway and stands. Bob steps onto the start of the walkway two seconds later and strolls forward along the walkway at a constant rate of 4 feet per second. Two seconds after that, Cy reaches the start of the wa... | 52 | Clearly we have people moving at speeds of $6,8$ and $10$ feet/second. Notice that out of the three people, Cy is at the largest disadvantage to begin with and since all speeds are close, it is hardest for him to catch up. Furthermore, Bob is clearly the farthest along. Thus it is reasonable to assume that there is som... | 4.875 | [
4,
6,
5,
6,
5,
5,
4,
4
] |
The complex number $z$ is equal to $9+bi$, where $b$ is a positive real number and $i^{2}=-1$. Given that the imaginary parts of $z^{2}$ and $z^{3}$ are the same, what is $b$ equal to? | 15 | Squaring, we find that $(9 + bi)^2 = 81 + 18bi - b^2$. Cubing and ignoring the real parts of the result, we find that $(81 + 18bi - b^2)(9 + bi) = \ldots + (9\cdot 18 + 81)bi - b^3i$.
Setting these two equal, we get that $18bi = 243bi - b^3i$, so $b(b^2 - 225) = 0$ and $b = -15, 0, 15$. Since $b > 0$, the solution is ... | 4.625 | [
4,
4,
4,
6,
4,
4,
6,
5
] |
Three planets orbit a star circularly in the same plane. Each moves in the same direction and moves at constant speed. Their periods are 60, 84, and 140 years. The three planets and the star are currently collinear. What is the fewest number of years from now that they will all be collinear again? | 105 | Denote the planets $A, B, C$ respectively. Let $a(t), b(t), c(t)$ denote the angle which each of the respective planets makes with its initial position after $t$ years. These are given by $a(t) = \frac{t \pi}{30}$, $b(t) = \frac{t \pi}{42}$, $c(t) = \frac{t \pi}{70}$.
In order for the planets and the central star to b... | 5.5 | [
6,
5,
5,
6,
5,
6,
6,
5
] |
The formula for converting a Fahrenheit temperature $F$ to the corresponding Celsius temperature $C$ is $C = \frac{5}{9}(F-32).$ An integer Fahrenheit temperature is converted to Celsius, rounded to the nearest integer, converted back to Fahrenheit, and again rounded to the nearest integer.
For how many integer Fahren... | 539 | Examine $F - 32$ modulo 9.
If $F - 32 \equiv 0 \pmod{9}$, then we can define $9x = F - 32$. This shows that $F = \left[\frac{9}{5}\left[\frac{5}{9}(F-32)\right] + 32\right] \Longrightarrow F = \left[\frac{9}{5}(5x) + 32\right] \Longrightarrow F = 9x + 32$. This case works.
If $F - 32 \equiv 1 \pmod{9}$, then we can def... | 5.625 | [
5,
6,
6,
6,
6,
6,
5,
5
] |
A frog is placed at the origin on the number line, and moves according to the following rule: in a given move, the frog advances to either the closest point with a greater integer coordinate that is a multiple of 3, or to the closest point with a greater integer coordinate that is a multiple of 13. A move sequence is a... | 169 | Another way would be to use a table representing the number of ways to reach a certain number
$\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c|c|c|c|c} 0 & 3 & 6 & 9 & 12 & 13 & 15 & 18 & 21 & 24 & 26 & 27 & 30 & 33 & 36 \\ \hline 1 & 1 & 1 & 1 & 1 & 5 & 6 & 6 & 6 & 6 & 29 & 35 & 35 & 35 & 35 \\ \end{tabular}$
How we came with ... | 5.75 | [
6,
6,
5,
6,
6,
5,
6,
6
] |
Let $N = \sum_{k = 1}^{1000} k ( \lceil \log_{\sqrt{2}} k \rceil - \lfloor \log_{\sqrt{2}} k \rfloor )$
Find the remainder when $N$ is divided by 1000. ($\lfloor{k}\rfloor$ is the greatest integer less than or equal to $k$, and $\lceil{k}\rceil$ is the least integer greater than or equal to $k$.) | 477 | The ceiling of a number minus the floor of a number is either equal to zero (if the number is an integer); otherwise, it is equal to 1. Thus, we need to find when or not $\log_{\sqrt{2}} k$ is an integer.
The change of base formula shows that $\frac{\log k}{\log \sqrt{2}} = \frac{2 \log k}{\log 2}$. For the $\log 2$ t... | 6.25 | [
6,
6,
6,
6,
7,
6,
6,
7
] |
The polynomial $P(x)$ is cubic. What is the largest value of $k$ for which the polynomials $Q_1(x) = x^2 + (k-29)x - k$ and $Q_2(x) = 2x^2+ (2k-43)x + k$ are both factors of $P(x)$? | 30 | We can see that $Q_1$ and $Q_2$ must have a root in common for them to both be factors of the same cubic.
Let this root be $a$.
We then know that $a$ is a root of $Q_{2}(x)-2Q_{1}(x) = 2x^{2}+2kx-43x+k-2x^{2}-2kx+58x+2k = 15x+3k = 0$ , so $x = \frac{-k}{5}$.
We then know that $\frac{-k}{5}$ is a root of $Q_{1}$ so w... | 5.625 | [
5,
6,
6,
5,
6,
6,
6,
5
] |
In right triangle $ABC$ with right angle $C$, $CA = 30$ and $CB = 16$. Its legs $CA$ and $CB$ are extended beyond $A$ and $B$. Points $O_1$ and $O_2$ lie in the exterior of the triangle and are the centers of two circles with equal radii. The circle with center $O_1$ is tangent to the hypotenuse and to the extension of... | 737 | Let the radius of the circle be $r$. It can be seen that $\Delta FHO_{1}$ and $\Delta O_{2}GJ$ are similar to $\Delta ACB$, and the length of the hypotenuses are $\frac{17}{8}r$ and $\frac {17}{15}r$, respectively. Then, the entire length of $HJ$ is going to be $(\frac{17}{8}+\frac{17}{15}+2)r = \frac{631}{120}r$. The ... | 6.75 | [
7,
6,
7,
6,
7,
7,
7,
7
] |
In a 6 x 4 grid (6 rows, 4 columns), 12 of the 24 squares are to be shaded so that there are two shaded squares in each row and three shaded squares in each column. Let $N$ be the number of shadings with this property. Find the remainder when $N$ is divided by 1000. | 860 | We start by showing that every group of $6$ rows can be grouped into $3$ complementary pairs. We proceed with proof by contradiction. Without loss of generality, assume that the first row has columns $1$ and $2$ shaded. Note how if there is no complement to this, then all the other five rows must have at least one squa... | 6.375 | [
7,
7,
6,
7,
6,
6,
6,
6
] |
For each positive integer $p$, let $b(p)$ denote the unique positive integer $k$ such that $|k-\sqrt{p}| < \frac{1}{2}$. For example, $b(6) = 2$ and $b(23) = 5$. If $S = \sum_{p=1}^{2007} b(p),$ find the remainder when $S$ is divided by 1000. | 955 | $\left(k- \frac 12\right)^2=k^2-k+\frac 14$ and $\left(k+ \frac 12\right)^2=k^2+k+ \frac 14$. Therefore $b(p)=k$ if and only if $p$ is in this range, or $k^2-k<p\leq k^2+k$. There are $2k$ numbers in this range, so the sum of $b(p)$ over this range is $(2k)k=2k^2$. $44<\sqrt{2007}<45$, so all numbers $1$ to $44$ have t... | 6.875 | [
6,
8,
7,
7,
7,
7,
7,
6
] |
In isosceles triangle $\triangle ABC$, $A$ is located at the origin and $B$ is located at $(20,0)$. Point $C$ is in the first quadrant with $AC = BC$ and angle $BAC = 75^{\circ}$. If triangle $ABC$ is rotated counterclockwise about point $A$ until the image of $C$ lies on the positive $y$-axis, the area of the region c... | 875 | Call the points of the intersections of the triangles $D$, $E$, and $F$ as noted in the diagram (the points are different from those in the diagram for solution 1). $\overline{AD}$ bisects $\angle EDE'$.
Through HL congruency, we can find that $\triangle AED$ is congruent to $\triangle AE'D$. This divides the region $... | 7.5 | [
7,
7,
8,
7,
7,
8,
8,
8
] |
A square pyramid with base $ABCD$ and vertex $E$ has eight edges of length $4$. A plane passes through the midpoints of $AE$, $BC$, and $CD$. The plane's intersection with the pyramid has an area that can be expressed as $\sqrt{p}$. Find $p$. | 80 | [asy]import three; import math; pointpen = black; pathpen = black+linewidth(0.7); currentprojection = perspective(2.5,-12,4); triple A=(-2,2,0), B=(2,2,0), C=(2,-2,0), D=(-2,-2,0), E=(0,0,2*2^.5), P=(A+E)/2, Q=(B+C)/2, R=(C+D)/2, Y=(-3/2,-3/2,2^.5/2),X=(3/2,3/2,2^.5/2), H=(4,2,0), I=(-2,-4,0); draw(A--B--C--D--A--E--B-... | 6.125 | [
6,
6,
6,
7,
6,
6,
6,
6
] |
A sequence is defined over non-negative integral indexes in the following way: $a_{0}=a_{1}=3$, $a_{n+1}a_{n-1}=a_{n}^{2}+2007$.
Find the greatest integer that does not exceed $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.$ | 224 | We are given that
$a_{n+1}a_{n-1}= a_{n}^{2}+2007$, $a_{n-1}^{2}+2007 = a_{n}a_{n-2}$.
Add these two equations to get
$a_{n-1}(a_{n-1}+a_{n+1}) = a_{n}(a_{n}+a_{n-2})$
$\frac{a_{n+1}+a_{n-1}}{a_{n}}= \frac{a_{n}+a_{n-2}}{a_{n-1}}$.
This is an invariant. Defining $b_{i}= \frac{a_{i}}{a_{i-1}}$ for each $i \ge 2$, the ... | 7 | [
7,
7,
7,
7,
7,
7,
7,
7
] |
Let $ABC$ be an equilateral triangle, and let $D$ and $F$ be points on sides $BC$ and $AB$, respectively, with $FA = 5$ and $CD = 2$. Point $E$ lies on side $CA$ such that angle $DEF = 60^{\circ}$. The area of triangle $DEF$ is $14\sqrt{3}$. The two possible values of the length of side $AB$ are $p \pm q \sqrt{r}$, whe... | 989 | First of all, assume $EC=x,BD=m, ED=a, EF=b$, then we can find $BF=m-3, AE=2+m-x$ It is not hard to find $ab*sin60^{\circ}*\frac{1}{2}=14\sqrt{3}, ab=56$, we apply LOC on $\triangle{DEF}, \triangle{BFD}$, getting that $(m-3)^2+m^2-m(m-3)=a^2+b^2-ab$, leads to $a^2+b^2=m^2-3m+65$ Apply LOC on $\triangle{CED}, \triangle{... | 7.75 | [
8,
8,
8,
8,
7,
8,
8,
7
] |
A mathematical organization is producing a set of commemorative license plates. Each plate contains a sequence of five characters chosen from the four letters in AIME and the four digits in 2007. No character may appear in a sequence more times than it appears among the four letters in AIME or the four digits in 2007. ... | 372 | There are 7 different characters that can be picked, with 0 being the only number that can be repeated twice.
If $0$ appears 0 or 1 times amongst the sequence, there are $\frac{7!}{(7-5)!} = 2520$ sequences possible.
If $0$ appears twice in the sequence, there are ${5\choose2} = 10$ places to place the $0$s. There are ... | 4.5 | [
4,
5,
4,
6,
4,
5,
4,
4
] |
Find the number of ordered triples $(a,b,c)$ where $a$, $b$, and $c$ are positive integers, $a$ is a factor of $b$, $a$ is a factor of $c$, and $a+b+c=100$. | 200 | Denote $x = \frac{b}{a}$ and $y = \frac{c}{a}$. The last condition reduces to $a(1 + x + y) = 100$. Therefore, $1 + x + y$ is equal to one of the 9 factors of $100 = 2^25^2$.
Subtracting the one, we see that $x + y = \{0,1,3,4,9,19,24,49,99\}$. There are exactly $n - 1$ ways to find pairs of $(x,y)$ if $x + y = n$. Th... | 4.5 | [
5,
4,
5,
4,
5,
4,
4,
5
] |
Square $ABCD$ has side length $13$, and points $E$ and $F$ are exterior to the square such that $BE=DF=5$ and $AE=CF=12$. Find $EF^{2}$. [asy]unitsize(0.2 cm); pair A, B, C, D, E, F; A = (0,13); B = (13,13); C = (13,0); D = (0,0); E = A + (12*12/13,5*12/13); F = D + (5*5/13,-5*12/13); draw(A--B--C--D--cycle); draw(A--E... | 578 | Drawing $EF$, it clearly passes through the center of $ABCD$. Letting this point be $P$, we note that $AEBP$ and $CFDP$ are congruent cyclic quadrilaterals, and that $AP=BP=CP=DP=\frac{13}{\sqrt{2}}.$ Now, from Ptolemy's, $13\cdot EP=\frac{13}{\sqrt{2}}(12+5)\implies EP+\frac{17\sqrt{2}}{2}$. Since $EF=EP+FP=2\cdot EP$... | 5.25 | [
5,
5,
6,
5,
5,
5,
6,
5
] |
The workers in a factory produce widgets and whoosits. For each product, production time is constant and identical for all workers, but not necessarily equal for the two products. In one hour, $100$ workers can produce $300$ widgets and $200$ whoosits. In two hours, $60$ workers can produce $240$ widgets and $300$ whoo... | 450 | Suppose that it takes $x$ hours for one worker to create one widget, and $y$ hours for one worker to create one whoosit.
Therefore, we can write that (note that two hours is similar to having twice the number of workers, and so on):
$100 = 300x + 200y$
$2(60) = 240x + 300y$
$3(50) = 150x + my$
Solve the system of equ... | 5.125 | [
5,
5,
6,
5,
5,
5,
5,
5
] |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.