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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The graph of the equation $9x+223y=2007$ is drawn on graph paper with each square representing one unit in each direction. How many of the $1$ by $1$ graph paper squares have interiors lying entirely below the graph and entirely in the first quadrant? | 888 | We know that the number of squares intersected in an $m\times{n}$ rectangle is $m + n -\mbox{gcd}(m,n)$. So if we apply that here, we get that the number of intersected squares is:
$9 + 223 - 1 = 231$.
Now just subtract that from the total number of squares and divide by 2, since we want the number of squares below th... | 4.5 | [
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An integer is called parity-monotonic if its decimal representation $a_{1}a_{2}a_{3}\cdots a_{k}$ satisfies $a_{i}<a_{i+1}$ if $a_{i}$ is odd, and $a_{i}>a_{i+1}$ if $a_{i}$ is even. How many four-digit parity-monotonic integers are there? | 640 | This problem can be solved via recursion since we are "building a string" of numbers with some condition. We want to create a new string by adding a new digit at the front so we avoid complications($0$ can't be at the front and no digit is less than $9$). There are $4$ options to add no matter what(try some examples if... | 5.125 | [
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Given a real number $x,$ let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x.$ For a certain integer $k,$ there are exactly $70$ positive integers $n_{1}, n_{2}, \ldots, n_{70}$ such that $k=\lfloor\sqrt[3]{n_{1}}\rfloor = \lfloor\sqrt[3]{n_{2}}\rfloor = \cdots = \lfloor\sqrt[3]{n_{70}}\rfloor$... | 553 | Obviously $k$ is positive. Then, we can let $n_1$ equal $k^3$ and similarly let $n_i$ equal $k^3 + (i - 1)k$.
The wording of this problem (which uses "exactly") tells us that $k^3+69k<(k+1)^3 = k^3 + 3k^2 + 3k+1 \leq k^3+70k$. Taking away $k^3$ from our inequality results in $69k<3k^2+3k+1\leq 70k$. Since $69k$, $3k^2... | 5.875 | [
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A rectangular piece of paper measures 4 units by 5 units. Several lines are drawn parallel to the edges of the paper. A rectangle determined by the intersections of some of these lines is called basic if
(i) all four sides of the rectangle are segments of drawn line segments, and
(ii) no segments of drawn lines lie i... | 896 | Denote the number of horizontal lines drawn as $x$, and the number of vertical lines drawn as $y$. The number of basic rectangles is $(x - 1)(y - 1)$. $5x + 4y = 2007 \Longrightarrow y = \frac{2007 - 5x}{4}$. Substituting, we find that $(x - 1)\left(-\frac 54x + \frac{2003}4\right)$.
FOIL this to get a quadratic, $-\fr... | 6.375 | [
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Rectangle $ABCD$ is given with $AB=63$ and $BC=448.$ Points $E$ and $F$ lie on $AD$ and $BC$ respectively, such that $AE=CF=84.$ The inscribed circle of triangle $BEF$ is tangent to $EF$ at point $P,$ and the inscribed circle of triangle $DEF$ is tangent to $EF$ at point $Q.$ Find $PQ.$ | 259 | Several Pythagorean triples exist amongst the numbers given. $BE = DF = \sqrt{63^2 + 84^2} = 21\sqrt{3^2 + 4^2} = 105$. Also, the length of $EF = \sqrt{63^2 + (448 - 2\cdot84)^2} = 7\sqrt{9^2 + 40^2} = 287$.
Use the Two Tangent Theorem on $\triangle BEF$. Since both circles are inscribed in congruent triangles, they a... | 6.625 | [
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Let $S$ be a set with six elements. Let $\mathcal{P}$ be the set of all subsets of $S.$ Subsets $A$ and $B$ of $S$, not necessarily distinct, are chosen independently and at random from $\mathcal{P}$. The probability that $B$ is contained in one of $A$ or $S-A$ is $\frac{m}{n^{r}},$ where $m$, $n$, and $r$ are positive... | 710 | Let $|S|$ denote the number of elements in a general set $S$. We use complementary counting.
There is a total of $2^6$ elements in $P$, so the total number of ways to choose $A$ and $B$ is $(2^6)^2 = 2^{12}$.
Note that the number of $x$-element subset of $S$ is $\binom{6}{x}$. In general, for $0 \le |A| \le 6$, in or... | 6.25 | [
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Two long cylindrical tubes of the same length but different diameters lie parallel to each other on a flat surface. The larger tube has radius $72$ and rolls along the surface toward the smaller tube, which has radius $24$. It rolls over the smaller tube and continues rolling along the flat surface until it comes to re... | 179 | If it weren’t for the small tube, the larger tube would travel $144\pi$. Consider the distance from which the larger tube first contacts the smaller tube, until when it completely loses contact with the smaller tube.
Drawing the radii as shown in the diagram, notice that the hypotenuse of the right triangle in the dia... | 7.125 | [
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The increasing geometric sequence $x_{0},x_{1},x_{2},\ldots$ consists entirely of integral powers of $3.$ Given that
$\sum_{n=0}^{7}\log_{3}(x_{n}) = 308$ and $56 \leq \log_{3}\left ( \sum_{n=0}^{7}x_{n}\right ) \leq 57,$
find $\log_{3}(x_{14}).$ | 91 | All these integral powers of $3$ are all different, thus in base $3$ the sum of these powers would consist of $1$s and $0$s. Thus the largest value $x_7$ must be $3^{56}$ in order to preserve the givens. Then we find by the given that $x_7x_6x_5\dots x_0 = 3^{308}$, and we know that the exponents of $x_i$ are in an ari... | 6.75 | [
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A triangular array of squares has one square in the first row, two in the second, and in general, $k$ squares in the $k$th row for $1 \leq k \leq 11.$ With the exception of the bottom row, each square rests on two squares in the row immediately below (illustrated in the given diagram). In each square of the eleventh ro... | 640 | Label each of the bottom squares as $x_0, x_1 \ldots x_9, x_{10}$.
Through induction, we can find that the top square is equal to ${10\choose0}x_0 + {10\choose1}x_1 + {10\choose2}x_2 + \ldots {10\choose10}x_{10}$. (This also makes sense based on a combinatorial argument: the number of ways a number can "travel" to the... | 6.25 | [
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Let $f(x)$ be a polynomial with real coefficients such that $f(0) = 1,$ $f(2)+f(3)=125,$ and for all $x$, $f(x)f(2x^{2})=f(2x^{3}+x).$ Find $f(5).$ | 676 | If the leading term of $f(x)$ is $ax^m$, then the leading term of $f(x)f(2x^2) = ax^m \cdot a(2x^2)^m = 2^ma^2x^{3m}$, and the leading term of $f(2x^3 + x) = 2^max^{3m}$. Hence $2^ma^2 = 2^ma$, and $a = 1$. Because $f(0) = 1$, the product of all the roots of $f(x)$ is $\pm 1$. If $f(\lambda) = 0$, then $f(2\lambda^3 + ... | 7 | [
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Four circles $\omega,$ $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ with the same radius are drawn in the interior of triangle $ABC$ such that $\omega_{A}$ is tangent to sides $AB$ and $AC$, $\omega_{B}$ to $BC$ and $BA$, $\omega_{C}$ to $CA$ and $CB$, and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ a... | 389 | Consider a 13-14-15 triangle. $A=84.$ [By Heron's Formula or by 5-12-13 and 9-12-15 right triangles.]
The inradius is $r=\frac{A}{s}=\frac{84}{21}=4$, where $s$ is the semiperimeter. Scale the triangle with the inradius by a linear scale factor, $u.$
The circumradius is $R=\frac{abc}{4rs}=\frac{13\cdot 14\cdot 15}{4\... | 7 | [
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Of the students attending a school party, $60\%$ of the students are girls, and $40\%$ of the students like to dance. After these students are joined by $20$ more boy students, all of whom like to dance, the party is now $58\%$ girls. How many students now at the party like to dance? | 252 | Let $p$ denote the total number of people at the party. Then, because we know the proportions of boys to $p$ both before and after 20 boys arrived, we can create the following equation: \[0.4p+20 = 0.42(p+20)\] Solving for p gives us $p=580$, so the solution is $0.4p+20 = \boxed{252}$ | 4.25 | [
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Square $AIME$ has sides of length $10$ units. Isosceles triangle $GEM$ has base $EM$, and the area common to triangle $GEM$ and square $AIME$ is $80$ square units. Find the length of the altitude to $EM$ in $\triangle GEM$. | 25 | Note that if the altitude of the triangle is at most $10$, then the maximum area of the intersection of the triangle and the square is $5\cdot10=50$. This implies that vertex G must be located outside of square $AIME$.
[asy] pair E=(0,0), M=(10,0), I=(10,10), A=(0,10); draw(A--I--M--E--cycle); pair G=(5,25); draw(G--E-... | 5.75 | [
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Ed and Sue bike at equal and constant rates. Similarly, they jog at equal and constant rates, and they swim at equal and constant rates. Ed covers $74$ kilometers after biking for $2$ hours, jogging for $3$ hours, and swimming for $4$ hours, while Sue covers $91$ kilometers after jogging for $2$ hours, swimming for $3$... | 314 | Let the biking rate be $b$, swimming rate be $s$, jogging rate be $j$, all in km/h.
We have $2b + 3j + 4s = 74,2j + 3s + 4b = 91$. Subtracting the second from twice the first gives $4j + 5s = 57$. Mod 4, we need $s\equiv1\pmod{4}$. Thus, $(j,s) = (13,1),(8,5),(3,9)$.
$(13,1)$ and $(3,9)$ give non-integral $b$, but $(8... | 5.75 | [
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There exist unique positive integers $x$ and $y$ that satisfy the equation $x^2 + 84x + 2008 = y^2$. Find $x + y$. | 80 | We see that $y^2 \equiv x^2 + 4 \pmod{6}$. By quadratic residues, we find that either $x \equiv 0, 3 \pmod{6}$. Also, $y^2 \equiv (x+42)^2 + 244 \equiv (x+2)^2 \pmod{4}$, so $x \equiv 0, 2 \mod{4}$. Combining, we see that $x \equiv 0 \mod{6}$.
Testing $x = 6$ and other multiples of $6$, we quickly find that $x = 18, y... | 4.25 | [
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A right circular cone has base radius $r$ and height $h$. The cone lies on its side on a flat table. As the cone rolls on the surface of the table without slipping, the point where the cone's base meets the table traces a circular arc centered at the point where the vertex touches the table. The cone first returns to i... | 14 | The path is a circle with radius equal to the slant height of the cone, which is $\sqrt {r^{2} + h^{2}}$. Thus, the length of the path is $2\pi\sqrt {r^{2} + h^{2}}$.
Also, the length of the path is 17 times the circumference of the base, which is $34r\pi$. Setting these equal gives $\sqrt {r^{2} + h^{2}} = 17r$, or $... | 6.625 | [
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A triangular array of numbers has a first row consisting of the odd integers $1,3,5,\ldots,99$ in increasing order. Each row below the first has one fewer entry than the row above it, and the bottom row has a single entry. Each entry in any row after the top row equals the sum of the two entries diagonally above it in ... | 17 | The result above is fairly intuitive if we write out several rows and then divide all numbers in row $r$ by $2^{r-1}$ (we can do this because dividing by a power of 2 doesn't affect divisibility by $67$). The second row will be $2, 4, 6, \cdots , 98$, the third row will be $3, 5, \cdots, 97$, and so forth. Clearly, onl... | 5.75 | [
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Let $S_i$ be the set of all integers $n$ such that $100i\leq n < 100(i + 1)$. For example, $S_4$ is the set ${400,401,402,\ldots,499}$. How many of the sets $S_0, S_1, S_2, \ldots, S_{999}$ do not contain a perfect square? | 708 | The difference between consecutive squares is $(x + 1)^2 - x^2 = 2x + 1$, which means that all squares above $50^2 = 2500$ are more than $100$ apart.
Then the first $26$ sets ($S_0,\cdots S_{25}$) each have at least one perfect square because the differences between consecutive squares in them are all less than $100$.... | 4.625 | [
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Find the positive integer $n$ such that
\[\arctan\frac {1}{3} + \arctan\frac {1}{4} + \arctan\frac {1}{5} + \arctan\frac {1}{n} = \frac {\pi}{4}.\] | 47 | Adding a series of angles is the same as multiplying the complex numbers whose arguments they are. In general, $\arctan\frac{1}{n}$, is the argument of $n+i$. The sum of these angles is then just the argument of the product
\[(3+i)(4+i)(5+i)(n+i)\]
and expansion give us $(48n-46)+(48+46n)i$. Since the argument of this... | 5.875 | [
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Ten identical crates each of dimensions $3\mathrm{ft}\times 4\mathrm{ft}\times 6\mathrm{ft}$. The first crate is placed flat on the floor. Each of the remaining nine crates is placed, in turn, flat on top of the previous crate, and the orientation of each crate is chosen at random. Let $\frac {m}{n}$ be the probability... | 190 | Only the heights matter, and each crate is either 3, 4, or 6 feet tall with equal probability. We have the following:
\begin{align*}3a + 4b + 6c &= 41\\ a + b + c &= 10\end{align*}
Subtracting 3 times the second from the first gives $b + 3c = 11$, or $(b,c) = (2,3),(5,2),(8,1),(11,0)$. The last doesn't work, obviously... | 6.25 | [
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Let $ABCD$ be an isosceles trapezoid with $\overline{AD}||\overline{BC}$ whose angle at the longer base $\overline{AD}$ is $\dfrac{\pi}{3}$. The diagonals have length $10\sqrt {21}$, and point $E$ is at distances $10\sqrt {7}$ and $30\sqrt {7}$ from vertices $A$ and $D$, respectively. Let $F$ be the foot of the altitud... | 32 | Extend $\overline {AB}$ through $B$, to meet $\overline {DC}$ (extended through $C$) at $G$. $ADG$ is an equilateral triangle because of the angle conditions on the base.
If $\overline {GC} = x$ then $\overline {CD} = 40\sqrt{7}-x$, because $\overline{AD}$ and therefore $\overline{GD}$ $= 40\sqrt{7}$.
By simple angle... | 6.875 | [
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Consider sequences that consist entirely of $A$'s and $B$'s and that have the property that every run of consecutive $A$'s has even length, and every run of consecutive $B$'s has odd length. Examples of such sequences are $AA$, $B$, and $AABAA$, while $BBAB$ is not such a sequence. How many such sequences have length 1... | 172 | Let $a_n$ and $b_n$ denote, respectively, the number of sequences of length $n$ ending in $A$ and $B$. If a sequence ends in an $A$, then it must have been formed by appending two $A$s to the end of a string of length $n-2$. If a sequence ends in a $B,$ it must have either been formed by appending one $B$ to a string o... | 6.125 | [
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On a long straight stretch of one-way single-lane highway, cars all travel at the same speed and all obey the safety rule: the distance from the back of the car ahead to the front of the car behind is exactly one car length for each 15 kilometers per hour of speed or fraction thereof (Thus the front of a car traveling ... | 375 | Disclaimer: This is for the people who may not understand calculus, and is also how I did it. First, we assume several things. First, we assume the speeds must be multiples of 15 to maximize cars, because any less will be a waste. Second, we start with one car in front of the photoelectric eye. We first set the speed o... | 7 | [
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Let
\[p(x,y) = a_0 + a_1x + a_2y + a_3x^2 + a_4xy + a_5y^2 + a_6x^3 + a_7x^2y + a_8xy^2 + a_9y^3.\]
Suppose that
\[p(0,0) = p(1,0) = p( - 1,0) = p(0,1) = p(0, - 1)\\ = p(1,1) = p(1, - 1) = p(2,2) = 0.\]
There is a point $\left(\frac {a}{c},\frac {b}{c}\right)$ for which $p\left(\frac {a}{c},\frac {b}{c}\right) = 0$ f... | 40 | Consider the cross section of $z = p(x, y)$ on the plane $z = 0$. We realize that we could construct the lines/curves in the cross section such that their equations multiply to match the form of $p(x, y)$ and they go over the eight given points. A simple way to do this would be to use the equations $x = 0$, $x = 1$, an... | 7.125 | [
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Let $\overline{AB}$ be a diameter of circle $\omega$. Extend $\overline{AB}$ through $A$ to $C$. Point $T$ lies on $\omega$ so that line $CT$ is tangent to $\omega$. Point $P$ is the foot of the perpendicular from $A$ to line $CT$. Suppose $\overline{AB} = 18$, and let $m$ denote the maximum possible length of segment ... | 432 | Proceed as follows for Solution 1.
Once you approach the function $k=(2x-27)/x^2$, find the maximum value by setting $dk/dx=0$.
Simplifying $k$ to take the derivative, we have $2/x-27/x^2$, so $dk/dx=-2/x^2+54/x^3$. Setting $dk/dx=0$, we have $2/x^2=54/x^3$.
Solving, we obtain $x=27$ as the critical value. Hence, $k... | 6.25 | [
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A square piece of paper has sides of length $100$. From each corner a wedge is cut in the following manner: at each corner, the two cuts for the wedge each start at a distance $\sqrt{17}$ from the corner, and they meet on the diagonal at an angle of $60^{\circ}$ (see the figure below). The paper is then folded up along... | 871 | In the final pyramid, let $ABCD$ be the smaller square and let $A'B'C'D'$ be the larger square such that $AA'$, etc. are edges.
It is obvious from the diagram that $\angle A'AB = \angle A'AD = 105^\circ$.
Let $AB$ and $AD$ be the positive $x$ and $y$ axes in a 3-d coordinate system such that $A'$ has a positive $z$ c... | 6.625 | [
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Let $N = 100^2 + 99^2 - 98^2 - 97^2 + 96^2 + \cdots + 4^2 + 3^2 - 2^2 - 1^2$, where the additions and subtractions alternate in pairs. Find the remainder when $N$ is divided by $1000$. | 100 | Since we want the remainder when $N$ is divided by $1000$, we may ignore the $100^2$ term. Then, applying the difference of squares factorization to consecutive terms,
\begin{align*} N &= (99-98)(99+98) - (97-96)(97+96) + (95-94)(95 + 94) + \cdots + (3-2)(3+2) - 1 \\ &= \underbrace{197 - 193}_4 + \underbrace{189 - 185}... | 4.25 | [
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Rudolph bikes at a constant rate and stops for a five-minute break at the end of every mile. Jennifer bikes at a constant rate which is three-quarters the rate that Rudolph bikes, but Jennifer takes a five-minute break at the end of every two miles. Jennifer and Rudolph begin biking at the same time and arrive at the $... | 620 | Let $r$ be the time Rudolph takes disregarding breaks and $\frac{4}{3}r$ be the time Jennifer takes disregarding breaks. We have the equation \[r+5\left(49\right)=\frac{4}{3}r+5\left(24\right)\] \[125=\frac13r\] \[r=375.\] Thus, the total time they take is $375 + 5(49) = \boxed{620}$ minutes. | 5 | [
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A block of cheese in the shape of a rectangular solid measures $10$ cm by $13$ cm by $14$ cm. Ten slices are cut from the cheese. Each slice has a width of $1$ cm and is cut parallel to one face of the cheese. The individual slices are not necessarily parallel to each other. What is the maximum possible volume in cubic... | 729 | Let the lengths of the three sides of the rectangular solid after the cutting be $a,b,c$, so that the desired volume is $abc$. Note that each cut reduces one of the dimensions by one, so that after ten cuts, $a+b+c = 10 + 13 + 14 - 10 = 27$. By AM-GM, $\frac{a+b+c}{3} = 9 \ge \sqrt[3]{abc} \Longrightarrow abc \le \boxe... | 4.875 | [
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There exist $r$ unique nonnegative integers $n_1 > n_2 > \cdots > n_r$ and $r$ unique integers $a_k$ ($1\le k\le r$) with each $a_k$ either $1$ or $- 1$ such that \[a_13^{n_1} + a_23^{n_2} + \cdots + a_r3^{n_r} = 2008.\] Find $n_1 + n_2 + \cdots + n_r$. | 21 | In base $3$, we find that $\overline{2008}_{10} = \overline{2202101}_{3}$. In other words,
$2008 = 2 \cdot 3^{6} + 2 \cdot 3^{5} + 2 \cdot 3^3 + 1 \cdot 3^2 + 1 \cdot 3^0$
In order to rewrite as a sum of perfect powers of $3$, we can use the fact that $2 \cdot 3^k = 3^{k+1} - 3^k$:
$2008 = (3^7 - 3^6) + (3^6-3^5) + (3... | 6 | [
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In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$, let $BC = 1000$ and $AD = 2008$. Let $\angle A = 37^\circ$, $\angle D = 53^\circ$, and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD}$, respectively. Find the length $MN$. | 504 | Obviously, these angles are random--the only special thing about them is that they add up to 90. So we might as well let the given angles equal 45 and 45, and now the answer is trivially $\boxed{504}$. (The trapezoid is isosceles, and you see two 45-45-90 triangles;from there you can get the answer.) | 3.875 | [
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The sequence $\{a_n\}$ is defined by \[a_0 = 1,a_1 = 1, \text{ and } a_n = a_{n - 1} + \frac {a_{n - 1}^2}{a_{n - 2}}\text{ for }n\ge2.\] The sequence $\{b_n\}$ is defined by \[b_0 = 1,b_1 = 3, \text{ and } b_n = b_{n - 1} + \frac {b_{n - 1}^2}{b_{n - 2}}\text{ for }n\ge2.\] Find $\frac {b_{32}}{a_{32}}$. | 561 | Rearranging the definitions, we have \[\frac{a_n}{a_{n-1}} = \frac{a_{n-1}}{a_{n-2}} + 1,\quad \frac{b_n}{b_{n-1}} = \frac{b_{n-1}}{b_{n-2}} + 1\] from which it follows that $\frac{a_n}{a_{n-1}} = 1+ \frac{a_{n-1}}{a_{n-2}} = \cdots = (n-1) + \frac{a_{1}}{a_0} = n$ and $\frac{b_n}{b_{n-1}} = (n-1) + \frac{b_{1}}{b_0} =... | 6.375 | [
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Let $r$, $s$, and $t$ be the three roots of the equation \[8x^3 + 1001x + 2008 = 0.\] Find $(r + s)^3 + (s + t)^3 + (t + r)^3$. | 753 | Write $(r+s)^3+(s+t)^3+(t+r)^3=-(r^3+s^3+t^3)$ and let $f(x)=8x^3+1001x+2008$. Then \[f(r)+f(s)+f(t)=8(r^3+s^3+t^3)+1001(r+s+t)+6024=8(r^3+s^3+t^3)+6024=0.\] Solving for $r^3+s^3+t^3$ and negating the result yields the answer $\boxed{753}.$ | 6 | [
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Let $a = \pi/2008$. Find the smallest positive integer $n$ such that \[2[\cos(a)\sin(a) + \cos(4a)\sin(2a) + \cos(9a)\sin(3a) + \cdots + \cos(n^2a)\sin(na)]\] is an integer. | 251 | By the product-to-sum identities, we have that $2\cos a \sin b = \sin (a+b) - \sin (a-b)$. Therefore, this reduces to a telescoping series: \begin{align*} \sum_{k=1}^{n} 2\cos(k^2a)\sin(ka) &= \sum_{k=1}^{n} [\sin(k(k+1)a) - \sin((k-1)ka)]\\ &= -\sin(0) + \sin(2a)- \sin(2a) + \sin(6a) - \cdots - \sin((n-1)na) + \sin(n(... | 6.625 | [
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A particle is located on the coordinate plane at $(5,0)$. Define a move for the particle as a counterclockwise rotation of $\pi/4$ radians about the origin followed by a translation of $10$ units in the positive $x$-direction. Given that the particle's position after $150$ moves is $(p,q)$, find the greatest integer le... | 19 | Let $T:\begin{pmatrix}x\\y\end{pmatrix}\rightarrow R(\frac{\pi}{4})\begin{pmatrix}x\\y\end{pmatrix}+\begin{pmatrix}10\\0\end{pmatrix}$. We assume that the rotation matrix $R(\frac{\pi}{4}) = R$ here. Then we have
$T^{150}\begin{pmatrix}5\\0\end{pmatrix}=R(R(...R(R\begin{pmatrix}5\\0\end{pmatrix}+\begin{pmatrix}10\\0\en... | 6.5 | [
6,
7,
6,
7,
6,
7,
7,
6
] |
The diagram below shows a $4\times4$ rectangular array of points, each of which is $1$ unit away from its nearest neighbors.
[asy] unitsize(0.25inch); defaultpen(linewidth(0.7)); int i, j; for(i = 0; i < 4; ++i) for(j = 0; j < 4; ++j) dot(((real)i, (real)j)); [/asy]
Define a growing path to be a sequence of distinct p... | 240 | We label our points using coordinates $0 \le x,y \le 3$, with the bottom-left point being $(0,0)$. By the Pythagorean Theorem, the distance between two points is $\sqrt{d_x^2 + d_y^2}$ where $0 \le d_x, d_y \le 3$; these yield the possible distances (in decreasing order) \[\sqrt{18},\ \sqrt{13},\ \sqrt{10},\ \sqrt{9},\... | 6.625 | [
7,
6,
7,
8,
6,
6,
6,
7
] |
In triangle $ABC$, $AB = AC = 100$, and $BC = 56$. Circle $P$ has radius $16$ and is tangent to $\overline{AC}$ and $\overline{BC}$. Circle $Q$ is externally tangent to $P$ and is tangent to $\overline{AB}$ and $\overline{BC}$. No point of circle $Q$ lies outside of $\triangle ABC$. The radius of circle $Q$ can be expr... | 254 | Refer to the above diagram. Let the larger circle have center $O_1$, the smaller have center $O_2$, and the incenter be $I$. We can easily calculate that the area of $\triangle ABC = 2688$, and $s = 128$ and $R = 21$, where $R$ is the inradius.
Now, Line $\overline{AI}$ is the perpendicular bisector of $\overline{BC}$... | 7.125 | [
7,
8,
7,
7,
7,
7,
7,
7
] |
There are two distinguishable flagpoles, and there are $19$ flags, of which $10$ are identical blue flags, and $9$ are identical green flags. Let $N$ be the number of distinguishable arrangements using all of the flags in which each flagpole has at least one flag and no two green flags on either pole are adjacent. Find... | 310 | Split the problem into two cases:
Case 1 - Both poles have blue flags: There are 9 ways to place the 10 blue flags on the poles. In each of these configurations, there are 12 spots that a green flag could go. (either in between two blues or at the tops or bottoms of the poles) Then, since there are 9 green flags, all ... | 6.5 | [
6,
6,
6,
6,
7,
7,
7,
7
] |
A regular hexagon with center at the origin in the complex plane has opposite pairs of sides one unit apart. One pair of sides is parallel to the imaginary axis. Let $R$ be the region outside the hexagon, and let $S = \left\lbrace\frac{1}{z}|z \in R\right\rbrace$. Then the area of $S$ has the form $a\pi + \sqrt{b}$, wh... | 29 | We can describe the line parallel to the imaginary axis $x=\frac{1}{2}$ using polar coordinates as $r(\theta)=\dfrac{1}{2\cos{\theta}},$
which rearranges to $z=\left(\dfrac{1}{2\cos{\theta}}\right)(cis{\theta})\implies \frac{1}{z}=2\cos{\theta}cis(-\theta).$
Denote the area of $S$ as $[S].$ Now, dividing the hexagon ... | 7.25 | [
7,
7,
7,
7,
8,
7,
8,
7
] |
Let $a$ and $b$ be positive real numbers with $a\ge b$. Let $\rho$ be the maximum possible value of $\frac {a}{b}$ for which the system of equations \[a^2 + y^2 = b^2 + x^2 = (a - x)^2 + (b - y)^2\] has a solution in $(x,y)$ satisfying $0\le x < a$ and $0\le y < b$. Then $\rho^2$ can be expressed as a fraction $\frac {... | 7 | Notice that the given equation implies
$a^2 + y^2 = b^2 + x^2 = 2(ax + by)$
We have $2by \ge y^2$, so $2ax \le a^2 \implies x \le \frac {a}{2}$.
Then, notice $b^2 + x^2 = a^2 + y^2 \ge a^2$, so $b^2 \ge \frac {3}{4}a^2 \implies \rho^2 \le \frac {4}{3}$.
The solution $(a, b, x, y) = \left(1, \frac {\sqrt {3}}{2}, \fr... | 6.5 | [
7,
5,
7,
6,
7,
7,
7,
6
] |
Find the largest integer $n$ satisfying the following conditions:
(i) $n^2$ can be expressed as the difference of two consecutive cubes;
(ii) $2n + 79$ is a perfect square. | 181 | Let us generate numbers $1$ to $1000$ for the second condition, for squares. We know for $N$ to be integer, the squares must be odd. So we generate $N = 1, 21, 45, 73, 105, 141, 181, 381, 441, 721, 801$. $N$ cannot exceed $1000$ since it is AIME problem. Now take the first criterion, let $a$ be the smaller consecutive ... | 6.25 | [
5,
7,
6,
6,
7,
6,
6,
7
] |
Call a $3$-digit number geometric if it has $3$ distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric numbers. | 840 | The smallest geometric number is $124$ because $123$ and any number containing a zero does not work. $964$ is the largest geometric number because the middle digit cannot be 8 or 7. Subtracting the numbers gives $\boxed{840}.$
~IceMatrix
~Shreyas S | 3.75 | [
4,
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3,
4,
4,
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4,
3
] |
There is a complex number $z$ with imaginary part $164$ and a positive integer $n$ such that
\[\frac {z}{z + n} = 4i.\]
Find $n$. | 697 | \[\frac {z}{z+n}=4i\]
\[1-\frac {n}{z+n}=4i\]
\[1-4i=\frac {n}{z+n}\]
\[\frac {1}{1-4i}=\frac {z+n}{n}\]
\[\frac {1+4i}{17}=\frac {z}{n}+1\]
Since their imaginary part has to be equal,
\[\frac {4i}{17}=\frac {164i}{n}\]
\[n=\frac {(164)(17)}{4}=697\]
\[n = \boxed{697}.\] | 5.5 | [
4,
6,
6,
6,
5,
6,
6,
5
] |
A coin that comes up heads with probability $p > 0$ and tails with probability $1 - p > 0$ independently on each flip is flipped $8$ times. Suppose that the probability of three heads and five tails is equal to $\frac {1}{25}$ of the probability of five heads and three tails. Let $p = \frac {m}{n}$, where $m$ and $n$ a... | 11 | Rewrite it as : $(P)^3$$(1-P)^5=\frac {1}{25}$ $(P)^5$$(1-P)^3$
This can be simplified as $24P^2 -50P + 25 = 0$
This can be factored into $(4P-5)(6P-5)$
This yields two solutions: $\frac54$ (ignored because it would result in $1-p<0$ ) or $\frac56$
Therefore, the answer is $5+6$ = $\boxed{011}$
~IceMatrix
~Shreya... | 6 | [
6,
7,
6,
6,
5,
5,
7,
6
] |
In parallelogram $ABCD$, point $M$ is on $\overline{AB}$ so that $\frac {AM}{AB} = \frac {17}{1000}$ and point $N$ is on $\overline{AD}$ so that $\frac {AN}{AD} = \frac {17}{2009}$. Let $P$ be the point of intersection of $\overline{AC}$ and $\overline{MN}$. Find $\frac {AC}{AP}$. | 177 | One of the ways to solve this problem is to make this parallelogram a straight line. So the whole length of the line is $APC$($AMC$ or $ANC$), and $ABC$ is $1000x+2009x=3009x.$
$AP$($AM$ or $AN$) is $17x.$
So the answer is $3009x/17x = \boxed{177}$ | 4.375 | [
4,
4,
5,
4,
5,
4,
4,
5
] |
Triangle $ABC$ has $AC = 450$ and $BC = 300$. Points $K$ and $L$ are located on $\overline{AC}$ and $\overline{AB}$ respectively so that $AK = CK$, and $\overline{CL}$ is the angle bisector of angle $C$. Let $P$ be the point of intersection of $\overline{BK}$ and $\overline{CL}$, and let $M$ be the point on line $BK$ f... | 72 | Using the diagram above, we can solve this problem by using mass points. By angle bisector theorem: \[\frac{BL}{CB}=\frac{AL}{CA}\implies\frac{BL}{300}=\frac{AL}{450}\implies 3BL=2AL\] So, we can weight $A$ as $2$ and $B$ as $3$ and $L$ as $5$. Since $K$ is the midpoint of $A$ and $C$, the weight of $A$ is equal to the... | 6 | [
6,
6,
6,
6,
6,
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7,
5
] |
How many positive integers $N$ less than $1000$ are there such that the equation $x^{\lfloor x\rfloor} = N$ has a solution for $x$? | 412 | For a positive integer $k$, we find the number of positive integers $N$ such that $x^{\lfloor x\rfloor}=N$ has a solution with ${\lfloor x\rfloor}=k$. Then $x=\sqrt[k]{N}$, and because $k \le x < k+1$, we have $k^k \le x^k < (k+1)^k$, and because $(k+1)^k$ is an integer, we get $k^k \le x^k \le (k+1)^k-1$. The number o... | 6 | [
6,
6,
6,
6,
6,
6,
6,
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] |
The sequence $(a_n)$ satisfies $a_1 = 1$ and $5^{(a_{n + 1} - a_n)} - 1 = \frac {1}{n + \frac {2}{3}}$ for $n \geq 1$. Let $k$ be the least integer greater than $1$ for which $a_k$ is an integer. Find $k$. | 41 | We notice that by multiplying the equation from an arbitrary $a_n$ all the way to $a_1$, we get: \[5^{a_n-a_1}=\dfrac{n+\tfrac23}{1+\tfrac23}\] This simplifies to \[5^{a_n}=3n+2.\] We can now test powers of $5$.
$5$ - that gives us $n=1$, which is useless.
$25$ - that gives a non-integer $n$.
$125$ - that gives $n=\box... | 6 | [
6,
7,
6,
6,
5,
6,
6,
6
] |
Let $S = \{2^0,2^1,2^2,\ldots,2^{10}\}$. Consider all possible positive differences of pairs of elements of $S$. Let $N$ be the sum of all of these differences. Find the remainder when $N$ is divided by $1000$. | 398 | Find the positive differences in all $55$ pairs and you will get $\boxed{398}$. | 3.5 | [
4,
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3,
4,
3,
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3
] |
A game show offers a contestant three prizes A, B and C, each of which is worth a whole number of dollars from $$ 1$ to $$ 9999$ inclusive. The contestant wins the prizes by correctly guessing the price of each prize in the order A, B, C. As a hint, the digits of the three prices are given. On a particular day, the dig... | 420 | [Clarification: You are supposed to find the number of all possible tuples of prices, $(A, B, C)$, that could have been on that day.]
Since we have three numbers, consider the number of ways we can put these three numbers together in a string of 7 digits. For example, if $A=113, B=13, C=31$, then the string is
\[11313... | 5.5 | [
5,
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6,
5,
6,
5,
6,
5
] |
The Annual Interplanetary Mathematics Examination (AIME) is written by a committee of five Martians, five Venusians, and five Earthlings. At meetings, committee members sit at a round table with chairs numbered from $1$ to $15$ in clockwise order. Committee rules state that a Martian must occupy chair $1$ and an Earthl... | 346 | The arrangements must follow the pattern MVEMVE....... where each MVE consists of some Martians followed by some Venusians followed by some Earthlings, for $1, 2, 3, 4,$ or $5$ MVE's. If there are $k$ MVE's, then by stars and bars, there are ${4 \choose k-1}$ choices for the Martians in each block, and the same goes fo... | 6.375 | [
7,
6,
6,
7,
7,
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6,
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] |
Consider the set of all triangles $OPQ$ where $O$ is the origin and $P$ and $Q$ are distinct points in the plane with nonnegative integer coordinates $(x,y)$ such that $41x + y = 2009$. Find the number of such distinct triangles whose area is a positive integer. | 600 | Let the two points $P$ and $Q$ be defined with coordinates; $P=(x_1,y_1)$ and $Q=(x_2,y_2)$
We can calculate the area of the parallelogram with the determinant of the matrix of the coordinates of the two points(shoelace theorem).
$\det \left(\begin{array}{c} P \\ Q\end{array}\right)=\det \left(\begin{array}{cc}x_1 &y_... | 6 | [
6,
6,
6,
6,
6,
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6,
6
] |
In right $\triangle ABC$ with hypotenuse $\overline{AB}$, $AC = 12$, $BC = 35$, and $\overline{CD}$ is the altitude to $\overline{AB}$. Let $\omega$ be the circle having $\overline{CD}$ as a diameter. Let $I$ be a point outside $\triangle ABC$ such that $\overline{AI}$ and $\overline{BI}$ are both tangent to circle $\o... | 11 | This solution is not a real solution and is solving the problem with a ruler and compass.
Draw $AC = 4.8, BC = 14, AB = 14.8$. Then, drawing the tangents and intersecting them, we get that $IA$ is around $6.55$ and $IB$ is around $18.1$. We then find the ratio to be around $\frac{39.45}{14.8}$. Using long division, we... | 6.25 | [
6,
6,
6,
6,
7,
6,
6,
7
] |
The terms of the sequence $(a_i)$ defined by $a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}$ for $n \ge 1$ are positive integers. Find the minimum possible value of $a_1 + a_2$. | 90 | If $a_{n} \ne \frac {2009}{a_{n+1}}$, then either \[a_{n} = \frac {a_{n}}{1} < \frac {a_{n} + 2009}{1 + a_{n+1}} < \frac {2009}{a_{n+1}}\]
or
\[\frac {2009}{a_{n+1}} < \frac {2009 + a_{n}}{a_{n+1} + 1} < \frac {a_{n}}{1} = a_{n}\]
All the integers between $a_{n}$ and $\frac {2009}{a_{n+1}}$ would be included in the s... | 6.125 | [
6,
5,
6,
6,
7,
7,
6,
6
] |
For $t = 1, 2, 3, 4$, define $S_t = \sum_{i = 1}^{350}a_i^t$, where $a_i \in \{1,2,3,4\}$. If $S_1 = 513$ and $S_4 = 4745$, find the minimum possible value for $S_2$. | 905 | Because the order of the $a$'s doesn't matter, we simply need to find the number of $1$s $2$s $3$s and $4$s that minimize $S_2$. So let $w, x, y,$ and $z$ represent the number of $1$s, $2$s, $3$s, and $4$s respectively. Then we can write three equations based on these variables. Since there are a total of $350$ $a$s, w... | 5.625 | [
4,
6,
6,
6,
6,
6,
6,
5
] |
In triangle $ABC$, $AB = 10$, $BC = 14$, and $CA = 16$. Let $D$ be a point in the interior of $\overline{BC}$. Let points $I_B$ and $I_C$ denote the incenters of triangles $ABD$ and $ACD$, respectively. The circumcircles of triangles $BI_BD$ and $CI_CD$ meet at distinct points $P$ and $D$. The maximum possible area of ... | 150 | Proceed as in Solution 2 until you find $\angle CPB = 150$. The locus of points $P$ that give $\angle CPB = 150$ is a fixed arc from $B$ to $C$ ($P$ will move along this arc as $D$ moves along $BC$) and we want to maximise the area of [$\triangle BPC$]. This means we want $P$ to be farthest distance away from $BC$ as p... | 6.5 | [
6,
7,
7,
7,
7,
6,
6,
6
] |
Before starting to paint, Bill had $130$ ounces of blue paint, $164$ ounces of red paint, and $188$ ounces of white paint. Bill painted four equally sized stripes on a wall, making a blue stripe, a red stripe, a white stripe, and a pink stripe. Pink is a mixture of red and white, not necessarily in equal amounts. When ... | 114 | Just like in solution 1, we note that all colors will be used equally, except for the pink stripe. This must bring red and white down to $130$ each, so $34$ red and $58$ white are used, making for a total of $92$ for the pink stripe. Thus, the other stripes also use $92$. The answer is $130+164+188-4(92)=\boxed{114}$ | 3.875 | [
4,
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4,
4,
4,
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4
] |
Suppose that $a$, $b$, and $c$ are positive real numbers such that $a^{\log_3 7} = 27$, $b^{\log_7 11} = 49$, and $c^{\log_{11}25} = \sqrt{11}$. Find \[a^{(\log_3 7)^2} + b^{(\log_7 11)^2} + c^{(\log_{11} 25)^2}.\] | 469 | We know from the first three equations that $\log_a27$ = $\log_37$, $\log_b49$ = $\log_711$, and $\log_c\sqrt{11}$ = $\log_{11}25$. Substituting, we find
$a^{(\log_a27)(\log_37)} + b^{(\log_b49)(\log_711)} + c^{(\log_c\sqrt {11})(\log_{11}25)}$.
We know that $x^{\log_xy} =y$, so we find
$27^{\log_37} + 49^{\log_711} +... | 6.375 | [
7,
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6,
6,
6,
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] |
In rectangle $ABCD$, $AB=100$. Let $E$ be the midpoint of $\overline{AD}$. Given that line $AC$ and line $BE$ are perpendicular, find the greatest integer less than $AD$. | 141 | Let $x$ be the ratio of $BC$ to $AB$. On the coordinate plane, plot $A=(0,0)$, $B=(100,0)$, $C=(100,100x)$, and $D=(0,100x)$. Then $E=(0,50x)$. Furthermore, the slope of $\overline{AC}$ is $x$ and the slope of $\overline{BE}$ is $-x/2$. They are perpendicular, so they multiply to $-1$, that is, \[x\cdot-\frac{x}{2}=-1,... | 4 | [
4,
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4,
4,
4,
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] |
A group of children held a grape-eating contest. When the contest was over, the winner had eaten $n$ grapes, and the child in $k$-th place had eaten $n+2-2k$ grapes. The total number of grapes eaten in the contest was $2009$. Find the smallest possible value of $n$. | 89 | If the first child ate $n=2m$ grapes, then the maximum number of grapes eaten by all the children together is $2m + (2m-2) + (2m-4) + \cdots + 4 + 2 = m(m+1)$. Similarly, if the first child ate $2m-1$ grapes, the maximum total number of grapes eaten is $(2m-1)+(2m-3)+\cdots+3+1 = m^2$.
For $m=44$ the value $m(m+1)=44\... | 5.75 | [
5,
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6,
7,
5,
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6,
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] |
Equilateral triangle $T$ is inscribed in circle $A$, which has radius $10$. Circle $B$ with radius $3$ is internally tangent to circle $A$ at one vertex of $T$. Circles $C$ and $D$, both with radius $2$, are internally tangent to circle $A$ at the other two vertices of $T$. Circles $B$, $C$, and $D$ are all externally ... | 32 | Let $X$ be the intersection of the circles with centers $B$ and $E$, and $Y$ be the intersection of the circles with centers $C$ and $E$. Since the radius of $B$ is $3$, $AX =4$. Assume $AE$ = $p$. Then $EX$ and $EY$ are radii of circle $E$ and have length $4+p$. $AC = 8$, and angle $CAE = 60$ degrees because we are gi... | 6.875 | [
7,
7,
7,
7,
7,
6,
7,
7
] |
Let $m$ be the number of five-element subsets that can be chosen from the set of the first $14$ natural numbers so that at least two of the five numbers are consecutive. Find the remainder when $m$ is divided by $1000$. | 750 | We can use complementary counting. We can choose a five-element subset in ${14\choose 5}$ ways. We will now count those where no two numbers are consecutive. We will show a bijection between this set, and the set of 10-element strings that contain 5 $A$s and 5 $B$s, thereby showing that there are ${10\choose 5}$ such s... | 4.875 | [
5,
5,
4,
5,
5,
4,
6,
5
] |
Define $n!!$ to be $n(n-2)(n-4)\cdots 3\cdot 1$ for $n$ odd and $n(n-2)(n-4)\cdots 4\cdot 2$ for $n$ even. When $\sum_{i=1}^{2009} \frac{(2i-1)!!}{(2i)!!}$ is expressed as a fraction in lowest terms, its denominator is $2^ab$ with $b$ odd. Find $\dfrac{ab}{10}$. | 401 | Using the steps of the previous solution we get $c = \sum_{i=1}^{2009} {2i\choose i} \cdot 2^{2\cdot 2009 - 2i}$ and if you do the small cases(like $1, 2, 3, 4, 5, 6$) you realize that you can "thin-slice" the problem and simply look at the cases where $i=2009, 2008$(they're nearly identical in nature but one has $4$ w... | 7.25 | [
7,
7,
7,
7,
8,
8,
7,
7
] |
Dave rolls a fair six-sided die until a six appears for the first time. Independently, Linda rolls a fair six-sided die until a six appears for the first time. Let $m$ and $n$ be relatively prime positive integers such that $\dfrac mn$ is the probability that the number of times Dave rolls his die is equal to or within... | 41 | Let $p$ be the probability that the number of times Dave rolls his die is equal to or within one of the number of times Linda rolls her die. (We will call this event "a win", and the opposite event will be "a loss".)
Let both players roll their first die.
With probability $\frac 1{36}$, both throw a six and we win.
... | 5.625 | [
6,
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] |
Let $m$ be the number of solutions in positive integers to the equation $4x+3y+2z=2009$, and let $n$ be the number of solutions in positive integers to the equation $4x+3y+2z=2000$. Find the remainder when $m-n$ is divided by $1000$. | 0 | It is actually reasonably easy to compute $m$ and $n$ exactly.
First, note that if $4x+3y+2z=2009$, then $y$ must be odd. Let $y=2y'-1$. We get $4x + 6y' - 3 + 2z = 2009$, which simplifies to $2x + 3y' + z = 1006$. For any pair of positive integers $(x,y')$ such that $2x + 3y' < 1006$ we have exactly one $z$ such that... | 5.375 | [
6,
6,
5,
5,
5,
5,
5,
6
] |
Four lighthouses are located at points $A$, $B$, $C$, and $D$. The lighthouse at $A$ is $5$ kilometers from the lighthouse at $B$, the lighthouse at $B$ is $12$ kilometers from the lighthouse at $C$, and the lighthouse at $A$ is $13$ kilometers from the lighthouse at $C$. To an observer at $A$, the angle determined by ... | 96 | Let $O$ be the intersection of $BC$ and $AD$. By the Angle Bisector Theorem, $\frac {5}{BO}$ = $\frac {13}{CO}$, so $BO$ = $5x$ and $CO$ = $13x$, and $BO$ + $OC$ = $BC$ = $12$, so $x$ = $\frac {2}{3}$, and $OC$ = $\frac {26}{3}$. Let $P$ be the foot of the altitude from $D$ to $OC$. It can be seen that triangle $DOP$ i... | 6.875 | [
8,
8,
6,
6,
6,
7,
7,
7
] |
For certain pairs $(m,n)$ of positive integers with $m\geq n$ there are exactly $50$ distinct positive integers $k$ such that $|\log m - \log k| < \log n$. Find the sum of all possible values of the product $mn$. | 125 | We have $\log m - \log k = \log \left( \frac mk \right)$, hence we can rewrite the inequality as follows: \[- \log n < \log \left( \frac mk \right) < \log n\] We can now get rid of the logarithms, obtaining: \[\frac 1n < \frac mk < n\] And this can be rewritten in terms of $k$ as \[\frac mn < k < mn\]
From $k<mn$ it f... | 6.5 | [
6,
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7,
6,
7,
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] |
From the set of integers $\{1,2,3,\dots,2009\}$, choose $k$ pairs $\{a_i,b_i\}$ with $a_i<b_i$ so that no two pairs have a common element. Suppose that all the sums $a_i+b_i$ are distinct and less than or equal to $2009$. Find the maximum possible value of $k$. | 803 | Suppose that we have a valid solution with $k$ pairs. As all $a_i$ and $b_i$ are distinct, their sum is at least $1+2+3+\cdots+2k=k(2k+1)$. On the other hand, as the sum of each pair is distinct and at most equal to $2009$, the sum of all $a_i$ and $b_i$ is at most $2009 + (2009-1) + \cdots + (2009-(k-1)) = \frac{k(401... | 7.125 | [
7,
7,
7,
7,
7,
7,
7,
8
] |
Let $A$ and $B$ be the endpoints of a semicircular arc of radius $2$. The arc is divided into seven congruent arcs by six equally spaced points $C_1$, $C_2$, $\dots$, $C_6$. All chords of the form $\overline {AC_i}$ or $\overline {BC_i}$ are drawn. Let $n$ be the product of the lengths of these twelve chords. Find the ... | 672 | Let $O$ be the midpoint of $A$ and $B$. Assume $C_1$ is closer to $A$ instead of $B$. $\angle AOC_1$ = $\frac {\pi}{7}$. Using the Law of Cosines,
$\overline {AC_1}^2$ = $8 - 8 \cos \frac {\pi}{7}$, $\overline {AC_2}^2$ = $8 - 8 \cos \frac {2\pi}{7}$, . . . $\overline {AC_6}^2$ = $8 - 8 \cos \frac {6\pi}{7}$
So $n$ = ... | 6.75 | [
7,
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7,
6,
7,
7,
6,
7
] |
The sequence $(a_n)$ satisfies $a_0=0$ and $a_{n + 1} = \frac85a_n + \frac65\sqrt {4^n - a_n^2}$ for $n\geq 0$. Find the greatest integer less than or equal to $a_{10}$. | 983 | We can now simply start to compute the values $b_i$ by hand:
\begin{align*} b_1 & = \frac 35 \\ b_2 & = \frac 45\cdot \frac 35 + \frac 35 \sqrt{1 - \left(\frac 35\right)^2} = \frac{24}{25} \\ b_3 & = \frac 45\cdot \frac {24}{25} + \frac 35 \sqrt{1 - \left(\frac {24}{25}\right)^2} = \frac{96}{125} + \frac 35\cdot\frac 7... | 7 | [
7,
7,
7,
7,
7,
7,
7,
7
] |
Let $\overline{MN}$ be a diameter of a circle with diameter 1. Let $A$ and $B$ be points on one of the semicircular arcs determined by $\overline{MN}$ such that $A$ is the midpoint of the semicircle and $MB=\frac{3}5$. Point $C$ lies on the other semicircular arc. Let $d$ be the length of the line segment whose endpoin... | 14 | By Pythagoras in $\triangle BMN,$ we get $BN=\dfrac{4}{5}.$
Since cross ratios are preserved upon projecting, note that $(M,Y;X,N)\stackrel{C}{=}(M,B;A,N).$ By definition of a cross ratio, this becomes \[\dfrac{XM}{NY}:\dfrac{NM}{NY}=\dfrac{AM}{AB}:\dfrac{MN}{NB}.\] Let $MY=a,YX=b,XN=c$ such that $a+b+c=1.$ We know th... | 7.125 | [
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7,
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] |
Maya lists all the positive divisors of $2010^2$. She then randomly selects two distinct divisors from this list. Let $p$ be the probability that exactly one of the selected divisors is a perfect square. The probability $p$ can be expressed in the form $\frac {m}{n}$, where $m$ and $n$ are relatively prime positive int... | 107 | $2010^2 = 2^2\cdot3^2\cdot5^2\cdot67^2$. Thus there are $(2+1)^4$ divisors, $(1+1)^4$ of which are squares (the exponent of each prime factor must either be $0$ or $2$). Therefore the probability is \[\frac {2\cdot2^4\cdot(3^4 - 2^4)}{3^4(3^4 - 1)} = \frac {26}{81} \Longrightarrow 26+ 81 = \boxed{107}.\] | 5.375 | [
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5,
6,
5
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Find the remainder when $9 \times 99 \times 999 \times \cdots \times \underbrace{99\cdots9}_{\text{999 9's}}$ is divided by $1000$. | 109 | Note that $999\equiv 9999\equiv\dots \equiv\underbrace{99\cdots9}_{\text{999 9's}}\equiv -1 \pmod{1000}$ (see modular arithmetic). That is a total of $999 - 3 + 1 = 997$ integers, so all those integers multiplied out are congruent to $- 1\pmod{1000}$. Thus, the entire expression is congruent to $- 1\times9\times99 = - ... | 4.875 | [
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Suppose that $y = \frac34x$ and $x^y = y^x$. The quantity $x + y$ can be expressed as a rational number $\frac {r}{s}$, where $r$ and $s$ are relatively prime positive integers. Find $r + s$. | 529 | Substitute $y = \frac34x$ into $x^y = y^x$ and solve. \[x^{\frac34x} = \left(\frac34x\right)^x\] \[x^{\frac34x} = \left(\frac34\right)^x \cdot x^x\] \[x^{-\frac14x} = \left(\frac34\right)^x\] \[x^{-\frac14} = \frac34\] \[x = \frac{256}{81}\] \[y = \frac34x = \frac{192}{81}\] \[x + y = \frac{448}{81}\] \[448 + 81 = \box... | 5.125 | [
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Jackie and Phil have two fair coins and a third coin that comes up heads with probability $\frac47$. Jackie flips the three coins, and then Phil flips the three coins. Let $\frac {m}{n}$ be the probability that Jackie gets the same number of heads as Phil, where $m$ and $n$ are relatively prime positive integers. Find ... | 515 | This can be solved quickly and easily with generating functions.
Let $x^n$ represent flipping $n$ heads.
The generating functions for these coins are $(1+x)$,$(1+x)$,and $(3+4x)$ in order.
The product is $3+10x+11x^2+4x^3$. ($ax^n$ means there are $a$ ways to get $n$ heads, eg there are $10$ ways to get $1$ head, an... | 6.125 | [
6,
6,
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Positive integers $a$, $b$, $c$, and $d$ satisfy $a > b > c > d$, $a + b + c + d = 2010$, and $a^2 - b^2 + c^2 - d^2 = 2010$. Find the number of possible values of $a$. | 501 | Since $a+b$ must be greater than $1005$, it follows that the only possible value for $a-b$ is $1$ (otherwise the quantity $a^2 - b^2$ would be greater than $2010$). Therefore the only possible ordered pairs for $(a,b)$ are $(504, 503)$, $(505, 504)$, ... , $(1004, 1003)$, so $a$ has $\boxed{501}$ possible values. | 4.875 | [
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Let $P(x)$ be a quadratic polynomial with real coefficients satisfying $x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3$ for all real numbers $x$, and suppose $P(11) = 181$. Find $P(16)$. | 406 | Let $Q(x) = P(x) - (x^2-2x+2)$, then $0\le Q(x) \le (x-1)^2$ (note this is derived from the given inequality chain). Therefore, $0\le Q(x+1) \le x^2 \Rightarrow Q(x+1) = Ax^2$ for some real value A.
$Q(11) = 10^2A \Rightarrow P(11)-(11^2-22+2)=100A \Rightarrow 80=100A \Rightarrow A=\frac{4}{5}$.
$Q(16)=15^2A=180 \Right... | 6.125 | [
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Define an ordered triple $(A, B, C)$ of sets to be minimally intersecting if $|A \cap B| = |B \cap C| = |C \cap A| = 1$ and $A \cap B \cap C = \emptyset$. For example, $(\{1,2\},\{2,3\},\{1,3,4\})$ is a minimally intersecting triple. Let $N$ be the number of minimally intersecting ordered triples of sets for which each... | 760 | Let each pair of two sets have one element in common. Label the common elements as $x$, $y$, $z$. Set $A$ will have elements $x$ and $y$, set $B$ will have $y$ and $z$, and set $C$ will have $x$ and $z$. There are $7 \cdot 6 \cdot 5 = 210$ ways to choose values of $x$, $y$ and $z$. There are $4$ unpicked numbers, and e... | 6.625 | [
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For a real number $a$, let $\lfloor a \rfloor$ denote the greatest integer less than or equal to $a$. Let $\mathcal{R}$ denote the region in the coordinate plane consisting of points $(x,y)$ such that $\lfloor x \rfloor ^2 + \lfloor y \rfloor ^2 = 25$. The region $\mathcal{R}$ is completely contained in a disk of radiu... | 132 | The desired region consists of 12 boxes, whose lower-left corners are integers solutions of $x^2 + y^2 = 25$, namely $(\pm5,0), (0,\pm5), (\pm3,\pm4), (\pm4,\pm3).$ Since the points themselves are symmetric about $(0,0)$, the boxes are symmetric about $\left(\frac12,\frac12\right)$. The distance from $\left(\frac12,\fr... | 6.375 | [
6,
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Let $(a,b,c)$ be the real solution of the system of equations $x^3 - xyz = 2$, $y^3 - xyz = 6$, $z^3 - xyz = 20$. The greatest possible value of $a^3 + b^3 + c^3$ can be written in the form $\frac {m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 158 | Add the three equations to get $a^3 + b^3 + c^3 = 28 + 3abc$. Now, let $abc = p$. $a = \sqrt [3]{p + 2}$, $b = \sqrt [3]{p + 6}$ and $c = \sqrt [3]{p + 20}$, so $p = abc = (\sqrt [3]{p + 2})(\sqrt [3]{p + 6})(\sqrt [3]{p + 20})$. Now cube both sides; the $p^3$ terms cancel out. Solve the remaining quadratic to get $p =... | 6.625 | [
6,
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Let $N$ be the number of ways to write $2010$ in the form $2010 = a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$, where the $a_i$'s are integers, and $0 \le a_i \le 99$. An example of such a representation is $1\cdot 10^3 + 3\cdot 10^2 + 67\cdot 10^1 + 40\cdot 10^0$. Find $N$. | 202 | If we choose $a_3$ and $a_1$ such that $(10^3)(a_3) + (10)(a_1) \leq 2010$ there is a unique choice of $a_2$ and $a_0$ that makes the equality hold. So $N$ is just the number of combinations of $a_3$ and $a_1$ we can pick. If $a_3 = 0$ or $a_3 = 1$ we can let $a_1$ be anything from $0$ to $99$. If $a_3 = 2$ then $a_1 =... | 4.625 | [
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Let $\mathcal{R}$ be the region consisting of the set of points in the coordinate plane that satisfy both $|8 - x| + y \le 10$ and $3y - x \ge 15$. When $\mathcal{R}$ is revolved around the line whose equation is $3y - x = 15$, the volume of the resulting solid is $\frac {m\pi}{n\sqrt {p}}$, where $m$, $n$, and $p$ are... | 365 | [asy]size(280); import graph; real min = 2, max = 12; pen dark = linewidth(1); real P(real x) { return x/3 + 5; } real Q(real x) { return 10 - abs(x - 8); } path p = (2,P(2))--(8,P(8))--(12,P(12)), q = (2,Q(2))--(12,Q(12)); pair A = (8,10), B = (4.5,6.5), C= (9.75,8.25), F=foot(A,B,C), G=2*F-A; fill(A--B--C--cycle,rgb(... | 7.125 | [
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Let $m \ge 3$ be an integer and let $S = \{3,4,5,\ldots,m\}$. Find the smallest value of $m$ such that for every partition of $S$ into two subsets, at least one of the subsets contains integers $a$, $b$, and $c$ (not necessarily distinct) such that $ab = c$.
Note: a partition of $S$ is a pair of sets $A$, $B$ such that... | 243 | We claim that $243$ is the minimal value of $m$. Let the two partitioned sets be $A$ and $B$; we will try to partition $3, 9, 27, 81,$ and $243$ such that the $ab=c$ condition is not satisfied. Without loss of generality, we place $3$ in $A$. Then $9$ must be placed in $B$, so $81$ must be placed in $A$, and $27$ must ... | 6.5 | [
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Rectangle $ABCD$ and a semicircle with diameter $AB$ are coplanar and have nonoverlapping interiors. Let $\mathcal{R}$ denote the region enclosed by the semicircle and the rectangle. Line $\ell$ meets the semicircle, segment $AB$, and segment $CD$ at distinct points $N$, $U$, and $T$, respectively. Line $\ell$ divides ... | 69 | Once we establish that $\Delta ANO$ is equilateral, we have \[[{\rm Sector } BON] = 2[{\rm Sector } AON], [BCT'U]=2[ADT'U]\] \[\Rightarrow [\overset{\large\frown}{NB} CT'UO]=2[\overset{\large\frown}{NA} DT'UO]\] On the other hand, \[[\overset{\large\frown}{NB} CT]=2[\overset{\large\frown}{NA} DT]\] Therefore, $[UT'T]=[... | 6.875 | [
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For each positive integer n, let $f(n) = \sum_{k = 1}^{100} \lfloor \log_{10} (kn) \rfloor$. Find the largest value of $n$ for which $f(n) \le 300$.
Note: $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$. | 109 | Observe that $f$ is strictly increasing in $n$. We realize that we need $100$ terms to add up to around $300$, so we need some sequence of $2$s, $3$s, and then $4$s.
It follows that $n \approx 100$ (alternatively, use binary search to get to this, with $n\le 1000$). Manually checking shows that $f(109) = 300$ and $f(1... | 5.5 | [
6,
6,
5,
5,
6,
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5
] |
In $\triangle{ABC}$ with $AB = 12$, $BC = 13$, and $AC = 15$, let $M$ be a point on $\overline{AC}$ such that the incircles of $\triangle{ABM}$ and $\triangle{BCM}$ have equal radii. Then $\frac{AM}{CM} = \frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$. | 45 | Let $CM=x, AM=rx, BM=d$. $x+rx=15\Rightarrow x=\frac{15}{1+r}$.
Similar to Solution 1, we have \[r=\frac{[AMB]}{[CMB]}=\frac{12+rx+d}{13+x+d} \Rightarrow d=\frac{13r-12}{1-r}\] as well as \[12^2\cdot x + 13^2 rx=15x\cdot rx+15d^2 (\text{via Stewart's Theorem})\] \[\frac{(12^2 + 13^2r) \cdot 15}{1+r} - \frac{15r\cdot 1... | 6.875 | [
7,
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7,
7,
7,
7
] |
Let $N$ be the greatest integer multiple of $36$ all of whose digits are even and no two of whose digits are the same. Find the remainder when $N$ is divided by $1000$. | 640 | If an integer is divisible by $36$, it must also be divisible by $9$ since $9$ is a factor of $36$. It is a well-known fact that, if $N$ is divisible by $9$, the sum of the digits of $N$ is a multiple of $9$. Hence, if $N$ contains all the even digits, the sum of the digits would be $0 + 2 + 4 + 6 + 8 = 20$, which is n... | 5.125 | [
5,
5,
5,
5,
4,
6,
6,
5
] |
A point $P$ is chosen at random in the interior of a unit square $S$. Let $d(P)$ denote the distance from $P$ to the closest side of $S$. The probability that $\frac{1}{5}\le d(P)\le\frac{1}{3}$ is equal to $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 281 | First, let's figure out $d(P) \geq \frac{1}{3}$ which is\[\left(\frac{3}{5}\right)^2=\frac{9}{25}.\]Then, $d(P) \geq \frac{1}{5}$ is a square inside $d(P) \geq \frac{1}{3}$, so\[\left(\frac{1}{3}\right)^2=\frac{1}{9}.\]Therefore, the probability that $\frac{1}{5}\le d(P)\le\frac{1}{3}$ is\[\frac{9}{25}-\frac{1}{9}=\fra... | 5.375 | [
5,
5,
5,
6,
5,
5,
6,
6
] |
Let $K$ be the product of all factors $(b-a)$ (not necessarily distinct) where $a$ and $b$ are integers satisfying $1\le a < b \le 20$. Find the greatest positive integer $n$ such that $2^n$ divides $K$. | 150 | In general, there are $20-n$ pairs of integers $(a, b)$ that differ by $n$ because we can let $b$ be any integer from $n+1$ to $20$ and set $a$ equal to $b-n$. Thus, the product is $(1^{19})(2^{18})\cdots(19^1)$ (or alternatively, $19! \cdot 18! \cdots 1!$.)
When we count the number of factors of $2$, we have 4 groups... | 6.375 | [
6,
7,
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6,
6,
6,
7,
7
] |
Dave arrives at an airport which has twelve gates arranged in a straight line with exactly $100$ feet between adjacent gates. His departure gate is assigned at random. After waiting at that gate, Dave is told the departure gate has been changed to a different gate, again at random. Let the probability that Dave walks $... | 52 | There are $12 \cdot 11 = 132$ possible situations ($12$ choices for the initially assigned gate, and $11$ choices for which gate Dave's flight was changed to). We are to count the situations in which the two gates are at most $400$ feet apart.
If we number the gates $1$ through $12$, then gates $1$ and $12$ have four ... | 3.875 | [
4,
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4,
4,
4,
4
] |
Positive numbers $x$, $y$, and $z$ satisfy $xyz = 10^{81}$ and $(\log_{10}x)(\log_{10} yz) + (\log_{10}y) (\log_{10}z) = 468$. Find $\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}$. | 75 | Let $a=\log_{10}x$, $b=\log_{10}y,$ and $c=\log_{10}z$.
We have $a+b+c=81$ and $a(b+c)+bc=ab+ac+bc=468$. Since these two equations look a lot like Vieta's for a cubic, create the polynomial $x^3-81x^2+468x=0$ (leave the constant term as $0$ to make things easy). Dividing by $x$ yields $x^2-81x+468=0$.
Now we use the ... | 6.25 | [
6,
7,
6,
7,
6,
6,
6,
6
] |
Find the smallest positive integer $n$ with the property that the polynomial $x^4 - nx + 63$ can be written as a product of two nonconstant polynomials with integer coefficients. | 8 | You can factor the polynomial into two quadratic factors or a linear and a cubic factor.
For two quadratic factors, let $x^2+ax+b$ and $x^2+cx+d$ be the two quadratics, so that
\[(x^2 + ax + b )(x^2 + cx + d) = x^4 + (a + c)x^3 + (b + d + ac)x^2 + (ad + bc)x + bd.\]
Therefore, again setting coefficients equal, $a + c... | 4.75 | [
5,
5,
5,
5,
5,
4,
4,
5
] |
Let $P(z)=z^3+az^2+bz+c$, where a, b, and c are real. There exists a complex number $w$ such that the three roots of $P(z)$ are $w+3i$, $w+9i$, and $2w-4$, where $i^2=-1$. Find $|a+b+c|$. | 136 | Set $w=x+yi$, so $x_1 = x+(y+3)i$, $x_2 = x+(y+9)i$, $x_3 = 2x-4+2yi$.
Since $a,b,c\in{R}$, the imaginary part of $a,b,c$ must be $0$.
Start with a, since it's the easiest one to do: $y+3+y+9+2y=0, y=-3$,
and therefore: $x_1 = x$, $x_2 = x+6i$, $x_3 = 2x-4-6i$.
Now, do the part where the imaginary part of c is 0 si... | 6.375 | [
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6,
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6,
7
] |
Let $N$ be the number of ordered pairs of nonempty sets $\mathcal{A}$ and $\mathcal{B}$ that have the following properties:
$\mathcal{A} \cup \mathcal{B} = \{1,2,3,4,5,6,7,8,9,10,11,12\}$,
$\mathcal{A} \cap \mathcal{B} = \emptyset$,
The number of elements of $\mathcal{A}$ is not an element of $\mathcal{A}$,
The number ... | 772 | Regardless of the size $n$ of $A$ (ignoring the case when $n = 6$), $n$ must not be in $A$ and $12 - n$ must be in $A$.
There are $10$ remaining elements whose placements have yet to be determined. Note that the actual value of $n$ does not matter; there is always $1$ necessary element, $1$ forbidden element, and $10$... | 6.375 | [
6,
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7,
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6
] |
Let $ABCDEF$ be a regular hexagon. Let $G$, $H$, $I$, $J$, $K$, and $L$ be the midpoints of sides $AB$, $BC$, $CD$, $DE$, $EF$, and $AF$, respectively. The segments $\overline{AH}$, $\overline{BI}$, $\overline{CJ}$, $\overline{DK}$, $\overline{EL}$, and $\overline{FG}$ bound a smaller regular hexagon. Let the ratio of ... | 11 | Let $BC=2$ (without loss of generality).
Note that $\angle BMH$ is the vertical angle to an angle of regular hexagon, and so has degree $120^\circ$.
Because $\triangle ABH$ and $\triangle BCI$ are rotational images of one another, we get that $\angle{MBH}=\angle{HAB}$ and hence $\triangle ABH \sim \triangle BMH \sim ... | 5.875 | [
6,
6,
6,
6,
6,
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5,
6
] |
Find the number of second-degree polynomials $f(x)$ with integer coefficients and integer zeros for which $f(0)=2010$. | 163 | We use Burnside's Lemma. The set being acted upon is the set of integer triples $(a,r,s)$ such that $ars=2010$. Because $r$ and $s$ are indistinguishable, the permutation group consists of the identity and the permutation that switches $r$ and $s$. In cycle notation, the group consists of $(a)(r)(s)$ and $(a)(r \: s)$.... | 6.5 | [
6,
7,
7,
7,
7,
6,
6,
6
] |
Define a T-grid to be a $3\times3$ matrix which satisfies the following two properties:
Exactly five of the entries are $1$'s, and the remaining four entries are $0$'s.
Among the eight rows, columns, and long diagonals (the long diagonals are $\{a_{13},a_{22},a_{31}\}$ and $\{a_{11},a_{22},a_{33}\}$, no more than one o... | 68 | The T-grid can be considered as a tic-tac-toe board: five $1$'s (or X's) and four $0$'s (or O's).
There are only $\dbinom{9}{5} = 126$ ways to fill the board with five $1$'s and four $0$'s. Now we just need to subtract the number of bad grids. Bad grids are ones with more than one person winning, or where someone has ... | 6.25 | [
6,
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6,
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6,
7,
6
] |
Two noncongruent integer-sided isosceles triangles have the same perimeter and the same area. The ratio of the lengths of the bases of the two triangles is $8: 7$. Find the minimum possible value of their common perimeter. | 676 | Let the first triangle have sides $16n,a,a$, so the second has sides $14n,a+n,a+n$. The height of the first triangle is $\frac{7}{8}$ the height of the second triangle. Therefore, we have \[a^2-64n^2=\frac{49}{64}((a+n)^2-49n^2).\] Multiplying this, we get \[64a^2-4096n^2=49a^2+98an-2352n^2,\] which simplifies to \[15a... | 6.625 | [
7,
7,
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] |
The $52$ cards in a deck are numbered $1, 2, \cdots, 52$. Alex, Blair, Corey, and Dylan each picks a card from the deck without replacement and with each card being equally likely to be picked, The two persons with lower numbered cards form a team, and the two persons with higher numbered cards form another team. Let $... | 263 | Once the two cards are drawn, there are $\dbinom{50}{2} = 1225$ ways for the other two people to draw. Alex and Dylan are the team with higher numbers if Blair and Corey both draw below $a$, which occurs in $\dbinom{a-1}{2}$ ways. Alex and Dylan are the team with lower numbers if Blair and Corey both draw above $a+9$, ... | 6.875 | [
7,
7,
7,
7,
7,
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6
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Triangle $ABC$ with right angle at $C$, $\angle BAC < 45^\circ$ and $AB = 4$. Point $P$ on $\overline{AB}$ is chosen such that $\angle APC = 2\angle ACP$ and $CP = 1$. The ratio $\frac{AP}{BP}$ can be represented in the form $p + q\sqrt{r}$, where $p$, $q$, $r$ are positive integers and $r$ is not divisible by the squa... | 7 | Let $\angle{ACP}$ be equal to $x$. Then by Law of Sines, $PB = -\frac{\cos{x}}{\cos{3x}}$ and $AP = \frac{\sin{x}}{\sin{3x}}$. We then obtain $\cos{3x} = 4\cos^3{x} - 3\cos{x}$ and $\sin{3x} = 3\sin{x} - 4\sin^3{x}$. Solving, we determine that $\sin^2{x} = \frac{4 \pm \sqrt{2}}{8}$. Plugging this in gives that $\frac{A... | 6.625 | [
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