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 |
|---|---|---|---|---|
Find all positive integers $n$ such that there exists a sequence of positive integers $a_1$, $a_2$,$\ldots$, $a_n$ satisfying: \[a_{k+1}=\frac{a_k^2+1}{a_{k-1}+1}-1\] for every $k$ with $2\leq k\leq n-1$.
[i] | n=1,2,3,4 |
To solve the given problem, we need to determine all positive integers \( n \) such that there exists a sequence of positive integers \( a_1, a_2, \ldots, a_n \) satisfying the recurrence relation:
\[
a_{k+1} = \frac{a_k^2 + 1}{a_{k-1} + 1} - 1
\]
for every \( k \) where \( 2 \leq k \leq n-1 \).
### Step-by-step So... | 6 | [
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] |
For each interger $n\geq 4$, we consider the $m$ subsets $A_1, A_2,\dots, A_m$ of $\{1, 2, 3,\dots, n\}$, such that
$A_1$ has exactly one element, $A_2$ has exactly two elements,...., $A_m$ has exactly $m$ elements and none of these subsets is contained in any other set. Find the maximum value of $m$. | \left\lfloor \frac{-1 + \sqrt{1 + 8n}}{2} \right\rfloor |
To find the maximum value of \( m \), we need to evaluate the constraints given by the problem, specifically that none of the subsets \( A_1, A_2, \ldots, A_m \) is contained in any other subset. Each subset \( A_i \) has exactly \( i \) elements.
Let's consider the key points to derive the maximum value of \( m \):
... | 6.125 | [
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6,
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6,
6
] |
For what real values of $k>0$ is it possible to dissect a $1 \times k$ rectangle into two similar, but noncongruent, polygons? | $k \ne 1$ |
Given a \(1 \times k\) rectangle, we want to determine for which real values of \(k > 0\) it is possible to dissect the rectangle into two similar, but noncongruent, polygons.
First, let's understand the requirements: two polygons are similar if their corresponding angles are equal and their corresponding sides are i... | 5.75 | [
6,
6,
6,
6,
5,
7,
5,
5
] |
For an arbitrary positive integer $m$, not divisible by $3$, consider the permutation $x \mapsto 3x \pmod{m}$ on the set $\{ 1,2,\dotsc ,m-1\}$. This permutation can be decomposed into disjointed cycles; for instance, for $m=10$ the cycles are $(1\mapsto 3\to 9,\mapsto 7,\mapsto 1)$, $(2\mapsto 6\mapsto 8\mapsto 4\maps... | m \equiv 2, 5, 7, 10 \pmod{12} |
Given a positive integer \( m \), not divisible by 3, we are interested in the permutation \( x \mapsto 3x \pmod{m} \) on the set \(\{ 1, 2, \dotsc, m-1 \}\). The task is to determine for which integers \( m \) the number of cycles in this permutation is odd.
### Understanding the Problem
For a permutation \( \sigma... | 6.75 | [
6,
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7,
7,
7,
7,
6,
7
] |
Solve in integers the system of equations: $$x^2-y^2=z$$
$$3xy+(x-y)z=z^2$$ | {(2, 1, 3), (1, 2, -3), (1, 0, 1), (0, 1, -1), (0, 0, 0)} |
To solve the system of equations in integers:
\[
x^2 - y^2 = z
\]
\[
3xy + (x-y)z = z^2
\]
we begin by analyzing the first equation. The expression \( x^2 - y^2 = z \) can be factored as:
\[
(x+y)(x-y) = z
\]
Substituting this into the second equation, we have:
\[
3xy + (x-y)z = z^2
\]
Given the expression for \( z \... | 5.875 | [
6,
6,
6,
6,
6,
6,
6,
5
] |
Solve the following system of equations in integer numbers:
$$\begin{cases} x^2 = yz + 1 \\ y^2 = zx + 1 \\ z^2 = xy + 1 \end{cases}$$ | (1, 0, -1) |
To solve the given system of equations in integer numbers:
\[
\begin{cases}
x^2 = yz + 1 \\
y^2 = zx + 1 \\
z^2 = xy + 1
\end{cases}
\]
we need to find integer solutions \((x, y, z)\).
### Analysis
First, consider the symmetry of the problem; each equation is structurally similar, suggesting potential symmetry i... | 6.25 | [
6,
6,
6,
7,
6,
7,
6,
6
] |
Solve in integers the equation
\[ x^2+xy+y^2 = \left(\frac{x+y}{3}+1\right)^3. \] | (-1,1)(3,3)(19,-1)(53, -17) |
To solve the integer equation
\[
x^2 + xy + y^2 = \left(\frac{x+y}{3} + 1\right)^3,
\]
let us explore potential integer solutions by breaking down the equation and using substitution techniques for easier handling.
### Step 1: Simplify the Equation
The right-hand side of the equation contains a cubic term involvi... | 6.625 | [
7,
6,
7,
7,
6,
7,
6,
7
] |
Can three persons, having one double motorcycle, overcome the distance of $70$ km in $3$ hours? Pedestrian speed is $5$ km / h and motorcycle speed is $50$ km / h. | \text{No} |
Consider the scenario where three persons need to cover a distance of 70 km within 3 hours using one double-seater motorcycle. The speeds involved are pedestrian speed of 5 km/h and motorcycle speed of 50 km/h.
To determine whether this is possible, let us calculate the maximum possible distance covered by any combin... | 3.875 | [
4,
3,
4,
4,
4,
4,
4,
4
] |
Let $OX, OY$ and $OZ$ be three rays in the space, and $G$ a point "[i]between these rays[/i]" (i. e. in the interior of the part of the space bordered by the angles $Y OZ, ZOX$ and $XOY$). Consider a plane passing through $G$ and meeting the rays $OX, OY$ and $OZ$ in the points $A, B, C$, respectively. There are infini... |
To solve for the plane that minimizes the volume of the tetrahedron \( OABC \), where the plane meets the rays \( OX, OY, \) and \( OZ \) at points \( A, B, \) and \( C \) respectively, we need to strategically place these intersection points. To achieve the minimum volume for the tetrahedron \( OABC \), we should mak... | 6.625 | [
7,
6,
7,
6,
7,
7,
7,
6
] | |
Let $n$ be a given positive integer. Solve the system
\[x_1 + x_2^2 + x_3^3 + \cdots + x_n^n = n,\]
\[x_1 + 2x_2 + 3x_3 + \cdots + nx_n = \frac{n(n+1)}{2}\]
in the set of nonnegative real numbers. | (x_1, x_2, \ldots, x_n) = (1, 1, \ldots, 1) |
To solve the given system of equations:
\[
x_1 + x_2^2 + x_3^3 + \cdots + x_n^n = n,
\]
\[
x_1 + 2x_2 + 3x_3 + \cdots + nx_n = \frac{n(n+1)}{2},
\]
with \( x_i \geq 0 \) for all \( i \) and \( x_i \) being nonnegative real numbers, we will demonstrate that the only solution is
\[
(x_1, x_2, \ldots, x_n) = (1, 1, \... | 6 | [
6,
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6,
6,
6,
6,
6,
6
] |
Are there integers $m$ and $n$ such that
\[5m^2 - 6mn + 7n^2 = 1985 \ ?\] | \text{No} |
To determine whether there are integers \( m \) and \( n \) such that
\[
5m^2 - 6mn + 7n^2 = 1985,
\]
we begin by analyzing the quadratic form. We can rewrite the equation as:
\[
5m^2 - 6mn + 7n^2.
\]
First, let's complete the square with respect to \( m \) in the expression \( 5m^2 - 6mn \):
1. Factor out the co... | 4.75 | [
4,
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4,
6,
6,
4,
4,
5
] |
An infinite sequence $ \,x_{0},x_{1},x_{2},\ldots \,$ of real numbers is said to be [b]bounded[/b] if there is a constant $ \,C\,$ such that $ \, \vert x_{i} \vert \leq C\,$ for every $ \,i\geq 0$. Given any real number $ \,a > 1,\,$ construct a bounded infinite sequence $ x_{0},x_{1},x_{2},\ldots \,$ such that
\[ \ver... | 1 |
To solve this problem, we need to construct a bounded sequence of real numbers \( x_0, x_1, x_2, \ldots \) such that for any two distinct nonnegative integers \( i \) and \( j \), the condition
\[
|x_i - x_j| \cdot |i - j|^a \geq 1
\]
is satisfied, given \( a > 1 \).
### Step-by-step Solution
1. **Defining the Seq... | 6.5 | [
8,
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6,
6,
7,
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8,
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] |
Solve in the set of real numbers the equation \[ 3x^3 \minus{} [x] \equal{} 3,\] where $ [x]$ denotes the integer part of $ x.$ | x = \sqrt [3]{\frac {4}{3}} | To solve the equation \( 3x^3 - [x] = 3 \), where \([x]\) represents the integer part of \(x\), let's outline the steps systematically.
### Step 1: Analyze the Equation
Given the equation
\[
3x^3 - [x] = 3,
\]
we need to analyze how \([x]\) (the greatest integer less than or equal to \(x\)) interacts with \(3x^3\).
... | 4.75 | [
5,
4,
4,
6,
5,
4,
5,
5
] |
Consider the following transformation of the Cartesian plane: choose a lattice point and rotate the plane $90^\circ$ counterclockwise about that lattice point. Is it possible, through a sequence of such transformations, to take the triangle with vertices $(0,0)$, $(1,0)$ and $(0,1)$ to the triangle with vertices $(0,0)... | $\text { No }$ |
To determine if it is possible to transform the triangle with vertices \((0,0)\), \((1,0)\), and \((0,1)\) into the triangle with vertices \((0,0)\), \((1,0)\), and \((1,1)\) through a sequence of 90° counterclockwise rotations about lattice points, we analyze the effects of such rotations on the plane.
### Step 1: U... | 6.5 | [
7,
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5,
7,
7,
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7,
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] |
Solve in positive real numbers: $n+ \lfloor \sqrt{n} \rfloor+\lfloor \sqrt[3]{n} \rfloor=2014$ | 1958 |
We are asked to solve the equation \( n+ \lfloor \sqrt{n} \rfloor + \lfloor \sqrt[3]{n} \rfloor = 2014 \) for positive real numbers \( n \).
To begin, we denote:
- \( x = \lfloor \sqrt{n} \rfloor \),
- \( y = \lfloor \sqrt[3]{n} \rfloor \).
Thus, we have:
\[
x \leq \sqrt{n} < x+1
\]
\[
y \leq \sqrt[3]{n} < y+1
\]
T... | 5.375 | [
5,
5,
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4,
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6,
6
] |
Three persons $A,B,C$, are playing the following game:
A $k$-element subset of the set $\{1, . . . , 1986\}$ is randomly chosen, with an equal probability of each choice, where $k$ is a fixed positive integer less than or equal to $1986$. The winner is $A,B$ or $C$, respectively, if the sum of the chosen numbers leave... |
Consider the set \( S = \{1, 2, \ldots, 1986\} \), and let \( A_k \) be the event of choosing a \( k \)-element subset from \( S \). We are interested in the sum of the elements of the chosen subset modulo \( 3 \) being \( 0 \), \( 1 \), or \( 2 \). The game is fair if each of these outcomes occurs with equal probabil... | 6.125 | [
7,
6,
6,
5,
6,
7,
6,
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] | |
Let $x$, $y$ and $z$ all exceed $1$ and let $w$ be a positive number such that $\log_x w = 24$, $\log_y w = 40$ and $\log_{xyz} w = 12$. Find $\log_z w$. | 60 | Converting all of the logarithms to exponentials gives $x^{24} = w, y^{40} =w,$ and $x^{12}y^{12}z^{12}=w.$ Thus, we have $y^{40} = x^{24} \Rightarrow z^3=y^2.$ We are looking for $\log_z w,$ which by substitution, is $\log_{y^{\frac{2}{3}}} y^{40} = 40 \div \frac{2}{3} =\boxed{60}.$
~coolmath2017
~Lucas | 4.75 | [
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] |
Let $f(x)=|x-p|+|x-15|+|x-p-15|$, where $0 < p < 15$. Determine the minimum value taken by $f(x)$ for $x$ in the interval $p \leq x\leq15$. | 15 | It is best to get rid of the absolute values first.
Under the given circumstances, we notice that $|x-p|=x-p$, $|x-15|=15-x$, and $|x-p-15|=15+p-x$.
Adding these together, we find that the sum is equal to $30-x$, which attains its minimum value (on the given interval $p \leq x \leq 15$) when $x=15$, giving a minimum ... | 4.25 | [
4,
4,
6,
4,
4,
4,
4,
4
] |
What is the product of the real roots of the equation $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$? | 20 | Begin by completing the square on both sides of the equation, which gives \[(x+9)^2-51=2\sqrt{(x+3)(x+15)}\] Now by substituting $y=x+9$, we get $y^2-51=2\sqrt{(y-6)(y+6)}$, or \[y^4-106y^2+2745=0\] The solutions in $y$ are then \[y=x+9=\pm3\sqrt{5},\pm\sqrt{61}\] Turns out, $\pm3\sqrt{5}$ are a pair of extraneous solu... | 5.75 | [
5,
6,
6,
6,
5,
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6,
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] |
A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is $6$ cm and that of $BC$ is $2$ cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle.
[asy] size(150); defaultp... | 26 | We use coordinates. Let the circle have center $(0,0)$ and radius $\sqrt{50}$; this circle has equation $x^2 + y^2 = 50$. Let the coordinates of $B$ be $(a,b)$. We want to find $a^2 + b^2$. $A$ and $C$ with coordinates $(a,b+6)$ and $(a+2,b)$, respectively, both lie on the circle. From this we obtain the system of equa... | 5.625 | [
6,
6,
6,
6,
5,
4,
6,
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] |
Suppose that the sum of the squares of two complex numbers $x$ and $y$ is $7$ and the sum of the cubes is $10$. What is the largest real value that $x + y$ can have? | 4 | Begin by assuming that $x$ and $y$ are roots of some polynomial of the form $w^2+bw+c$, such that by Vieta's Formulæ and some algebra (left as an exercise to the reader), $b^2-2c=7$ and $3bc-b^3=10$. Substituting $c=\frac{b^2-7}{2}$, we deduce that $b^3-21b-20=0$, whose roots are $-4$, $-1$, and $5$. Since $-b$ is the ... | 6 | [
5,
6,
6,
7,
5,
6,
7,
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] |
Let $a_n=6^{n}+8^{n}$. Determine the remainder upon dividing $a_ {83}$ by $49$. | 35 | $6^{83} + 8^{83} = (6+8)(6^{82}-6^{81}8+\ldots-8^{81}6+8^{82})$
Becuase $7|(6+8)$, we only consider $6^{82}-6^{81}8+\ldots-8^{81}6+8^{82} \pmod{7}$
$6^{82}-6^{81}8+\ldots-8^{81}6+8^{82} \equiv (-1)^{82} - (-1)^{81}+ \ldots - (-1)^1 + 1 = 83 \equiv 6 \pmod{7}$
$6^{83} + 8^{83} \equiv 14 \cdot 6 \equiv \boxed{035} \pmod... | 5.5 | [
6,
6,
5,
5,
5,
6,
6,
5
] |
Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent off to slay a troublesome dragon. Let $P$ be the probability that at least two of the three had been sitting next to each other. If $P$ is written as a fraction in l... | 57 | We simplify this problem by using complementary counting and fixing one knight in place. Then, either a knight can sit two spaces apart from the fixed knight, or a knight can sit more than two spaces apart from the fixed knight. The probability is then $\frac{24\left(23\right)-\left[2\left(20\right)+20\left(19\right)\r... | 6.125 | [
6,
6,
6,
7,
6,
6,
6,
6
] |
What is the largest $2$-digit prime factor of the integer $n = {200\choose 100}$? | 61 | Expanding the binomial coefficient, we get ${200 \choose 100}=\frac{200!}{100!100!}$. Let the required prime be $p$; then $10 \le p < 100$. If $p > 50$, then the factor of $p$ appears twice in the denominator. Thus, we need $p$ to appear as a factor at least three times in the numerator, so $3p<200$. The largest such p... | 5.5 | [
6,
4,
6,
6,
6,
6,
5,
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] |
Find the minimum value of $\frac{9x^2\sin^2 x + 4}{x\sin x}$ for $0 < x < \pi$. | 12 | We can rewrite the numerator to be a perfect square by adding $-\dfrac{12x \sin x}{x \sin x}$. Thus, we must also add back $12$.
This results in $\dfrac{(3x \sin x-2)^2}{x \sin x}+12$.
Thus, if $3x \sin x-2=0$, then the minimum is obviously $12$. We show this possible with the same methods in Solution 1; thus the ans... | 6.125 | [
6,
6,
6,
6,
6,
6,
7,
6
] |
The numbers $1447$, $1005$ and $1231$ have something in common: each is a $4$-digit number beginning with $1$ that has exactly two identical digits. How many such numbers are there? | 432 | Suppose that the two identical digits are both $1$. Since the thousands digit must be $1$, only one of the other three digits can be $1$. This means the possible forms for the number are
$11xy,\qquad 1x1y,\qquad1xy1$
Because the number must have exactly two identical digits, $x\neq y$, $x\neq1$, and $y\neq1$. Hence, t... | 4.375 | [
4,
4,
5,
4,
4,
4,
4,
6
] |
The solid shown has a square base of side length $s$. The upper edge is parallel to the base and has length $2s$. All other edges have length $s$. Given that $s=6\sqrt{2}$, what is the volume of the solid?
[asy] size(180); import three; pathpen = black+linewidth(0.65); pointpen = black; currentprojection = perspective(... | 288 | Draw an altitude from a vertex of the square base to the top edge. By using $30,60, 90$ triangle ratios, we obtain that the altitude has a length of $3 \sqrt{6}$, and that little portion that hangs out has a length of $3\sqrt2$. This is a triangular pyramid with a base of $3\sqrt6, 3\sqrt6, 3\sqrt2$, and a height of $3... | 6 | [
6,
6,
5,
6,
6,
7,
6,
6
] |
Diameter $AB$ of a circle has length a $2$-digit integer (base ten). Reversing the digits gives the length of the perpendicular chord $CD$. The distance from their intersection point $H$ to the center $O$ is a positive rational number. Determine the length of $AB$. | 65 | Let $AB=10x+y$ and $CD=10y+x$. It follows that $CO=\frac{AB}{2}=\frac{10x+y}{2}$ and $CH=\frac{CD}{2}=\frac{10y+x}{2}$. Scale up this triangle by 2 to ease the arithmetic. Applying the Pythagorean Theorem on $2CO$, $2OH$ and $2CH$, we deduce \[(2OH)^2=\left(10x+y\right)^2-\left(10y+x\right)^2=99(x+y)(x-y)\]
Because $O... | 5.625 | [
5,
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6,
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] |
For $\{1, 2, 3, \ldots, n\}$ and each of its non-empty subsets a unique alternating sum is defined as follows. Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers. For example, the alternating sum for $\{1, 2, 3, 6,9\}$ is $9-6+3-2+... | 448 | Consider a given subset $T$ of $S$ that contains $7$; then there is a subset $T'$ which contains all the elements of $T$ except for $7$, and only those elements . Since each element of $T'$ has one fewer element preceding it than it does in $T$, their signs are opposite. Thus the sum of the alternating sums of $T$ and ... | 5 | [
6,
5,
5,
4,
4,
5,
6,
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] |
In the adjoining figure, two circles with radii $8$ and $6$ are drawn with their centers $12$ units apart. At $P$, one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$.
[asy]size(160); defaultpen(linewidth(.8pt)+fontsize(... | 130 | Let $QP = PR = x$. Extend the line containing the centers of the two circles to meet $R$, and to meet the other side of the large circle at a point $S$.
The part of this line from $R$ to the point nearest to $R$ where it intersects the larger circle has length $6+(12-8)=10$. The length of the diameter of the larger ci... | 6 | [
5,
6,
6,
6,
6,
7,
7,
5
] |
The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$. Suppose that the radius of the circle is $5$, that $BC=6$, and that $AD$ is bisected by $BC$. Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$. It follows that the sine of the central angle o... | 175 | Let $O$ be the center of the circle. The locus of midpoints of chords with $A$ as a endpoint is a circle with diameter $\overline{AO}$. Additionally, this circle must be tangent to $\overline{BC}$. Let the center of this circle be $P$. Let $M$ be the midpoint of $BC$, $N$ be the foot of the perpendicular from $P$ to $\... | 6.875 | [
7,
6,
7,
7,
7,
7,
7,
7
] |
Find the value of $a_2+a_4+a_6+a_8+\ldots+a_{98}$ if $a_1$, $a_2$, $a_3\ldots$ is an arithmetic progression with common difference 1, and $a_1+a_2+a_3+\ldots+a_{98}=137$. | 93 | Since we are dealing with an arithmetic sequence, \[a_2+a_4+a_6+a_8+\ldots+a_{98} = 49a_{50}\] We can also figure out that \[a_1+a_2+a_3+\ldots+a_{98} = a_1 + 97a_{50} = 137\] \[a_1 = a_{50}-49 \Rightarrow 98a_{50}-49 = 137\] Thus, $49a_{50} = \frac{137 + 49}{2} = \boxed{093}$
~Hithere22702 | 3 | [
3,
3,
3,
3,
3,
3,
3,
3
] |
The integer $n$ is the smallest positive multiple of $15$ such that every digit of $n$ is either $8$ or $0$. Compute $\frac{n}{15}$. | 592 | Any multiple of 15 is a multiple of 5 and a multiple of 3.
Any multiple of 5 ends in 0 or 5; since $n$ only contains the digits 0 and 8, the units digit of $n$ must be 0.
The sum of the digits of any multiple of 3 must be divisible by 3. If $n$ has $a$ digits equal to 8, the sum of the digits of $n$ is $8a$. For this... | 4 | [
4,
4,
4,
4,
4,
4,
4,
4
] |
A point $P$ is chosen in the interior of $\triangle ABC$ such that when lines are drawn through $P$ parallel to the sides of $\triangle ABC$, the resulting smaller triangles $t_{1}$, $t_{2}$, and $t_{3}$ in the figure, have areas $4$, $9$, and $49$, respectively. Find the area of $\triangle ABC$.
[asy] size(200); pathp... | 144 | By the transversals that go through $P$, all four triangles are similar to each other by the $AA$ postulate. Also, note that the length of any one side of the larger triangle is equal to the sum of the sides of each of the corresponding sides on the smaller triangles. We use the identity $K = \dfrac{ab\sin C}{2}$ to sh... | 5.375 | [
5,
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5,
6,
6,
5,
5,
5
] |
Let $S$ be a list of positive integers--not necessarily distinct--in which the number $68$ appears. The average (arithmetic mean) of the numbers in $S$ is $56$. However, if $68$ is removed, the average of the remaining numbers drops to $55$. What is the largest number that can appear in $S$? | 649 | Suppose that $S$ has $n$ numbers other than the $68.$ We have the following table: \[\begin{array}{c|c|c|c} & & & \\ [-2.5ex] & \textbf{Count} & \textbf{Arithmetic Mean} & \textbf{Sum} \\ \hline & & & \\ [-2.5ex] \textbf{Initial} & n+1 & 56 & 56(n+1) \\ \hline & & & \\ [-2.5ex] \textbf{Final} & n & 55 & 55n \end{array}... | 5.25 | [
5,
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] |
Determine the value of $ab$ if $\log_8a+\log_4b^2=5$ and $\log_8b+\log_4a^2=7$. | 512 | $\log_8a+\log_4b^2=5$ and $\log_8b+\log_4a^2=7$ adding both we get \[\log_8{ab} +2\log_4{ab}=12\] then we see $\log_4 {ab}$ is 3/2 times $\log_8 {ab}$ putting $\log_8 {ab}$ as x we get x+3x=12 so x=3 and $\log_8 {ab}$ = 3 so ab= $8^3$=\[\boxed{512}\] ~ math_comb01 | 5 | [
5,
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5
] |
Three circles, each of radius $3$, are drawn with centers at $(14, 92)$, $(17, 76)$, and $(19, 84)$. A line passing through $(17,76)$ is such that the total area of the parts of the three circles to one side of the line is equal to the total area of the parts of the three circles to the other side of it. What is the ab... | 24 | Notice that any line that passes through the bottom circle's center cuts it in half, so all we really care about are the top two circles. Suppose $\ell$ is the desired line. Draw lines $\ell_1$ and $\ell_2$ both parallel to $\ell$ such that $\ell_1$ passes through $(14,92)$ and $\ell_2$ passes through $(19,84)$. Clearl... | 5.75 | [
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] |
The function f is defined on the set of integers and satisfies $f(n)=\begin{cases} n-3&\mbox{if}\ n\ge 1000\\ f(f(n+5))&\mbox{if}\ n<1000\end{cases}$
Find $f(84)$. | 997 | Assume that $f(84)$ is to be performed $n+1$ times. Then we have \[f(84)=f^{n+1}(84)=f(f^n(84+5))\] In order to find $f(84)$, we want to know the smallest value of \[f^n(84+5)\ge1000\] Because then \[f(84)=f(f^n(84+5))=(f^n(84+5))-3\] From which we'll get a numerical value for $f(84)$.
Notice that the value of $n$ we ... | 6.5 | [
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] |
The equation $z^6+z^3+1=0$ has complex roots with argument $\theta$ between $90^\circ$ and $180^\circ$ in the complex plane. Determine the degree measure of $\theta$. | 160 | The substitution $y=z^3$ simplifies the equation to $y^2+y+1 = 0$. Applying the quadratic formula gives roots $y=-\frac{1}{2}\pm \frac{\sqrt{3}i}{2}$, which have arguments of $120$ and $240,$ respectively. We can write them as $z^3 = \cos 240^\circ + i\sin 240^\circ$ and $z^3 = \cos 120^\circ + i\sin 120^\circ$. So we ... | 5.875 | [
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] |
In tetrahedron $ABCD$, edge $AB$ has length 3 cm. The area of face $ABC$ is $15\mbox{cm}^2$ and the area of face $ABD$ is $12 \mbox { cm}^2$. These two faces meet each other at a $30^\circ$ angle. Find the volume of the tetrahedron in $\mbox{cm}^3$. | 20 | It is clear that $DX=8$ and $CX=10$ where $X$ is the foot of the perpendicular from $D$ and $C$ to side $AB$. Thus $[DXC]=\frac{ab\sin{c}}{2}=20=5 \cdot h \rightarrow h = 4$ where h is the height of the tetrahedron from $D$. Hence, the volume of the tetrahedron is $\frac{bh}{3}=15\cdot \frac{4}{3}=\boxed{020}$ ~ Mathom... | 6.625 | [
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Mary told John her score on the American High School Mathematics Examination (AHSME), which was over $80$. From this, John was able to determine the number of problems Mary solved correctly. If Mary's score had been any lower, but still over $80$, John could not have determined this. What was Mary's score? (Recall that... | 119 | Given that Mary's score is $30+4c-w$, two other ways to get that score are $30+4(c+1)-(w+4)$ and $30+4(c-1)-(w-4)$. Since it is clear that $c>1$, we must have $w<4$. In order to minimize the score, assume that $w=3$. The number of problems left blank must be less than $5$ because of the $30+4(c+1)-(w+4)$ case. In order... | 4.875 | [
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A gardener plants three maple trees, four oaks, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let $\frac m n$ in lowest terms be the probability that no two birch trees are next to one another. Find $m+n$. | 106 | Let $b$, $n$ denote birch tree and not-birch tree, respectively. Notice that we only need $4$ $n$s to separate the $5$ $b$s. Specifically, \[b,n,b,n,b,n,b,n,b\] Since we have $7$ $n$s, we are placing the extra $3$ $n$s into the $6$ intervals beside the $b$s.
Now doing simple casework.
If all $3$ $n$s are in the same ... | 4.625 | [
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] |
A function $f$ is defined for all real numbers and satisfies $f(2+x)=f(2-x)$ and $f(7+x)=f(7-x)$ for all $x$. If $x=0$ is a root for $f(x)=0$, what is the least number of roots $f(x)=0$ must have in the interval $-1000\leq x \leq 1000$? | 401 | We notice that the function has reflectional symmetry across both $x=2$ and $x=7$. We also use the fact that $x=0$ is a root. This shows that $x=4$ and $x=14$ are also roots. We then apply the reflection across the other axis to form $x=\pm 10$ as roots. Continuing this shows that the roots are $0 \mod 10$ or $4 \mod 1... | 7 | [
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Find the value of $10\cot(\cot^{-1}3+\cot^{-1}7+\cot^{-1}13+\cot^{-1}21).$ | 15 | On the coordinate plane, let $O=(0,0)$, $A_1=(3,0)$, $A_2=(3,1)$, $B_1=(21,7)$, $B_2=(20,10)$, $C_1=(260,130)$, $C_2=(250,150)$, $D_1=(5250,3150)$, $D_2=(5100,3400)$, and $H=(5100,0)$. We see that $\cot^{-1}(\angle A_2OA_1)=3$, $\cot^{-1}(\angle B_2OB_1)=7$, $\cot^{-1}(\angle C_2OC_1)=13$, and $\cot^{-1}(\angle D_2OD_1... | 6.25 | [
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] |
What is the largest even integer that cannot be written as the sum of two odd composite numbers? | 38 | Let $n$ be an integer that cannot be written as the sum of two odd composite numbers. If $n>33$, then $n-9,n-15,n-21,n-25,n-27,$ and $n-33$ must all be prime (or $n-33=1$, which yields $n=34=9+25$ which does not work). Thus $n-9,n-15,n-21,n-27,$ and $n-33$ form a prime quintuplet. However, only one prime quintuplet exi... | 5.25 | [
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] |
Determine $x^2+y^2+z^2+w^2$ if
$\frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2}=1$
$\frac{x^2}{4^2-1}+\frac{y^2}{4^2-3^2}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1$
$\frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1$
$\frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\fra... | 36 | As in Solution 1, we have
$(t-1)(t-9)(t-25)(t-49)-x^2(t-9)(t-25)(t-49)-y^2(t-1)(t-25)(t-49)$ $-z^2(t-1)(t-9)(t-49)-w^2(t-1)(t-9)(t-25)$
$=(t-4)(t-16)(t-36)(t-64)$
Now the coefficient of $t^3$ on both sides must be equal. So instead of expanding it fully, we will find what the coefficients of the $t^4$ and $t^3$ terms ... | 6.625 | [
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] |
Let $x_1=97$, and for $n>1$, let $x_n=\frac{n}{x_{n-1}}$. Calculate the product $x_1x_2x_3x_4x_5x_6x_7x_8$. | 384 | Since $x_n=\frac{n}{x_{n-1}}$, $x_n \cdot x_{n - 1} = n$. Setting $n = 2, 4, 6$ and $8$ in this equation gives us respectively $x_1x_2 = 2$, $x_3x_4 = 4$, $x_5x_6 = 6$ and $x_7x_8 = 8$ so \[x_1x_2x_3x_4x_5x_6x_7x_8 = 2\cdot4\cdot6\cdot8 = \boxed{384}.\] Notice that the value of $x_1$ was completely unneeded! | 4.25 | [
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When a right triangle is rotated about one leg, the volume of the cone produced is $800\pi \;\textrm{ cm}^3$. When the triangle is rotated about the other leg, the volume of the cone produced is $1920\pi \;\textrm{ cm}^3$. What is the length (in cm) of the hypotenuse of the triangle? | 26 | Let $a$ and $b$ be the two legs of the equation. We can find $\frac{a}{b}$ by doing $\frac{1920\pi}{800\pi}$. This simplified is $\frac{12}{5}$. We can represent the two legs as $12x$ and $5x$ for $a$ and $b$ respectively.
Since the volume of the first cone is $800\pi$, we use the formula for the volume of a cone and ... | 5.25 | [
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] |
Find $c$ if $a$, $b$, and $c$ are positive integers which satisfy $c=(a + bi)^3 - 107i$, where $i^2 = -1$. | 198 | Expanding out both sides of the given equation we have $c + 107i = (a^3 - 3ab^2) + (3a^2b - b^3)i$. Two complex numbers are equal if and only if their real parts and imaginary parts are equal, so $c = a^3 - 3ab^2$ and $107 = 3a^2b - b^3 = (3a^2 - b^2)b$. Since $a, b$ are integers, this means $b$ is a divisor of 107, wh... | 5.125 | [
5,
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5,
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] |
A small square is constructed inside a square of area 1 by dividing each side of the unit square into $n$ equal parts, and then connecting the vertices to the division points closest to the opposite vertices. Find the value of $n$ if the the area of the small square is exactly $\frac1{1985}$. | 32 | Line Segment $DE = \frac{1}{n}$, so $EC = 1 - \frac{1}{n} = \frac{n-1}{n}$. Draw line segment $HE$ parallel to the corresponding sides of the small square, $HE$ has length $\frac{1}{\sqrt{1985}}$, as it is the same length as the sides of the square. Notice that $\triangle CEL$ is similar to $\triangle HDE$ by $AA$ simi... | 5.625 | [
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] |
A sequence of integers $a_1, a_2, a_3, \ldots$ is chosen so that $a_n = a_{n - 1} - a_{n - 2}$ for each $n \ge 3$. What is the sum of the first 2001 terms of this sequence if the sum of the first 1492 terms is 1985, and the sum of the first 1985 terms is 1492? | 986 | The problem gives us a sequence defined by a recursion, so let's calculate a few values to get a feel for how it acts. We aren't given initial values, so let $a_1 = a$ and $a_2 = b$. Then $a_3 = b - a$, $a_4 = (b - a) - b = -a$, $a_5 = -a - (b - a) = -b$, $a_6 = -b - (-a) = a - b$, $a_7 = (a - b) - (-b) = a$ and $a_8 =... | 6.25 | [
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] |
As shown in the figure, triangle $ABC$ is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are as indicated. Find the area of triangle $ABC$. | 315 | This problem can be done using mass points. Assign B a weight of 1 and realize that many of the triangles have the same altitude. After continuously using the formulas that state (The sum of the two weights) = (The middle weight), and (The weight $\times$ side) = (Other weight) $\times$ (The other side), the problem yi... | 4.375 | [
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] |
Assume that $a$, $b$, $c$, and $d$ are positive integers such that $a^5 = b^4$, $c^3 = d^2$, and $c - a = 19$. Determine $d - b$. | 757 | It follows from the givens that $a$ is a perfect fourth power, $b$ is a perfect fifth power, $c$ is a perfect square and $d$ is a perfect cube. Thus, there exist integers $s$ and $t$ such that $a = t^4$, $b = t^5$, $c = s^2$ and $d = s^3$. So $s^2 - t^4 = 19$. We can factor the left-hand side of this equation as a diff... | 5.125 | [
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] |
The sum of the following seven numbers is exactly 19: $a_1 = 2.56$, $a_2 = 2.61$, $a_3 = 2.65$, $a_4 = 2.71$, $a_5 = 2.79$, $a_6 = 2.82$, $a_7 = 2.86$. It is desired to replace each $a_i$ by an integer approximation $A_i$, $1\le i \le 7$, so that the sum of the $A_i$'s is also 19 and so that $M$, the maximum of the "er... | 61 | If any of the approximations $A_i$ is less than 2 or more than 3, the error associated with that term will be larger than 1, so the largest error will be larger than 1. However, if all of the $A_i$ are 2 or 3, the largest error will be less than 1. So in the best case, we write 19 as a sum of 7 numbers, each of which i... | 6.25 | [
7,
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] |
In a circle, parallel chords of lengths 2, 3, and 4 determine central angles of $\alpha$, $\beta$, and $\alpha + \beta$ radians, respectively, where $\alpha + \beta < \pi$. If $\cos \alpha$, which is a positive rational number, is expressed as a fraction in lowest terms, what is the sum of its numerator and denominator... | 49 | Using the first diagram above, \[\sin \frac{\alpha}{2} = \frac{1}{r}\] \[\sin \frac{\beta}{2} = \frac{1.5}{r}\] \[\sin(\frac{\alpha}{2}+\frac{\beta}{2})=\frac{2}{r}\] by the Pythagorean trig identities, \[\cos\frac{\alpha}{2}=\sqrt{1-\frac{1}{r^2}}\] \[\cos\frac{\beta}{2}=\sqrt{1-\frac{2.25}{r^2}}\] so by the composite... | 6 | [
6,
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6,
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] |
How many of the first 1000 positive integers can be expressed in the form
$\lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 6x \rfloor + \lfloor 8x \rfloor$,
where $x$ is a real number, and $\lfloor z \rfloor$ denotes the greatest integer less than or equal to $z$? | 600 | Because $2,4,6,8$ are all multiples of $2$, we can speed things up. We only need to check up to $\frac{12}{24}$, and the rest should repeat. As shown before, we hit 6 integers ($1,2,4,5,6,10$) from $\frac{1}{24}$ to $\frac{12}{24}$. Similarly, this should repeat 100 times, for $\boxed{600}$
~N828335 | 4.75 | [
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An ellipse has foci at $(9,20)$ and $(49,55)$ in the $xy$-plane and is tangent to the $x$-axis. What is the length of its major axis? | 85 | An ellipse is defined as the set of points where the sum of the distances from the foci to the point is fixed. The length of major axis is equal to the sum of these distances $(2a)$. Thus if we find the sum of the distances, we get the answer. Let k be this fixed sum; then we get, by the distance formula:
$k = \sqrt{(x... | 6.875 | [
7,
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7,
7,
7,
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] |
Let $A$, $B$, $C$ and $D$ be the vertices of a regular tetrahedron, each of whose edges measures $1$ meter. A bug, starting from vertex $A$, observes the following rule: at each vertex it chooses one of the three edges meeting at that vertex, each edge being equally likely to be chosen, and crawls along that edge to th... | 182 | We evaluate $P(7)$ recursively: \begin{alignat*}{6} P(0)&=1, \\ P(1)&=\frac13(1-P(0))&&=0, \\ P(2)&=\frac13(1-P(1))&&=\frac13, \\ P(3)&=\frac13(1-P(2))&&=\frac29, \\ P(4)&=\frac13(1-P(3))&&=\frac{7}{27}, \\ P(5)&=\frac13(1-P(4))&&=\frac{20}{81}, \\ P(6)&=\frac13(1-P(5))&&=\frac{61}{243},\\ P(7)&=\frac13(1-P(6))&&=\frac... | 5.75 | [
6,
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] |
The numbers in the sequence $101$, $104$, $109$, $116$,$\ldots$ are of the form $a_n=100+n^2$, where $n=1,2,3,\ldots$ For each $n$, let $d_n$ be the greatest common divisor of $a_n$ and $a_{n+1}$. Find the maximum value of $d_n$ as $n$ ranges through the positive integers. | 401 | We know that $a_n = 100+n^2$ and $a_{n+1} = 100+(n+1)^2 = 100+ n^2+2n+1$. Since we want to find the GCD of $a_n$ and $a_{n+1}$, we can use the Euclidean algorithm:
$a_{n+1}-a_n = 2n+1$
Now, the question is to find the GCD of $2n+1$ and $100+n^2$. We subtract $2n+1$ 100 times from $100+n^2$. This leaves us with $n^2-20... | 5.125 | [
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In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded $1$ point, the loser got $0$ points, and each of the two players earned $\frac{1}{2}$ point if the game was a tie. After the completion of the tournament, it was found that exactly half of the poin... | 25 | Suppose that there are $n$ players participating in the tournament. We break this up into a group of the weakest ten, and the other $n-10$ people. Note that the $10$ players who played each other generated a total of $\dbinom{10}{2} = 45$ points playing each other. Thus, they earned $45$ playing the $n-10$ other people... | 6.375 | [
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Three 12 cm $\times$12 cm squares are each cut into two pieces $A$ and $B$, as shown in the first figure below, by joining the midpoints of two adjacent sides. These six pieces are then attached to a regular hexagon, as shown in the second figure, so as to fold into a polyhedron. What is the volume (in $\mathrm{cm}^3$)... | 864 | Note that gluing two of the given polyhedra together along a hexagonal face (rotated $60^\circ$ from each other) yields a cube, so the volume is $\frac12 \cdot 12^3 = 864$, so our answer is $\boxed{864}$.
Image:
Sidenote: What if this simple and beautiful way was not realized? Taking a reference to the above left di... | 6.25 | [
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What is the sum of the solutions to the equation $\sqrt[4]{x} = \frac{12}{7 - \sqrt[4]{x}}$? | 337 | Let $y = \sqrt[4]{x}$. Then we have $y(7 - y) = 12$, or, by simplifying, \[y^2 - 7y + 12 = (y - 3)(y - 4) = 0.\]
This means that $\sqrt[4]{x} = y = 3$ or $4$.
Thus the sum of the possible solutions for $x$ is $4^4 + 3^4 = \boxed{337}$. | 3 | [
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Evaluate the product \[\left(\sqrt{5}+\sqrt{6}+\sqrt{7}\right)\left(\sqrt{5}+\sqrt{6}-\sqrt{7}\right)\left(\sqrt{5}-\sqrt{6}+\sqrt{7}\right)\left(-\sqrt{5}+\sqrt{6}+\sqrt{7}\right).\] | 104 | We group the first and last factors as well as the two middle factors, then apply the difference of squares repeatedly: \begin{align*} \left(\left(\sqrt{6} + \sqrt{7}\right)^2 - \sqrt{5}^2\right)\left(\sqrt{5}^2 - \left(\sqrt{6} - \sqrt{7}\right)^2\right) &= \left(13 + 2\sqrt{42} - 5\right)\left(5 - \left(13 - 2\sqrt{4... | 5.5 | [
6,
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If $\tan x+\tan y=25$ and $\cot x + \cot y=30$, what is $\tan(x+y)$? | 150 | Since $\cot$ is the reciprocal function of $\tan$:
$\cot x + \cot y = \frac{1}{\tan x} + \frac{1}{\tan y} = \frac{\tan x + \tan y}{\tan x \cdot \tan y} = 30$
Thus, $\tan x \cdot \tan y = \frac{\tan x + \tan y}{30} = \frac{25}{30} = \frac{5}{6}$
Using the tangent addition formula:
$\tan(x+y) = \frac{\tan x + \tan y}{1... | 5.25 | [
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Determine $3x_4+2x_5$ if $x_1$, $x_2$, $x_3$, $x_4$, and $x_5$ satisfy the system of equations below.
$2x_1+x_2+x_3+x_4+x_5=6$
$x_1+2x_2+x_3+x_4+x_5=12$
$x_1+x_2+2x_3+x_4+x_5=24$
$x_1+x_2+x_3+2x_4+x_5=48$
$x_1+x_2+x_3+x_4+2x_5=96$ | 181 | Adding all five equations gives us $6(x_1 + x_2 + x_3 + x_4 + x_5) = 6(1 + 2 + 4 + 8 + 16)$ so $x_1 + x_2 + x_3 + x_4 + x_5 = 31$. Subtracting this from the fourth given equation gives $x_4 = 17$ and subtracting it from the fifth given equation gives $x_5 = 65$, so our answer is $3\cdot17 + 2\cdot65 = \boxed{181}$. | 4.125 | [
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What is that largest positive integer $n$ for which $n^3+100$ is divisible by $n+10$?
~ pi_is_3.14
~ momeme | 890 | Let $n+10=k$, then $n=k-10$. Then $n^3+100 = k^3-30k^2+300k-900$ Therefore, $900$ must be divisible by $k$, which is largest when $k=900$ and $n=\boxed{890}$ | 3.375 | [
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The pages of a book are numbered $1_{}^{}$ through $n_{}^{}$. When the page numbers of the book were added, one of the page numbers was mistakenly added twice, resulting in an incorrect sum of $1986_{}^{}$. What was the number of the page that was added twice? | 33 | Denote the page number as $x$, with $x < n$. The sum formula shows that $\frac{n(n + 1)}{2} + x = 1986$. Since $x$ cannot be very large, disregard it for now and solve $\frac{n(n+1)}{2} = 1986$. The positive root for $n \approx \sqrt{3972} \approx 63$. Quickly testing, we find that $63$ is too large, but if we plug in ... | 3.875 | [
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The increasing sequence $1,3,4,9,10,12,13\cdots$ consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the $100^{\mbox{th}}$ term of this sequence. | 981 | Notice that the first term of the sequence is $1$, the second is $3$, the fourth is $9$, and so on. Thus the $64th$ term of the sequence is $729$. Now out of $64$ terms which are of the form $729$ + $'''S'''$, $32$ of them include $243$ and $32$ do not. The smallest term that includes $243$, i.e. $972$, is greater than... | 6 | [
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Let $S$ be the sum of the base $10$ logarithms of all the proper divisors (all divisors of a number excluding itself) of $1000000$. What is the integer nearest to $S$? | 141 | Note that we can just pair terms up such that the product is $10^{6}.$ Now, however, note that $10^{3}$ is not included. Therefore we first exclude. We have $\displaystyle\frac{49-1}{2} = 24$ pairs that all multiply to $10^{6}.$ Now we include $10^{3}$ so our current product is $24 \cdot 6 - 3.$ However we dont want to... | 5.75 | [
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In $\triangle ABC$, $AB= 425$, $BC=450$, and $AC=510$. An interior point $P$ is then drawn, and segments are drawn through $P$ parallel to the sides of the triangle. If these three segments are of an equal length $d$, find $d$. | 306 | Refer to the diagram in solution 2; let $a^2=[E'EP]$, $b^2=[D'DP]$, and $c^2=[F'FP]$. Now, note that $[E'BD]$, $[D'DP]$, and $[E'EP]$ are similar, so through some similarities we find that $\frac{E'P}{PD}=\frac{a}{b}\implies\frac{E'D}{PD}=\frac{a+b}{b}\implies[E'BD]=b^2\left(\frac{a+b}{b}\right)^2=(a+b)^2$. Similarly, ... | 6.375 | [
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In a parlor game, the magician asks one of the participants to think of a three digit number $(abc)$ where $a$, $b$, and $c$ represent digits in base $10$ in the order indicated. The magician then asks this person to form the numbers $(acb)$, $(bca)$, $(bac)$, $(cab)$, and $(cba)$, to add these five numbers, and to rev... | 358 | Let $m$ be the number $100a+10b+c$. Observe that $3194+m=222(a+b+c)$ so
\[m\equiv -3194\equiv -86\equiv 136\pmod{222}\]
This reduces $m$ to one of $136, 358, 580, 802$. But also $a+b+c=\frac{3194+m}{222}>\frac{3194}{222}>14$ so $a+b+c\geq 15$. Of the four options, only $m = \boxed{358}$ satisfies this inequality. | 5.375 | [
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5
] |
The polynomial $1-x+x^2-x^3+\cdots+x^{16}-x^{17}$ may be written in the form $a_0+a_1y+a_2y^2+\cdots +a_{16}y^{16}+a_{17}y^{17}$, where $y=x+1$ and the $a_i$'s are constants. Find the value of $a_2$. | 816 | Using the geometric series formula, $1 - x + x^2 + \cdots - x^{17} = \frac {1 - x^{18}}{1 + x} = \frac {1-x^{18}}{y}$. Since $x = y - 1$, this becomes $\frac {1-(y - 1)^{18}}{y}$. We want $a_2$, which is the coefficient of the $y^3$ term in $-(y - 1)^{18}$ (because the $y$ in the denominator reduces the degrees in the ... | 5.25 | [
5,
6,
5,
5,
5,
5,
5,
6
] |
Let the sum of a set of numbers be the sum of its elements. Let $S$ be a set of positive integers, none greater than 15. Suppose no two disjoint subsets of $S$ have the same sum. What is the largest sum a set $S$ with these properties can have? | 61 | By using the greedy algorithm, we obtain $\boxed{061}$, with $S=\{ 15,14,13,11,8\}$. We must now prove that no such set has sum greater than 61. Suppose such a set $S$ existed. Then $S$ must have more than 4 elements, otherwise its sum would be at most $15+14+13+12=54$.
$S$ can't have more than 5 elements. To see why, ... | 6.5 | [
6,
7,
7,
6,
7,
7,
7,
5
] |
In a sequence of coin tosses, one can keep a record of instances in which a tail is immediately followed by a head, a head is immediately followed by a head, and etc. We denote these by TH, HH, and etc. For example, in the sequence TTTHHTHTTTHHTTH of 15 coin tosses we observe that there are two HH, three HT, four TH, a... | 560 | Let's consider each of the sequences of two coin tosses as an operation instead; this operation takes a string and adds the next coin toss on (eg, THHTH + HT = THHTHT). We examine what happens to the last coin toss. Adding HH or TT is simply an identity for the last coin toss, so we will ignore them for now. However, a... | 6.75 | [
7,
7,
7,
6,
6,
7,
7,
7
] |
The shortest distances between an interior diagonal of a rectangular parallelepiped, $P$, and the edges it does not meet are $2\sqrt{5}$, $\frac{30}{\sqrt{13}}$, and $\frac{15}{\sqrt{10}}$. Determine the volume of $P$. | 750 | In the above diagram, we focus on the line that appears closest and is parallel to $BC$. All the blue lines are perpendicular lines to $BC$ and their other points are on $AB$, the main diagonal. The green lines are projections of the blue lines onto the bottom face; all of the green lines originate in the corner and re... | 7 | [
7,
7,
7,
7,
8,
6,
7,
7
] |
Let triangle $ABC$ be a right triangle in the xy-plane with a right angle at $C_{}$. Given that the length of the hypotenuse $AB$ is $60$, and that the medians through $A$ and $B$ lie along the lines $y=x+3$ and $y=2x+4$ respectively, find the area of triangle $ABC$. | 400 | We first seek to find the angle between the lines $y = x + 3$ and $y = 2x + 4$. [asy] import graph; size(150); Label f; f.p=fontsize(6); xaxis(-8,8,Ticks(f, 2.0)); yaxis(-8,8,Ticks(f, 2.0)); real f(real x) { return (x + 3); } real g( real x){ return (2x + 4); } draw(graph(f,-8,5),red+linewidth(1)); draw(graph(g,-6,2),b... | 6.625 | [
7,
7,
6,
6,
7,
7,
7,
6
] |
An ordered pair $(m,n)$ of non-negative integers is called "simple" if the addition $m+n$ in base $10$ requires no carrying. Find the number of simple ordered pairs of non-negative integers that sum to $1492$. | 300 | Since no carrying over is allowed, the range of possible values of any digit of $m$ is from $0$ to the respective digit in $1492$ (the values of $n$ are then fixed). Thus, the number of ordered pairs will be $(1 + 1)(4 + 1)(9 + 1)(2 + 1) = 2\cdot 5\cdot 10\cdot 3 = \boxed{300}$.
If you do not understand the above solu... | 4.375 | [
4,
4,
4,
4,
5,
4,
5,
5
] |
What is the largest possible distance between two points, one on the sphere of radius 19 with center $(-2,-10,5)$ and the other on the sphere of radius 87 with center $(12,8,-16)$? | 137 | The distance between the two centers of the spheres can be determined via the distance formula in three dimensions: $\sqrt{(12 - (-2))^2 + (8 - (-10))^2 + (-16 - 5)^2} = \sqrt{14^2 + 18^2 + 21^2} = 31$. The largest possible distance would be the sum of the two radii and the distance between the two centers, making it $... | 4.125 | [
4,
4,
4,
4,
4,
4,
5,
4
] |
By a proper divisor of a natural number we mean a positive integral divisor other than 1 and the number itself. A natural number greater than 1 will be called nice if it is equal to the product of its distinct proper divisors. What is the sum of the first ten nice numbers? | 182 | Let $p(n)$ denote the product of the distinct proper divisors of $n$. A number $n$ is nice in one of two instances:
It has exactly two distinct prime divisors.
If we let $n = pq$, where $p,q$ are the prime factors, then its proper divisors are $p$ and $q$, and $p(n) = p \cdot q = n$.
It is the cube of a prime number.
... | 6.25 | [
6,
6,
6,
6,
6,
6,
7,
7
] |
Find the area of the region enclosed by the graph of $|x-60|+|y|=\left|\frac{x}{4}\right|.$ | 480 | Since $|y|$ is nonnegative, $\left|\frac{x}{4}\right| \ge |x - 60|$. Solving this gives us two equations: $\frac{x}{4} \ge x - 60\ \mathrm{and} \ -\frac{x}{4} \le x - 60$. Thus, $48 \le x \le 80$. The maximum and minimum y value is when $|x - 60| = 0$, which is when $x = 60$ and $y = \pm 15$. Since the graph is symmetr... | 6 | [
6,
6,
6,
6,
6,
6,
6,
6
] |
Find $3x^2 y^2$ if $x$ and $y$ are integers such that $y^2 + 3x^2 y^2 = 30x^2 + 517$. | 588 | If we move the $x^2$ term to the left side, it is factorable with Simon's Favorite Factoring Trick:
\[(3x^2 + 1)(y^2 - 10) = 517 - 10\]
$507$ is equal to $3 \cdot 13^2$. Since $x$ and $y$ are integers, $3x^2 + 1$ cannot equal a multiple of three. $169$ doesn't work either, so $3x^2 + 1 = 13$, and $x^2 = 4$. This leaves... | 4.75 | [
5,
6,
4,
5,
5,
5,
4,
4
] |
Rectangle $ABCD$ is divided into four parts of equal area by five segments as shown in the figure, where $XY = YB + BC + CZ = ZW = WD + DA + AX$, and $PQ$ is parallel to $AB$. Find the length of $AB$ (in cm) if $BC = 19$ cm and $PQ = 87$ cm. | 193 | Since $XY = WZ$, $PQ = PQ$ and the areas of the trapezoids $PQZW$ and $PQYX$ are the same, then the heights of the trapezoids are the same. Thus both trapezoids have area $\frac{1}{2} \cdot \frac{19}{2}(XY + PQ) = \frac{19}{4}(XY + 87)$. This number is also equal to one quarter the area of the entire rectangle, which i... | 5.75 | [
6,
6,
6,
6,
6,
6,
5,
5
] |
Let $[r,s]$ denote the least common multiple of positive integers $r$ and $s$. Find the number of ordered triples $(a,b,c)$ of positive integers for which $[a,b] = 1000$, $[b,c] = 2000$, and $[c,a] = 2000$. | 70 | It's clear that we must have $a = 2^j5^k$, $b = 2^m 5^n$ and $c = 2^p5^q$ for some nonnegative integers $j, k, m, n, p, q$. Dealing first with the powers of 2: from the given conditions, $\max(j, m) = 3$, $\max(m, p) = \max(p, j) = 4$. Thus we must have $p = 4$ and at least one of $m, j$ equal to 3. This gives 7 possib... | 6.875 | [
7,
7,
7,
7,
6,
7,
7,
7
] |
What is the largest positive integer $n$ for which there is a unique integer $k$ such that $\frac{8}{15} < \frac{n}{n + k} < \frac{7}{13}$? | 112 | Notice that in order for $k$ to be unique, $\frac{n}{n + k+ 1} \le \frac{8}{15}$ and $\frac{n}{n+ k-1} \ge \frac{7}{13}$ must be true. Solving these inequalities for $k$ yields $\frac{7}{6}(k-1) \le n \le \frac{8}{7}(k+1)$.
Thus, we want to find $k$ such that $\frac{7}{6}(k-1)\le \frac{8}{7}(k+1)$. Solving this inequa... | 5.875 | [
6,
7,
6,
5,
6,
6,
5,
6
] |
Triangle $ABC$ has right angle at $B$, and contains a point $P$ for which $PA = 10$, $PB = 6$, and $\angle APB = \angle BPC = \angle CPA$. Find $PC$.
[asy] unitsize(0.2 cm); pair A, B, C, P; A = (0,14); B = (0,0); C = (21*sqrt(3),0); P = intersectionpoint(arc(B,6,0,180),arc(C,33,0,180)); draw(A--B--C--cycle); draw(A--P... | 33 | Let $PC = x$. Since $\angle APB = \angle BPC = \angle CPA$, each of them is equal to $120^\circ$. By the Law of Cosines applied to triangles $\triangle APB$, $\triangle BPC$ and $\triangle CPA$ at their respective angles $P$, remembering that $\cos 120^\circ = -\frac12$, we have
\[AB^2 = 36 + 100 + 60 = 196, BC^2 = 36 ... | 6 | [
6,
6,
6,
6,
6,
6,
6,
6
] |
Al walks down to the bottom of an escalator that is moving up and he counts 150 steps. His friend, Bob, walks up to the top of the escalator and counts 75 steps. If Al's speed of walking (in steps per unit time) is three times Bob's walking speed, how many steps are visible on the escalator at a given time? (Assume tha... | 120 | Let $e$ and $b$ be the speeds of the escalator and Bob, respectively.
When Al was on his way down, he took $150$ steps with a speed of $3b-e$ per step. When Bob was on his way up, he took $75$ steps with a speed of $b+e$ per step. Since Al and Bob were walking the same distance, we have \[150(3b-e)=75(b+e)\] Solving g... | 5.75 | [
6,
5,
6,
6,
5,
5,
6,
7
] |
Find the largest possible value of $k$ for which $3^{11}$ is expressible as the sum of $k$ consecutive positive integers. | 486 | Let us write down one such sum, with $m$ terms and first term $n + 1$:
$3^{11} = (n + 1) + (n + 2) + \ldots + (n + m) = \frac{1}{2} m(2n + m + 1)$.
Thus $m(2n + m + 1) = 2 \cdot 3^{11}$ so $m$ is a divisor of $2\cdot 3^{11}$. However, because $n \geq 0$ we have $m^2 < m(m + 1) \leq 2\cdot 3^{11}$ so $m < \sqrt{2\cdot ... | 5 | [
5,
5,
6,
5,
5,
4,
5,
5
] |
Let $m$ be the smallest integer whose cube root is of the form $n+r$, where $n$ is a positive integer and $r$ is a positive real number less than $1/1000$. Find $n$. | 19 | Note that the cube root is increasing for positive reals while its derivative is decreasing, so linear approximation gives \[\sqrt[3]{n^3+1} - n < \left.\frac{d\sqrt[3]{x}}{dx}\right|_{x=n^3} = \frac{1}{3n^2}\] and \[\sqrt[3]{n^3+1} - n > \left.\frac{d\sqrt[3]{x}}{dx}\right|_{x=n^3+1} = \frac{1}{3\sqrt[3]{(n^3+1)^2}}\]... | 5 | [
5,
6,
5,
6,
5,
4,
5,
4
] |
A given sequence $r_1, r_2, \dots, r_n$ of distinct real numbers can be put in ascending order by means of one or more "bubble passes". A bubble pass through a given sequence consists of comparing the second term with the first term, and exchanging them if and only if the second term is smaller, then comparing the thir... | 931 | If any of $r_1, \ldots, r_{19}$ is larger than $r_{20}$, one of these numbers will be compared with $r_{20}$ on the 19th step of the first bubble pass and $r_{20}$ will be moved back to the 19th position. Thus, $r_{20}$ must be the largest of the first 20 terms. In addition, $r_{20}$ must be larger than $r_{21}, r_{22}... | 6.875 | [
6,
7,
7,
7,
7,
8,
7,
6
] |
Compute \[\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}.\] | 373 | In both the numerator and the denominator, each factor is of the form $N^4+324=N^4+18^2$ for some positive integer $N.$
We factor $N^4+18^2$ by completing the square, then applying the difference of squares: \begin{align*} N^4+18^2&=\left(N^4+36N^2+18^2\right)-36N^2 \\ &=\left(N^2+18\right)^2-(6N)^2 \\ &=\left(N^2-6N+... | 6.5 | [
6,
6,
6,
7,
7,
7,
6,
7
] |
Squares $S_1$ and $S_2$ are inscribed in right triangle $ABC$, as shown in the figures below. Find $AC + CB$ if area $(S_1) = 441$ and area $(S_2) = 440$. | 462 | Let $\tan\angle ABC = x$. Now using the 1st square, $AC=21(1+x)$ and $CB=21(1+x^{-1})$. Using the second square, $AB=\sqrt{440}(1+x+x^{-1})$. We have $AC^2+CB^2=AB^2$, or \[441(x^2+x^{-2}+2x+2x^{-1}+2)=440(x^2+x^{-2}+2x+2x^{-1}+3).\] Rearranging and letting $u=x+x^{-1} \Rightarrow u^2 - 2 = x^2 + x^{-2}$ gives us $u^2+... | 5.875 | [
6,
5,
6,
6,
6,
6,
6,
6
] |
One commercially available ten-button lock may be opened by pressing -- in any order -- the correct five buttons. The sample shown below has $\{1,2,3,6,9\}$ as its combination. Suppose that these locks are redesigned so that sets of as many as nine buttons or as few as one button could serve as combinations. How many a... | 770 | Currently there are ${10 \choose 5}$ possible combinations. With any integer $x$ from $1$ to $9$, the number of ways to choose a set of $x$ buttons is $\sum^{9}_{k=1}{10 \choose k}$. Now we can use the identity $\sum^{n}_{k=0}{n \choose k}=2^{n}$. So the number of additional combinations is just $2^{10}-{10\choose 0}-{... | 4.75 | [
5,
5,
5,
4,
5,
4,
5,
5
] |
For any positive integer $k$, let $f_1(k)$ denote the square of the sum of the digits of $k$. For $n \ge 2$, let $f_n(k) = f_1(f_{n - 1}(k))$. Find $f_{1988}(11)$. | 169 | We see that $f_{1}(11)=4$
$f_2(11) = f_1(4)=16$
$f_3(11) = f_1(16)=49$
$f_4(11) = f_1(49)=169$
$f_5(11) = f_1(169)=256$
$f_6(11) = f_1(256)=169$
Note that this revolves between the two numbers. Since $1988$ is even, we thus have $f_{1988}(11) = f_{4}(11) = \boxed{169}$. | 4.375 | [
5,
4,
5,
4,
4,
4,
5,
4
] |
Find $(\log_2 x)^2$ if $\log_2 (\log_8 x) = \log_8 (\log_2 x)$. | 27 | Say that $\log_{2^3}x=a$ and $\log_2x=b$ so we have $\log_2a=\log_{2^3}b$. And we want $b^2$.
$\\ \log_2a=\frac13 \log_{2}b \ \ \text{\tiny{(step 1)}}\\ \frac{\log_2a}{\log_{2}b}=\log_ba=\frac13\\ b^{1/3}=a.$
Because $3a=b$ (as $2^{3a}=x$ and $2^b=x$ from our setup), we have that
$b^{1/3}=\frac{b}{3}\\ b^{-2/3}=\frac1... | 5.5 | [
6,
5,
5,
6,
5,
6,
5,
6
] |
Suppose that $|x_i| < 1$ for $i = 1, 2, \dots, n$. Suppose further that $|x_1| + |x_2| + \dots + |x_n| = 19 + |x_1 + x_2 + \dots + x_n|.$ What is the smallest possible value of $n$? | 20 | Since $|x_i| < 1$ then
\[|x_1| + |x_2| + \dots + |x_n| = 19 + |x_1 + x_2 + \dots + x_n| < n.\]
So $n \ge 20$. We now just need to find an example where $n = 20$: suppose $x_{2k-1} = \frac{19}{20}$ and $x_{2k} = -\frac{19}{20}$; then on the left hand side we have $\left|\frac{19}{20}\right| + \left|-\frac{19}{20}\right... | 4.5 | [
4,
5,
5,
5,
4,
5,
4,
4
] |
Let $m/n$, in lowest terms, be the probability that a randomly chosen positive divisor of $10^{99}$ is an integer multiple of $10^{88}$. Find $m + n$. | 634 | $10^{99} = 2^{99}5^{99}$, so it has $(99 + 1)(99 + 1) = 10000$ factors. Out of these, we only want those factors of $10^{99}$ which are divisible by $10^{88}$; it is easy to draw a bijection to the number of factors that $10^{11} = 2^{11}5^{11}$ has, which is $(11 + 1)(11 + 1) = 144$. Our probability is $\frac{m}{n} = ... | 3.75 | [
3,
4,
4,
4,
3,
4,
4,
4
] |
It is possible to place positive integers into the vacant twenty-one squares of the $5 \times 5$ square shown below so that the numbers in each row and column form arithmetic sequences. Find the number that must occupy the vacant square marked by the asterisk (*). | 142 | First, let $a =$ the number to be placed in the first column, fourth row. Let $b =$ the number to be placed in the second column, fifth row. We can determine the entire first column and fifth row in terms of $a$ and $b$:
$\begin{tabular}[b]{|c|c|c|c|c|}\hline 4a & & & & \\ \hline 3a & & & & \\ \hline 2a & & & & \\ \hli... | 5.25 | [
5,
6,
4,
6,
6,
5,
5,
5
] |
In triangle $ABC$, $\tan \angle CAB = 22/7$, and the altitude from $A$ divides $BC$ into segments of length 3 and 17. What is the area of triangle $ABC$? | 110 | Let $D$ be the intersection of the altitude with $\overline{BC}$, and $h$ be the length of the altitude. Without loss of generality, let $BD = 17$ and $CD = 3$. Then $\tan \angle DAB = \frac{17}{h}$ and $\tan \angle CAD = \frac{3}{h}$. Using the tangent sum formula,
\begin{align*} \tan CAB &= \tan (DAB + CAD)\\ \frac{2... | 4.875 | [
5,
5,
5,
5,
4,
5,
5,
5
] |
The function $f$, defined on the set of ordered pairs of positive integers, satisfies the following properties: \[f(x, x) = x,\; f(x, y) = f(y, x), {\rm \ and\ } (x+y)f(x, y) = yf(x, x+y).\] Calculate $f(14,52)$. | 364 | Since all of the function's properties contain a recursive definition except for the first one, we know that $f(x,x) = x$ in order to obtain an integer answer. So, we have to transform $f(14,52)$ to this form by exploiting the other properties. The second one doesn't help us immediately, so we will use the third one.
... | 6.25 | [
6,
6,
7,
6,
6,
6,
6,
7
] |
Find the smallest positive integer whose cube ends in $888$. | 192 | Let $x^3 = 1000a + 888$. We factor an $8$ out of the right hand side, and we note that $x$ must be of the form $x = 2y$, where $y$ is a positive integer. Then, this becomes $y^3 = 125a + 111$. Taking mod $5$, $25$, and $125$, we get $y^3 \equiv 1\pmod 5$, $y^3 \equiv 11\pmod{25}$, and $y^3 \equiv 111\pmod{125}$.
We ca... | 5.625 | [
5,
6,
5,
6,
6,
6,
5,
6
] |
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