question stringlengths 16 1.34k | answer stringlengths 26 6.77k ⌀ |
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The first four terms in an arithmetic sequence are $x+y$, $x-y$, $xy$, and $x/y$, in that order. What is the fifth term? Express your answer as a common fraction. | Since the difference of the first two terms is $-2y$, the third and fourth terms of the sequence must be $x-3y$ and $x-5y$. Thus \[
x-3y = xy \quad\text{and}\quad x-5y = \frac{x}{y},
\]so $xy - 5y^{2} = x.$ Combining these equations we obtain \[
(x - 3y) - 5y^{2}= x\quad\text{and, therefore, }\quad -3y - 5y^{2} = 0.
\]... |
The first four terms in an arithmetic sequence are $x+y$, $x-y$, $xy$, and $x/y$, in that order. What is the fifth term? Express your answer as a common fraction. | Since the difference of the first two terms is $-2y$, the third and fourth terms of the sequence must be $x-3y$ and $x-5y$. Thus \[
x-3y = xy \quad\text{and}\quad x-5y = \frac{x}{y},
\]so $xy - 5y^{2} = x.$ Combining these equations we obtain \[
(x - 3y) - 5y^{2}= x\quad\text{and, therefore, }\quad -3y - 5y^{2} = 0.
\]... |
The first four terms in an arithmetic sequence are $x+y$, $x-y$, $xy$, and $x/y$, in that order. What is the fifth term? Express your answer as a common fraction. | Since the difference of the first two terms is $-2y$, the third and fourth terms of the sequence must be $x-3y$ and $x-5y$. Thus \[
x-3y = xy \quad\text{and}\quad x-5y = \frac{x}{y},
\]so $xy - 5y^{2} = x.$ Combining these equations we obtain \[
(x - 3y) - 5y^{2}= x\quad\text{and, therefore, }\quad -3y - 5y^{2} = 0.
\]... |
A right cylindrical oil tank is $15$ feet tall and its circular bases have diameters of $4$ feet each. When the tank is lying flat on its side (not on one of the circular ends), the oil inside is $3$ feet deep. How deep, in feet, would the oil have been if the tank had been standing upright on one of its bases? Express... | Since the oil is $3$ feet deep, we want to find the ratio of the area of the part of the circle covered with oil (part under the horizontal line of the figure below) to the entire area of the circle.
[asy]
draw(Circle((0,0),2));
draw((-1.732,1)--(1.732,1));
draw((0,0)--(-1.732,1));
draw((0,0)--(1.732,1));
draw((0,0)--(... |
A right cylindrical oil tank is $15$ feet tall and its circular bases have diameters of $4$ feet each. When the tank is lying flat on its side (not on one of the circular ends), the oil inside is $3$ feet deep. How deep, in feet, would the oil have been if the tank had been standing upright on one of its bases? Express... | Since the oil is $3$ feet deep, we want to find the ratio of the area of the part of the circle covered with oil (part under the horizontal line of the figure below) to the entire area of the circle.
[asy]
draw(Circle((0,0),2));
draw((-1.732,1)--(1.732,1));
draw((0,0)--(-1.732,1));
draw((0,0)--(1.732,1));
draw((0,0)--(... |
A right cylindrical oil tank is $15$ feet tall and its circular bases have diameters of $4$ feet each. When the tank is lying flat on its side (not on one of the circular ends), the oil inside is $3$ feet deep. How deep, in feet, would the oil have been if the tank had been standing upright on one of its bases? Express... | Since the oil is $3$ feet deep, we want to find the ratio of the area of the part of the circle covered with oil (part under the horizontal line of the figure below) to the entire area of the circle.
[asy]
draw(Circle((0,0),2));
draw((-1.732,1)--(1.732,1));
draw((0,0)--(-1.732,1));
draw((0,0)--(1.732,1));
draw((0,0)--(... |
The sum of the squares of three consecutive positive even numbers is $12296$. Find the product of the three numbers divided by $8$. | If $n$ is the middle number of the three, the other two numbers are $n-2$ and $n+2$. Therefore, the squares are $n^2-4n+4$, $n^2$, and $n^2+4n+4$. Setting the sum of the three squares equal to $12296$, \begin{align*}
\left(n^2-4n+4\right)+\left(n^2\right)+\left(n^2+4n+4\right)&=12296\\
3n^2+8&=12296\\
3n^2&=12288\\
n^2... |
The sum of the squares of three consecutive positive even numbers is $12296$. Find the product of the three numbers divided by $8$. | If $n$ is the middle number of the three, the other two numbers are $n-2$ and $n+2$. Therefore, the squares are $n^2-4n+4$, $n^2$, and $n^2+4n+4$. Setting the sum of the three squares equal to $12296$, \begin{align*}
\left(n^2-4n+4\right)+\left(n^2\right)+\left(n^2+4n+4\right)&=12296\\
3n^2+8&=12296\\
3n^2&=12288\\
n^2... |
The sum of the squares of three consecutive positive even numbers is $12296$. Find the product of the three numbers divided by $8$. | If $n$ is the middle number of the three, the other two numbers are $n-2$ and $n+2$. Therefore, the squares are $n^2-4n+4$, $n^2$, and $n^2+4n+4$. Setting the sum of the three squares equal to $12296$, \begin{align*}
\left(n^2-4n+4\right)+\left(n^2\right)+\left(n^2+4n+4\right)&=12296\\
3n^2+8&=12296\\
3n^2&=12288\\
n^2... |
In triangle $ABC$, medians $AD$ and $CE$ intersect at $P$, $PE=1.5$, $PD=2$, and $DE=2.5$. What is the area of $AEDC$? | Note that $1.5^2 + 2^2 = 2.5^2,$ so $\triangle PED$ has a right angle at $P.$ (Alternatively, you could note that $(1.5, 2, 2.5)$ is half of the Pythagorean triple $(3,4,5).$) [asy]size(6cm);pair P=(0,0),D=(0,-2),E=(-1.5,0),C=(3,0),A=(0,4),B=extension(A,E,D,C);draw(A--B--C--cycle^^C--E^^A--D);draw(rightanglemark(E,P,D)... |
In triangle $ABC$, medians $AD$ and $CE$ intersect at $P$, $PE=1.5$, $PD=2$, and $DE=2.5$. What is the area of $AEDC$? | Note that $1.5^2 + 2^2 = 2.5^2,$ so $\triangle PED$ has a right angle at $P.$ (Alternatively, you could note that $(1.5, 2, 2.5)$ is half of the Pythagorean triple $(3,4,5).$) [asy]size(6cm);pair P=(0,0),D=(0,-2),E=(-1.5,0),C=(3,0),A=(0,4),B=extension(A,E,D,C);draw(A--B--C--cycle^^C--E^^A--D);draw(rightanglemark(E,P,D)... |
In triangle $ABC$, medians $AD$ and $CE$ intersect at $P$, $PE=1.5$, $PD=2$, and $DE=2.5$. What is the area of $AEDC$? | Note that $1.5^2 + 2^2 = 2.5^2,$ so $\triangle PED$ has a right angle at $P.$ (Alternatively, you could note that $(1.5, 2, 2.5)$ is half of the Pythagorean triple $(3,4,5).$) [asy]size(6cm);pair P=(0,0),D=(0,-2),E=(-1.5,0),C=(3,0),A=(0,4),B=extension(A,E,D,C);draw(A--B--C--cycle^^C--E^^A--D);draw(rightanglemark(E,P,D)... |
Let $\alpha$ and $\beta$ be the roots of $x^2 + px + 1 = 0,$ and let $\gamma$ and $\delta$ are the roots of $x^2 + qx + 1 = 0.$ Express
\[(\alpha - \gamma)(\beta - \gamma)(\alpha + \delta)(\beta + \delta)\]in terms of $p$ and $q.$ | Since $\alpha$ and $\beta$ are the roots of $x^2 + px + 1 = 0,$
\[(x - \alpha)(x - \beta) = x^2 + px + 1.\]Setting $x = \gamma,$ we get
\[(\gamma - \alpha)(\gamma - \beta) = \gamma^2 + p \gamma + 1.\]or $(\alpha - \gamma)(\beta - \gamma) = \gamma^2 + p \gamma + 1.$
Setting $x = -\delta,$ we get
\[(-\delta - \alpha)(-\... |
Let $\alpha$ and $\beta$ be the roots of $x^2 + px + 1 = 0,$ and let $\gamma$ and $\delta$ are the roots of $x^2 + qx + 1 = 0.$ Express
\[(\alpha - \gamma)(\beta - \gamma)(\alpha + \delta)(\beta + \delta)\]in terms of $p$ and $q.$ | Since $\alpha$ and $\beta$ are the roots of $x^2 + px + 1 = 0,$
\[(x - \alpha)(x - \beta) = x^2 + px + 1.\]Setting $x = \gamma,$ we get
\[(\gamma - \alpha)(\gamma - \beta) = \gamma^2 + p \gamma + 1.\]or $(\alpha - \gamma)(\beta - \gamma) = \gamma^2 + p \gamma + 1.$
Setting $x = -\delta,$ we get
\[(-\delta - \alpha)(-\... |
Let $\alpha$ and $\beta$ be the roots of $x^2 + px + 1 = 0,$ and let $\gamma$ and $\delta$ are the roots of $x^2 + qx + 1 = 0.$ Express
\[(\alpha - \gamma)(\beta - \gamma)(\alpha + \delta)(\beta + \delta)\]in terms of $p$ and $q.$ | Since $\alpha$ and $\beta$ are the roots of $x^2 + px + 1 = 0,$
\[(x - \alpha)(x - \beta) = x^2 + px + 1.\]Setting $x = \gamma,$ we get
\[(\gamma - \alpha)(\gamma - \beta) = \gamma^2 + p \gamma + 1.\]or $(\alpha - \gamma)(\beta - \gamma) = \gamma^2 + p \gamma + 1.$
Setting $x = -\delta,$ we get
\[(-\delta - \alpha)(-\... |
The sides of a triangle have lengths $11, 15,$ and $k,$ where $k$ is a positive integer. For how many values of $k$ is the triangle obtuse? | The longest side of the triangle either has length $15$ or has length $k.$ Take cases:
If the longest side has length $15,$ then $k \le 15.$ The triangle must be nondegenerate, which happens if and only if $15 < 11 + k,$ or $4 < k,$ by the triangle inequality. Now, for the triangle to be obtuse, we must have $15^2 > 1... |
The sides of a triangle have lengths $11, 15,$ and $k,$ where $k$ is a positive integer. For how many values of $k$ is the triangle obtuse? | The longest side of the triangle either has length $15$ or has length $k.$ Take cases:
If the longest side has length $15,$ then $k \le 15.$ The triangle must be nondegenerate, which happens if and only if $15 < 11 + k,$ or $4 < k,$ by the triangle inequality. Now, for the triangle to be obtuse, we must have $15^2 > 1... |
The sides of a triangle have lengths $11, 15,$ and $k,$ where $k$ is a positive integer. For how many values of $k$ is the triangle obtuse? | The longest side of the triangle either has length $15$ or has length $k.$ Take cases:
If the longest side has length $15,$ then $k \le 15.$ The triangle must be nondegenerate, which happens if and only if $15 < 11 + k,$ or $4 < k,$ by the triangle inequality. Now, for the triangle to be obtuse, we must have $15^2 > 1... |
Three points are chosen uniformly at random on a circle. What is the probability that no two of these points form an obtuse triangle with the circle's center? | Let us call the circle's center $O$. We first note that if $A$ and $B$ are points on the circle, then triangle $AOB$ is isosceles with $AO= BO$. Therefore, if $AOB$ is an obtuse triangle, then the obtuse angle must be at $O$. So $AOB$ is an obtuse triangle if and only if minor arc $AB$ has measure of more than $\pi/... |
Three points are chosen uniformly at random on a circle. What is the probability that no two of these points form an obtuse triangle with the circle's center? | Let us call the circle's center $O$. We first note that if $A$ and $B$ are points on the circle, then triangle $AOB$ is isosceles with $AO= BO$. Therefore, if $AOB$ is an obtuse triangle, then the obtuse angle must be at $O$. So $AOB$ is an obtuse triangle if and only if minor arc $AB$ has measure of more than $\pi/... |
Three points are chosen uniformly at random on a circle. What is the probability that no two of these points form an obtuse triangle with the circle's center? | Let us call the circle's center $O$. We first note that if $A$ and $B$ are points on the circle, then triangle $AOB$ is isosceles with $AO= BO$. Therefore, if $AOB$ is an obtuse triangle, then the obtuse angle must be at $O$. So $AOB$ is an obtuse triangle if and only if minor arc $AB$ has measure of more than $\pi/... |
For how many different digits $n$ is the three-digit number $14n$ divisible by $n$?
Note: $14n$ refers to a three-digit number with the unit digit of $n,$ not the product of $14$ and $n.$ | We have to account for each possible value of $n$ here. First of all, we can quickly find that for $n = 1, 2, 5,$ the resulting number $14n$ must be divisible by $n$, using their respective divisibility rules.
We see that for $n = 3$, we get $143.$ Since $1 + 4 + 3 = 8,$ which is not a multiple of $3,$ we can see that... |
For how many different digits $n$ is the three-digit number $14n$ divisible by $n$?
Note: $14n$ refers to a three-digit number with the unit digit of $n,$ not the product of $14$ and $n.$ | We have to account for each possible value of $n$ here. First of all, we can quickly find that for $n = 1, 2, 5,$ the resulting number $14n$ must be divisible by $n$, using their respective divisibility rules.
We see that for $n = 3$, we get $143.$ Since $1 + 4 + 3 = 8,$ which is not a multiple of $3,$ we can see that... |
For how many different digits $n$ is the three-digit number $14n$ divisible by $n$?
Note: $14n$ refers to a three-digit number with the unit digit of $n,$ not the product of $14$ and $n.$ | We have to account for each possible value of $n$ here. First of all, we can quickly find that for $n = 1, 2, 5,$ the resulting number $14n$ must be divisible by $n$, using their respective divisibility rules.
We see that for $n = 3$, we get $143.$ Since $1 + 4 + 3 = 8,$ which is not a multiple of $3,$ we can see that... |
What is the hundreds digit in the following product: $5 \times 6 \times 7 \times 8 \times 9 \times 10$ ? | The hundreds digit in the given expression is the same as the tens digit in the expression $5\times6\times7\times8\times9$, which is the same as the ones digit in the expression $6\times7\times4\times9$ (we divide out a 10 each time). $6\times7=42$ has a ones digit of 2 and $4\times9=36$ has a ones digit of 6, and $2\t... |
What is the hundreds digit in the following product: $5 \times 6 \times 7 \times 8 \times 9 \times 10$ ? | The hundreds digit in the given expression is the same as the tens digit in the expression $5\times6\times7\times8\times9$, which is the same as the ones digit in the expression $6\times7\times4\times9$ (we divide out a 10 each time). $6\times7=42$ has a ones digit of 2 and $4\times9=36$ has a ones digit of 6, and $2\t... |
What is the hundreds digit in the following product: $5 \times 6 \times 7 \times 8 \times 9 \times 10$ ? | The hundreds digit in the given expression is the same as the tens digit in the expression $5\times6\times7\times8\times9$, which is the same as the ones digit in the expression $6\times7\times4\times9$ (we divide out a 10 each time). $6\times7=42$ has a ones digit of 2 and $4\times9=36$ has a ones digit of 6, and $2\t... |
How many positive integers less than or equal to 100 have a prime factor that is greater than 4? | The easiest solution is to find the number of positive integers with only 2 and 3 as their prime factors. If the number has no factors of 3, the qualifying numbers are $2^0, 2^1, 2^2, 2^3, 2^4, 2^5, 2^6$ for 7 total. If there is one factor of 3, we have $2^0 \cdot 3^1, 2^1 \cdot 3^1, 2^2 \cdot 3^1, 2^3 \cdot 3^1, 2^4 \... |
How many positive integers less than or equal to 100 have a prime factor that is greater than 4? | The easiest solution is to find the number of positive integers with only 2 and 3 as their prime factors. If the number has no factors of 3, the qualifying numbers are $2^0, 2^1, 2^2, 2^3, 2^4, 2^5, 2^6$ for 7 total. If there is one factor of 3, we have $2^0 \cdot 3^1, 2^1 \cdot 3^1, 2^2 \cdot 3^1, 2^3 \cdot 3^1, 2^4 \... |
How many positive integers less than or equal to 100 have a prime factor that is greater than 4? | The easiest solution is to find the number of positive integers with only 2 and 3 as their prime factors. If the number has no factors of 3, the qualifying numbers are $2^0, 2^1, 2^2, 2^3, 2^4, 2^5, 2^6$ for 7 total. If there is one factor of 3, we have $2^0 \cdot 3^1, 2^1 \cdot 3^1, 2^2 \cdot 3^1, 2^3 \cdot 3^1, 2^4 \... |
What is the area of the portion of the circle defined by $x^2-12x+y^2=28$ that lies above the $x$-axis and to the right of the line $y=6-x$? | Completing the square, the equation of the circle can be rewritten in the form \[
(x^2-12x +36) +y^2=64,
\]or $(x-6)^2 +y^2 =8^2.$ The center of this circle is $(6,0)$, so both the $x$-axis and the line $y=6-x$ pass through the center of the circle: [asy]
size(8cm);
void axes(real x0, real x1, real y0, real y1)
{
d... |
What is the area of the portion of the circle defined by $x^2-12x+y^2=28$ that lies above the $x$-axis and to the right of the line $y=6-x$? | Completing the square, the equation of the circle can be rewritten in the form \[
(x^2-12x +36) +y^2=64,
\]or $(x-6)^2 +y^2 =8^2.$ The center of this circle is $(6,0)$, so both the $x$-axis and the line $y=6-x$ pass through the center of the circle: [asy]
size(8cm);
void axes(real x0, real x1, real y0, real y1)
{
d... |
What is the area of the portion of the circle defined by $x^2-12x+y^2=28$ that lies above the $x$-axis and to the right of the line $y=6-x$? | Completing the square, the equation of the circle can be rewritten in the form \[
(x^2-12x +36) +y^2=64,
\]or $(x-6)^2 +y^2 =8^2.$ The center of this circle is $(6,0)$, so both the $x$-axis and the line $y=6-x$ pass through the center of the circle: [asy]
size(8cm);
void axes(real x0, real x1, real y0, real y1)
{
d... |
Find the maximum of
\[\sqrt{x + 27} + \sqrt{13 - x} + \sqrt{x}\]for $0 \le x \le 13.$ | By Cauchy-Schwarz applied to $ \left( 1,\frac{1}{3},\frac{1}{2}\right) $ and $ (\sqrt{x+27},\sqrt{13-x},\sqrt{x}) $,
\[\left( 1 + \frac{1}{3} + \frac{1}{2} \right) ((x + 27) + 3(13 - x) + 2x) \ge (\sqrt{x + 27} + \sqrt{13 - x} + \sqrt{x})^2.\]Hence,
\[(\sqrt{x + 27} + \sqrt{13 - x} + \sqrt{x})^2 \le 121,\]so $\sqrt{x +... |
Find the maximum of
\[\sqrt{x + 27} + \sqrt{13 - x} + \sqrt{x}\]for $0 \le x \le 13.$ | By Cauchy-Schwarz applied to $ \left( 1,\frac{1}{3},\frac{1}{2}\right) $ and $ (\sqrt{x+27},\sqrt{13-x},\sqrt{x}) $,
\[\left( 1 + \frac{1}{3} + \frac{1}{2} \right) ((x + 27) + 3(13 - x) + 2x) \ge (\sqrt{x + 27} + \sqrt{13 - x} + \sqrt{x})^2.\]Hence,
\[(\sqrt{x + 27} + \sqrt{13 - x} + \sqrt{x})^2 \le 121,\]so $\sqrt{x +... |
Find the maximum of
\[\sqrt{x + 27} + \sqrt{13 - x} + \sqrt{x}\]for $0 \le x \le 13.$ | By Cauchy-Schwarz applied to $ \left( 1,\frac{1}{3},\frac{1}{2}\right) $ and $ (\sqrt{x+27},\sqrt{13-x},\sqrt{x}) $,
\[\left( 1 + \frac{1}{3} + \frac{1}{2} \right) ((x + 27) + 3(13 - x) + 2x) \ge (\sqrt{x + 27} + \sqrt{13 - x} + \sqrt{x})^2.\]Hence,
\[(\sqrt{x + 27} + \sqrt{13 - x} + \sqrt{x})^2 \le 121,\]so $\sqrt{x +... |
A $\textit{palindrome}$ is a positive integer which reads the same forward and backward, like $12321$ or $4884$.
How many $4$-digit palindromes are divisible by $3$? | Once we've picked the first two digits of a $4$-digit palindrome, the last two digits are automatically chosen by mirroring the first two. Thus, we can make exactly one $4$-digit palindrome for every $2$-digit number. For example, the $2$-digit number $57$ gives the palindrome $5775$.
For an integer to be divisible by... |
A $\textit{palindrome}$ is a positive integer which reads the same forward and backward, like $12321$ or $4884$.
How many $4$-digit palindromes are divisible by $3$? | Once we've picked the first two digits of a $4$-digit palindrome, the last two digits are automatically chosen by mirroring the first two. Thus, we can make exactly one $4$-digit palindrome for every $2$-digit number. For example, the $2$-digit number $57$ gives the palindrome $5775$.
For an integer to be divisible by... |
A $\textit{palindrome}$ is a positive integer which reads the same forward and backward, like $12321$ or $4884$.
How many $4$-digit palindromes are divisible by $3$? | Once we've picked the first two digits of a $4$-digit palindrome, the last two digits are automatically chosen by mirroring the first two. Thus, we can make exactly one $4$-digit palindrome for every $2$-digit number. For example, the $2$-digit number $57$ gives the palindrome $5775$.
For an integer to be divisible by... |
Let $S$ be the set of all ordered triple of integers $(a_1,a_2,a_3)$ with $1 \le a_1,a_2,a_3 \le 10$. Each ordered triple in $S$ generates a sequence according to the rule $a_n=a_{n-1}\cdot | a_{n-2}-a_{n-3} |$ for all $n\ge 4$. Find the number of such sequences for which $a_n=0$ for some $n$. | Let $a_1=x, a_2=y, a_3=z$. First note that if any absolute value equals 0, then $a_n=0$. Also note that if at any position, $a_n=a_{n-1}$, then $a_{n+2}=0$. Then, if any absolute value equals 1, then $a_n=0$. Therefore, if either $|y-x|$ or $|z-y|$ is less than or equal to 1, then that ordered triple meets the criteria... |
Let $S$ be the set of all ordered triple of integers $(a_1,a_2,a_3)$ with $1 \le a_1,a_2,a_3 \le 10$. Each ordered triple in $S$ generates a sequence according to the rule $a_n=a_{n-1}\cdot | a_{n-2}-a_{n-3} |$ for all $n\ge 4$. Find the number of such sequences for which $a_n=0$ for some $n$. | Let $a_1=x, a_2=y, a_3=z$. First note that if any absolute value equals 0, then $a_n=0$. Also note that if at any position, $a_n=a_{n-1}$, then $a_{n+2}=0$. Then, if any absolute value equals 1, then $a_n=0$. Therefore, if either $|y-x|$ or $|z-y|$ is less than or equal to 1, then that ordered triple meets the criteria... |
Let $S$ be the set of all ordered triple of integers $(a_1,a_2,a_3)$ with $1 \le a_1,a_2,a_3 \le 10$. Each ordered triple in $S$ generates a sequence according to the rule $a_n=a_{n-1}\cdot | a_{n-2}-a_{n-3} |$ for all $n\ge 4$. Find the number of such sequences for which $a_n=0$ for some $n$. | Let $a_1=x, a_2=y, a_3=z$. First note that if any absolute value equals 0, then $a_n=0$. Also note that if at any position, $a_n=a_{n-1}$, then $a_{n+2}=0$. Then, if any absolute value equals 1, then $a_n=0$. Therefore, if either $|y-x|$ or $|z-y|$ is less than or equal to 1, then that ordered triple meets the criteria... |
A positive integer $X$ is 2 more than a multiple of 3. Its units digit is the same as the units digit of a number that is 4 more than a multiple of 5. What is the smallest possible value of $X$? | If a positive integer is 4 more than a multiple of 5, then its units digit must be 4 or 9. We check positive integers ending in 4 or 9 until we find one which is 2 more than a multiple of 3: 4 is 1 more than a multiple of 3, 9 is a multiple of 3, and $\boxed{14}$ is 2 more than a multiple of 3. |
A positive integer $X$ is 2 more than a multiple of 3. Its units digit is the same as the units digit of a number that is 4 more than a multiple of 5. What is the smallest possible value of $X$? | If a positive integer is 4 more than a multiple of 5, then its units digit must be 4 or 9. We check positive integers ending in 4 or 9 until we find one which is 2 more than a multiple of 3: 4 is 1 more than a multiple of 3, 9 is a multiple of 3, and $\boxed{14}$ is 2 more than a multiple of 3. |
A positive integer $X$ is 2 more than a multiple of 3. Its units digit is the same as the units digit of a number that is 4 more than a multiple of 5. What is the smallest possible value of $X$? | If a positive integer is 4 more than a multiple of 5, then its units digit must be 4 or 9. We check positive integers ending in 4 or 9 until we find one which is 2 more than a multiple of 3: 4 is 1 more than a multiple of 3, 9 is a multiple of 3, and $\boxed{14}$ is 2 more than a multiple of 3. |
A PE class has 12 students, 6 girls and 6 boys. The coach has 4 jerseys in each of 3 colors to mark 3 teams for a soccer tournament. If the coach wants at least one girl and at least one boy on each team, how many ways can he give out the jerseys? (Jerseys of the same color are indistinguishable.) | In all, disregarding the gender rule, there are $$\binom{12}4\binom84=\frac{12\cdot11\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5}{4\cdot3\cdot2\cdot4\cdot3\cdot2}=34650$$ways of assigning the teams. We will count the number of ways a team can have all boys or all girls and subtract from this total.
There are 2 choices for t... |
A PE class has 12 students, 6 girls and 6 boys. The coach has 4 jerseys in each of 3 colors to mark 3 teams for a soccer tournament. If the coach wants at least one girl and at least one boy on each team, how many ways can he give out the jerseys? (Jerseys of the same color are indistinguishable.) | In all, disregarding the gender rule, there are $$\binom{12}4\binom84=\frac{12\cdot11\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5}{4\cdot3\cdot2\cdot4\cdot3\cdot2}=34650$$ways of assigning the teams. We will count the number of ways a team can have all boys or all girls and subtract from this total.
There are 2 choices for t... |
A PE class has 12 students, 6 girls and 6 boys. The coach has 4 jerseys in each of 3 colors to mark 3 teams for a soccer tournament. If the coach wants at least one girl and at least one boy on each team, how many ways can he give out the jerseys? (Jerseys of the same color are indistinguishable.) | In all, disregarding the gender rule, there are $$\binom{12}4\binom84=\frac{12\cdot11\cdot10\cdot9\cdot8\cdot7\cdot6\cdot5}{4\cdot3\cdot2\cdot4\cdot3\cdot2}=34650$$ways of assigning the teams. We will count the number of ways a team can have all boys or all girls and subtract from this total.
There are 2 choices for t... |
How many nonnegative integers can be written in the form\[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\]where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$?
$\textbf{(A) } 512 \qquad \textbf{(B) } 729 \qquad \textbf{(C) } 1094 \qquad \textbf{(D) } 3281 \qquad \textbf{(E)... | This looks like balanced ternary, in which all the integers with absolute values less than $\frac{3^n}{2}$ are represented in $n$ digits. There are 8 digits. Plugging in 8 into the formula for the balanced ternary gives a maximum bound of $|x|=3280.5$, which means there are 3280 positive integers, 0, and 3280 negative ... |
How many nonnegative integers can be written in the form\[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\]where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$?
$\textbf{(A) } 512 \qquad \textbf{(B) } 729 \qquad \textbf{(C) } 1094 \qquad \textbf{(D) } 3281 \qquad \textbf{(E)... | This looks like balanced ternary, in which all the integers with absolute values less than $\frac{3^n}{2}$ are represented in $n$ digits. There are 8 digits. Plugging in 8 into the formula for the balanced ternary gives a maximum bound of $|x|=3280.5$, which means there are 3280 positive integers, 0, and 3280 negative ... |
How many nonnegative integers can be written in the form\[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\]where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$?
$\textbf{(A) } 512 \qquad \textbf{(B) } 729 \qquad \textbf{(C) } 1094 \qquad \textbf{(D) } 3281 \qquad \textbf{(E)... | This looks like balanced ternary, in which all the integers with absolute values less than $\frac{3^n}{2}$ are represented in $n$ digits. There are 8 digits. Plugging in 8 into the formula for the balanced ternary gives a maximum bound of $|x|=3280.5$, which means there are 3280 positive integers, 0, and 3280 negative ... |
A regular tetrahedron is a pyramid with four faces, each of which is an equilateral triangle.
Let $ABCD$ be a regular tetrahedron and let $P$ be the unique point equidistant from points $A,B,C,D$. Extend $\overrightarrow{AP}$ to hit face $BCD$ at point $Q$. What is the ratio $PQ/AQ$ ? | Let's start with a picture: [asy]
import three;
triple d = (0,0,0);
triple b = (1,0,0);
triple c = (1/2,sqrt(3)/2,0);
triple a = (1/2,sqrt(3)/6,sqrt(6)/3);
triple p = (a+b+c+d)/4;
triple q = (d+b+c)/3;
draw(a--b--c--a); draw(c--d--b,dotted); draw(d--a,dotted); draw(a--q,dashed);
dot(a); dot(b); dot(c); dot(d); dot(p); ... |
A regular tetrahedron is a pyramid with four faces, each of which is an equilateral triangle.
Let $ABCD$ be a regular tetrahedron and let $P$ be the unique point equidistant from points $A,B,C,D$. Extend $\overrightarrow{AP}$ to hit face $BCD$ at point $Q$. What is the ratio $PQ/AQ$ ? | Let's start with a picture: [asy]
import three;
triple d = (0,0,0);
triple b = (1,0,0);
triple c = (1/2,sqrt(3)/2,0);
triple a = (1/2,sqrt(3)/6,sqrt(6)/3);
triple p = (a+b+c+d)/4;
triple q = (d+b+c)/3;
draw(a--b--c--a); draw(c--d--b,dotted); draw(d--a,dotted); draw(a--q,dashed);
dot(a); dot(b); dot(c); dot(d); dot(p); ... |
A regular tetrahedron is a pyramid with four faces, each of which is an equilateral triangle.
Let $ABCD$ be a regular tetrahedron and let $P$ be the unique point equidistant from points $A,B,C,D$. Extend $\overrightarrow{AP}$ to hit face $BCD$ at point $Q$. What is the ratio $PQ/AQ$ ? | Let's start with a picture: [asy]
import three;
triple d = (0,0,0);
triple b = (1,0,0);
triple c = (1/2,sqrt(3)/2,0);
triple a = (1/2,sqrt(3)/6,sqrt(6)/3);
triple p = (a+b+c+d)/4;
triple q = (d+b+c)/3;
draw(a--b--c--a); draw(c--d--b,dotted); draw(d--a,dotted); draw(a--q,dashed);
dot(a); dot(b); dot(c); dot(d); dot(p); ... |
Regular hexagon $ABCDEF$ is the base of right pyramid $\allowbreak PABCDEF$. If $PAD$ is an equilateral triangle with side length 8, then what is the volume of the pyramid? | [asy]
import three;
triple A = (1,0,0);
triple B = (0.5,sqrt(3)/2,0);
triple C = (-0.5,sqrt(3)/2,0);
triple D = (-1,0,0);
triple EE = (-0.5,-sqrt(3)/2,0);
triple F = (0.5,-sqrt(3)/2,0);
triple P = (0,0,1);
draw(F--A--B--C);
draw(C--D--EE--F,dashed);
draw(A--P--C);
draw(EE--P--D,dashed);
draw(B--P--F);
label("$A$",A,S... |
Regular hexagon $ABCDEF$ is the base of right pyramid $\allowbreak PABCDEF$. If $PAD$ is an equilateral triangle with side length 8, then what is the volume of the pyramid? | [asy]
import three;
triple A = (1,0,0);
triple B = (0.5,sqrt(3)/2,0);
triple C = (-0.5,sqrt(3)/2,0);
triple D = (-1,0,0);
triple EE = (-0.5,-sqrt(3)/2,0);
triple F = (0.5,-sqrt(3)/2,0);
triple P = (0,0,1);
draw(F--A--B--C);
draw(C--D--EE--F,dashed);
draw(A--P--C);
draw(EE--P--D,dashed);
draw(B--P--F);
label("$A$",A,S... |
Regular hexagon $ABCDEF$ is the base of right pyramid $\allowbreak PABCDEF$. If $PAD$ is an equilateral triangle with side length 8, then what is the volume of the pyramid? | [asy]
import three;
triple A = (1,0,0);
triple B = (0.5,sqrt(3)/2,0);
triple C = (-0.5,sqrt(3)/2,0);
triple D = (-1,0,0);
triple EE = (-0.5,-sqrt(3)/2,0);
triple F = (0.5,-sqrt(3)/2,0);
triple P = (0,0,1);
draw(F--A--B--C);
draw(C--D--EE--F,dashed);
draw(A--P--C);
draw(EE--P--D,dashed);
draw(B--P--F);
label("$A$",A,S... |
Two numbers $90$ and $m$ share exactly three positive divisors. What is the greatest of these three common divisors? | Recall that the common divisors of two integers are precisely the divisors of the greatest common divisor. So, for two numbers to have exactly three positive divisors in common, those divisors must be $1$, $p$, and $p^2$ such that $p$ is prime. We now look at the prime factorization of $90$: $90=2 \cdot 3^2 \cdot 5$. S... |
Two numbers $90$ and $m$ share exactly three positive divisors. What is the greatest of these three common divisors? | Recall that the common divisors of two integers are precisely the divisors of the greatest common divisor. So, for two numbers to have exactly three positive divisors in common, those divisors must be $1$, $p$, and $p^2$ such that $p$ is prime. We now look at the prime factorization of $90$: $90=2 \cdot 3^2 \cdot 5$. S... |
Two numbers $90$ and $m$ share exactly three positive divisors. What is the greatest of these three common divisors? | Recall that the common divisors of two integers are precisely the divisors of the greatest common divisor. So, for two numbers to have exactly three positive divisors in common, those divisors must be $1$, $p$, and $p^2$ such that $p$ is prime. We now look at the prime factorization of $90$: $90=2 \cdot 3^2 \cdot 5$. S... |
On the complex plane, the parallelogram formed by the points 0, $z,$ $\frac{1}{z},$ and $z + \frac{1}{z}$ has area $\frac{35}{37}.$ If the real part of $z$ is positive, let $d$ be the smallest possible value of $\left| z + \frac{1}{z} \right|.$ Compute $d^2.$ | Let $z = r (\cos \theta + i \sin \theta).$ Then
\[\frac{1}{z} = \frac{1}{r (\cos \theta + i \sin \theta)} = \frac{1}{r} (\cos (-\theta) + i \sin (-\theta)) = \frac{1}{r} (\cos \theta - i \sin \theta).\]By the shoelace formula, the area of the triangle formed by 0, $z = r \cos \theta + ir \sin \theta$ and $\frac{1}{z} ... |
On the complex plane, the parallelogram formed by the points 0, $z,$ $\frac{1}{z},$ and $z + \frac{1}{z}$ has area $\frac{35}{37}.$ If the real part of $z$ is positive, let $d$ be the smallest possible value of $\left| z + \frac{1}{z} \right|.$ Compute $d^2.$ | Let $z = r (\cos \theta + i \sin \theta).$ Then
\[\frac{1}{z} = \frac{1}{r (\cos \theta + i \sin \theta)} = \frac{1}{r} (\cos (-\theta) + i \sin (-\theta)) = \frac{1}{r} (\cos \theta - i \sin \theta).\]By the shoelace formula, the area of the triangle formed by 0, $z = r \cos \theta + ir \sin \theta$ and $\frac{1}{z} ... |
On the complex plane, the parallelogram formed by the points 0, $z,$ $\frac{1}{z},$ and $z + \frac{1}{z}$ has area $\frac{35}{37}.$ If the real part of $z$ is positive, let $d$ be the smallest possible value of $\left| z + \frac{1}{z} \right|.$ Compute $d^2.$ | Let $z = r (\cos \theta + i \sin \theta).$ Then
\[\frac{1}{z} = \frac{1}{r (\cos \theta + i \sin \theta)} = \frac{1}{r} (\cos (-\theta) + i \sin (-\theta)) = \frac{1}{r} (\cos \theta - i \sin \theta).\]By the shoelace formula, the area of the triangle formed by 0, $z = r \cos \theta + ir \sin \theta$ and $\frac{1}{z} ... |
Evaluate the product of $\sqrt[3]{2^2} + \sqrt[3]{2}$ with $2 - 2\sqrt[3]{2} - \sqrt[3]{2^2}$. | The given expression is equivalent to the product $$\left(\sqrt[3]{4} + \sqrt[3]{2}\right) \cdot \left(-\sqrt[3]{16} + \sqrt[3]{8} - \sqrt[3]{4}\right).$$If we let $a = \sqrt[3]{4}$ and $b = \sqrt[3]{2}$, then the above expression becomes $$(a+b)(-a^2 + ab - b^2) = -(a+b)(a^2 - ab + b^2) = -(a^3 + b^3).$$Thus, the expr... |
Evaluate the product of $\sqrt[3]{2^2} + \sqrt[3]{2}$ with $2 - 2\sqrt[3]{2} - \sqrt[3]{2^2}$. | The given expression is equivalent to the product $$\left(\sqrt[3]{4} + \sqrt[3]{2}\right) \cdot \left(-\sqrt[3]{16} + \sqrt[3]{8} - \sqrt[3]{4}\right).$$If we let $a = \sqrt[3]{4}$ and $b = \sqrt[3]{2}$, then the above expression becomes $$(a+b)(-a^2 + ab - b^2) = -(a+b)(a^2 - ab + b^2) = -(a^3 + b^3).$$Thus, the expr... |
Evaluate the product of $\sqrt[3]{2^2} + \sqrt[3]{2}$ with $2 - 2\sqrt[3]{2} - \sqrt[3]{2^2}$. | The given expression is equivalent to the product $$\left(\sqrt[3]{4} + \sqrt[3]{2}\right) \cdot \left(-\sqrt[3]{16} + \sqrt[3]{8} - \sqrt[3]{4}\right).$$If we let $a = \sqrt[3]{4}$ and $b = \sqrt[3]{2}$, then the above expression becomes $$(a+b)(-a^2 + ab - b^2) = -(a+b)(a^2 - ab + b^2) = -(a^3 + b^3).$$Thus, the expr... |
The polynomial
$$g(x) = x^3 - x^2 - (m^2 + m) x + 2m^2 + 4m + 2$$is divisible by $x-4$ and all of its zeroes are integers. Find all possible values of $m$. | Since $g(x)$ is divisible by $x-4$, we have $g(4)=0$. We also have
\begin{align*}
g(4) &= 4^3 - 4^2 - (m^2+m)(4) + 2m^2+4m+2 \\
&= 50 - 2m^2,
\end{align*}so $0=50-2m^2$. Thus $m$ can only be $5$ or $-5$. We check both possibilities.
If $m=5$, then $g(x)=x^3-x^2-30x+72=(x-4)(x^2+3x-18)=(x-4)(x+6)(x-3)$, so all zeroes a... |
The polynomial
$$g(x) = x^3 - x^2 - (m^2 + m) x + 2m^2 + 4m + 2$$is divisible by $x-4$ and all of its zeroes are integers. Find all possible values of $m$. | Since $g(x)$ is divisible by $x-4$, we have $g(4)=0$. We also have
\begin{align*}
g(4) &= 4^3 - 4^2 - (m^2+m)(4) + 2m^2+4m+2 \\
&= 50 - 2m^2,
\end{align*}so $0=50-2m^2$. Thus $m$ can only be $5$ or $-5$. We check both possibilities.
If $m=5$, then $g(x)=x^3-x^2-30x+72=(x-4)(x^2+3x-18)=(x-4)(x+6)(x-3)$, so all zeroes a... |
The polynomial
$$g(x) = x^3 - x^2 - (m^2 + m) x + 2m^2 + 4m + 2$$is divisible by $x-4$ and all of its zeroes are integers. Find all possible values of $m$. | Since $g(x)$ is divisible by $x-4$, we have $g(4)=0$. We also have
\begin{align*}
g(4) &= 4^3 - 4^2 - (m^2+m)(4) + 2m^2+4m+2 \\
&= 50 - 2m^2,
\end{align*}so $0=50-2m^2$. Thus $m$ can only be $5$ or $-5$. We check both possibilities.
If $m=5$, then $g(x)=x^3-x^2-30x+72=(x-4)(x^2+3x-18)=(x-4)(x+6)(x-3)$, so all zeroes a... |
If Greg rolls four fair six-sided dice, what is the probability that he rolls more 1's than 6's? | We notice that the probability that he rolls more 1's than 6's must equal the probability that he rolls more 6's than 1's. So, we can find the probability that Greg rolls the same number of 1's and 6's, subtract it from 1, and divide by 2 to find the probability that Greg rolls more 1's than 6's. There are three ways G... |
If Greg rolls four fair six-sided dice, what is the probability that he rolls more 1's than 6's? | We notice that the probability that he rolls more 1's than 6's must equal the probability that he rolls more 6's than 1's. So, we can find the probability that Greg rolls the same number of 1's and 6's, subtract it from 1, and divide by 2 to find the probability that Greg rolls more 1's than 6's. There are three ways G... |
If Greg rolls four fair six-sided dice, what is the probability that he rolls more 1's than 6's? | We notice that the probability that he rolls more 1's than 6's must equal the probability that he rolls more 6's than 1's. So, we can find the probability that Greg rolls the same number of 1's and 6's, subtract it from 1, and divide by 2 to find the probability that Greg rolls more 1's than 6's. There are three ways G... |
A bored student walks down a hall that contains a row of closed lockers, numbered $1$ to $1024$. He opens the locker numbered 1, and then alternates between skipping and opening each locker thereafter. When he reaches the end of the hall, the student turns around and starts back. He opens the first closed locker he enc... | On his first pass, he opens all of the odd lockers. So there are only even lockers closed. Then he opens the lockers that are multiples of $4$, leaving only lockers $2 \pmod{8}$ and $6 \pmod{8}$. Then he goes ahead and opens all lockers $2 \pmod {8}$, leaving lockers either $6 \pmod {16}$ or $14 \pmod {16}$. He then go... |
A bored student walks down a hall that contains a row of closed lockers, numbered $1$ to $1024$. He opens the locker numbered 1, and then alternates between skipping and opening each locker thereafter. When he reaches the end of the hall, the student turns around and starts back. He opens the first closed locker he enc... | On his first pass, he opens all of the odd lockers. So there are only even lockers closed. Then he opens the lockers that are multiples of $4$, leaving only lockers $2 \pmod{8}$ and $6 \pmod{8}$. Then he goes ahead and opens all lockers $2 \pmod {8}$, leaving lockers either $6 \pmod {16}$ or $14 \pmod {16}$. He then go... |
A bored student walks down a hall that contains a row of closed lockers, numbered $1$ to $1024$. He opens the locker numbered 1, and then alternates between skipping and opening each locker thereafter. When he reaches the end of the hall, the student turns around and starts back. He opens the first closed locker he enc... | On his first pass, he opens all of the odd lockers. So there are only even lockers closed. Then he opens the lockers that are multiples of $4$, leaving only lockers $2 \pmod{8}$ and $6 \pmod{8}$. Then he goes ahead and opens all lockers $2 \pmod {8}$, leaving lockers either $6 \pmod {16}$ or $14 \pmod {16}$. He then go... |
The least common multiple of $1!+2!$, $2!+3!$, $3!+4!$, $4!+5!$, $5!+6!$, $6!+7!$, $7!+8!$, and $8!+9!$ can be expressed in the form $a\cdot b!$, where $a$ and $b$ are integers and $b$ is as large as possible. What is $a+b$? | Note that we can factor $n!+(n+1)!$ as $n!\cdot [1+(n+1)] = n!\cdot(n+2)$. Thus, we have \begin{align*}
1!+2! &= 1!\cdot 3 \\
2!+3! &= 2!\cdot 4 \\
3!+4! &= 3!\cdot 5 \\
4!+5! &= 4!\cdot 6 \\
5!+6! &= 5!\cdot 7 \\
6!+7! &= 6!\cdot 8 \\
7!+8! &= 7!\cdot 9 \\
8!+9! &= 8!\cdot 10
\end{align*}The last two numbers are $9\cd... |
The least common multiple of $1!+2!$, $2!+3!$, $3!+4!$, $4!+5!$, $5!+6!$, $6!+7!$, $7!+8!$, and $8!+9!$ can be expressed in the form $a\cdot b!$, where $a$ and $b$ are integers and $b$ is as large as possible. What is $a+b$? | Note that we can factor $n!+(n+1)!$ as $n!\cdot [1+(n+1)] = n!\cdot(n+2)$. Thus, we have \begin{align*}
1!+2! &= 1!\cdot 3 \\
2!+3! &= 2!\cdot 4 \\
3!+4! &= 3!\cdot 5 \\
4!+5! &= 4!\cdot 6 \\
5!+6! &= 5!\cdot 7 \\
6!+7! &= 6!\cdot 8 \\
7!+8! &= 7!\cdot 9 \\
8!+9! &= 8!\cdot 10
\end{align*}The last two numbers are $9\cd... |
The least common multiple of $1!+2!$, $2!+3!$, $3!+4!$, $4!+5!$, $5!+6!$, $6!+7!$, $7!+8!$, and $8!+9!$ can be expressed in the form $a\cdot b!$, where $a$ and $b$ are integers and $b$ is as large as possible. What is $a+b$? | Note that we can factor $n!+(n+1)!$ as $n!\cdot [1+(n+1)] = n!\cdot(n+2)$. Thus, we have \begin{align*}
1!+2! &= 1!\cdot 3 \\
2!+3! &= 2!\cdot 4 \\
3!+4! &= 3!\cdot 5 \\
4!+5! &= 4!\cdot 6 \\
5!+6! &= 5!\cdot 7 \\
6!+7! &= 6!\cdot 8 \\
7!+8! &= 7!\cdot 9 \\
8!+9! &= 8!\cdot 10
\end{align*}The last two numbers are $9\cd... |
The product of the proper positive integer factors of $n$ can be written as $n^{(ax+b)/c}$, where $x$ is the number of positive divisors $n$ has, $c$ is a positive integer, and the greatest common factor of the three integers $a$, $b$, and $c$ is $1$. What is $a+b+c$? | Recall that by pairing up divisors of $n$, we can show that the product of the positive integer factors of $n$ is $n^\frac{x}{2}$. We divide this formula by $n$ to get the product of the proper positive integer factors of $n$, and we obtain $\frac{n^\frac{x}{2}}{n} = n^{\frac{x}{2}-1} = n^\frac{x-2}{2}$. Therefore, $a ... |
The product of the proper positive integer factors of $n$ can be written as $n^{(ax+b)/c}$, where $x$ is the number of positive divisors $n$ has, $c$ is a positive integer, and the greatest common factor of the three integers $a$, $b$, and $c$ is $1$. What is $a+b+c$? | Recall that by pairing up divisors of $n$, we can show that the product of the positive integer factors of $n$ is $n^\frac{x}{2}$. We divide this formula by $n$ to get the product of the proper positive integer factors of $n$, and we obtain $\frac{n^\frac{x}{2}}{n} = n^{\frac{x}{2}-1} = n^\frac{x-2}{2}$. Therefore, $a ... |
The product of the proper positive integer factors of $n$ can be written as $n^{(ax+b)/c}$, where $x$ is the number of positive divisors $n$ has, $c$ is a positive integer, and the greatest common factor of the three integers $a$, $b$, and $c$ is $1$. What is $a+b+c$? | Recall that by pairing up divisors of $n$, we can show that the product of the positive integer factors of $n$ is $n^\frac{x}{2}$. We divide this formula by $n$ to get the product of the proper positive integer factors of $n$, and we obtain $\frac{n^\frac{x}{2}}{n} = n^{\frac{x}{2}-1} = n^\frac{x-2}{2}$. Therefore, $a ... |
For how many integers $n$ where $2 \le n \le 100$ is $\binom{n}{2}$ odd? | $\binom{n}{2} = \frac{n(n-1)}{2}$. In order for this fraction to be odd, neither $n$ nor $n-1$ can be divisible by $4$, because only one of $n$ and $n-1$ can be even. There are $25$ integers where $n$ is divisible by $4$, namely the multiples of $4$ from $4$ to $100$. There are $24$ integers where $n-1$ is divisible by... |
For how many integers $n$ where $2 \le n \le 100$ is $\binom{n}{2}$ odd? | $\binom{n}{2} = \frac{n(n-1)}{2}$. In order for this fraction to be odd, neither $n$ nor $n-1$ can be divisible by $4$, because only one of $n$ and $n-1$ can be even. There are $25$ integers where $n$ is divisible by $4$, namely the multiples of $4$ from $4$ to $100$. There are $24$ integers where $n-1$ is divisible by... |
For how many integers $n$ where $2 \le n \le 100$ is $\binom{n}{2}$ odd? | $\binom{n}{2} = \frac{n(n-1)}{2}$. In order for this fraction to be odd, neither $n$ nor $n-1$ can be divisible by $4$, because only one of $n$ and $n-1$ can be even. There are $25$ integers where $n$ is divisible by $4$, namely the multiples of $4$ from $4$ to $100$. There are $24$ integers where $n-1$ is divisible by... |
All triangles have the same value, and all circles have the same value. What is the sum of two circles? \begin{align*}
\Delta + \bigcirc + \Delta + \bigcirc + \Delta + \bigcirc + \Delta + \Delta&= 21\\
\bigcirc + \Delta+\bigcirc+\Delta+\bigcirc + \Delta + \bigcirc + \bigcirc &= 16\\
\bigcirc + \bigcirc &= \ \, ?
\end{a... | Replace a triangle with the letter $a$ and a circle with the letter $b.$ The two given equations become \begin{align*}
5a+3b&=21\\
3a+5b&=16
\end{align*}Multiplying the first equation by $3,$ we get $15a+9b=63.$ Multiplying the second equation by $5,$ we get $15a+25b=80.$ Subtracting this last equation from the second-... |
All triangles have the same value, and all circles have the same value. What is the sum of two circles? \begin{align*}
\Delta + \bigcirc + \Delta + \bigcirc + \Delta + \bigcirc + \Delta + \Delta&= 21\\
\bigcirc + \Delta+\bigcirc+\Delta+\bigcirc + \Delta + \bigcirc + \bigcirc &= 16\\
\bigcirc + \bigcirc &= \ \, ?
\end{a... | Replace a triangle with the letter $a$ and a circle with the letter $b.$ The two given equations become \begin{align*}
5a+3b&=21\\
3a+5b&=16
\end{align*}Multiplying the first equation by $3,$ we get $15a+9b=63.$ Multiplying the second equation by $5,$ we get $15a+25b=80.$ Subtracting this last equation from the second-... |
All triangles have the same value, and all circles have the same value. What is the sum of two circles? \begin{align*}
\Delta + \bigcirc + \Delta + \bigcirc + \Delta + \bigcirc + \Delta + \Delta&= 21\\
\bigcirc + \Delta+\bigcirc+\Delta+\bigcirc + \Delta + \bigcirc + \bigcirc &= 16\\
\bigcirc + \bigcirc &= \ \, ?
\end{a... | Replace a triangle with the letter $a$ and a circle with the letter $b.$ The two given equations become \begin{align*}
5a+3b&=21\\
3a+5b&=16
\end{align*}Multiplying the first equation by $3,$ we get $15a+9b=63.$ Multiplying the second equation by $5,$ we get $15a+25b=80.$ Subtracting this last equation from the second-... |
Find the number of real solutions to $\sin 6 \pi x = x.$ | Since $|\sin 6 \pi x| \le 1$ for all $x,$ any points of intersection must lie in the interval $x \in [-1,1].$
[asy]
unitsize(2 cm);
real func(real x) {
return(sin(6*pi*x));
}
draw(xscale(2)*graph(func,-1,1),red);
draw((-2,-1)--(2,1),blue);
draw((-2.2,0)--(2.2,0));
draw((0,-1)--(0,1));
label("$-1$", (-2,0), S, UnF... |
Find the number of real solutions to $\sin 6 \pi x = x.$ | Since $|\sin 6 \pi x| \le 1$ for all $x,$ any points of intersection must lie in the interval $x \in [-1,1].$
[asy]
unitsize(2 cm);
real func(real x) {
return(sin(6*pi*x));
}
draw(xscale(2)*graph(func,-1,1),red);
draw((-2,-1)--(2,1),blue);
draw((-2.2,0)--(2.2,0));
draw((0,-1)--(0,1));
label("$-1$", (-2,0), S, UnF... |
Find the number of real solutions to $\sin 6 \pi x = x.$ | Since $|\sin 6 \pi x| \le 1$ for all $x,$ any points of intersection must lie in the interval $x \in [-1,1].$
[asy]
unitsize(2 cm);
real func(real x) {
return(sin(6*pi*x));
}
draw(xscale(2)*graph(func,-1,1),red);
draw((-2,-1)--(2,1),blue);
draw((-2.2,0)--(2.2,0));
draw((0,-1)--(0,1));
label("$-1$", (-2,0), S, UnF... |
Six points are equally spaced around a circle of radius 1. Three of these points are the vertices of a triangle that is neither equilateral nor isosceles. What is the area of this triangle? | The only possible scalene (not equilateral or isosceles) triangle, up to congruence, that can be made from the given points is shown below: [asy] markscalefactor /= 2;size(4cm); draw(unitcircle); for(int i=0; i<6; ++i) dot(dir(60*i)); draw(dir(120)--dir(60)--dir(-60)--cycle); dot((0,0)); draw((0,0)--dir(60),dotted); dr... |
Six points are equally spaced around a circle of radius 1. Three of these points are the vertices of a triangle that is neither equilateral nor isosceles. What is the area of this triangle? | The only possible scalene (not equilateral or isosceles) triangle, up to congruence, that can be made from the given points is shown below: [asy] markscalefactor /= 2;size(4cm); draw(unitcircle); for(int i=0; i<6; ++i) dot(dir(60*i)); draw(dir(120)--dir(60)--dir(-60)--cycle); dot((0,0)); draw((0,0)--dir(60),dotted); dr... |
Six points are equally spaced around a circle of radius 1. Three of these points are the vertices of a triangle that is neither equilateral nor isosceles. What is the area of this triangle? | The only possible scalene (not equilateral or isosceles) triangle, up to congruence, that can be made from the given points is shown below: [asy] markscalefactor /= 2;size(4cm); draw(unitcircle); for(int i=0; i<6; ++i) dot(dir(60*i)); draw(dir(120)--dir(60)--dir(-60)--cycle); dot((0,0)); draw((0,0)--dir(60),dotted); dr... |
Find the volume of the region in space defined by
\[|x + y + z| + |x + y - z| \le 8\]and $x,$ $y,$ $z \ge 0.$ | Let $a$ and $b$ be real numbers. If $a \ge b,$ then
\[|a + b| + |a - b| = (a + b) + (a - b) = 2a.\]If $a \le b,$ then
\[|a + b| + |a - b| = (a + b) + (b - a) = 2b.\]In either case, $|a + b| + |a - b| = 2 \max\{a,b\}.$
Thus, the condition $|x + y + z| + |x + y - z| \le 8$ is equivalent to
\[2 \max \{x + y, z\} \le 8,\... |
Find the volume of the region in space defined by
\[|x + y + z| + |x + y - z| \le 8\]and $x,$ $y,$ $z \ge 0.$ | Let $a$ and $b$ be real numbers. If $a \ge b,$ then
\[|a + b| + |a - b| = (a + b) + (a - b) = 2a.\]If $a \le b,$ then
\[|a + b| + |a - b| = (a + b) + (b - a) = 2b.\]In either case, $|a + b| + |a - b| = 2 \max\{a,b\}.$
Thus, the condition $|x + y + z| + |x + y - z| \le 8$ is equivalent to
\[2 \max \{x + y, z\} \le 8,\... |
Find the volume of the region in space defined by
\[|x + y + z| + |x + y - z| \le 8\]and $x,$ $y,$ $z \ge 0.$ | Let $a$ and $b$ be real numbers. If $a \ge b,$ then
\[|a + b| + |a - b| = (a + b) + (a - b) = 2a.\]If $a \le b,$ then
\[|a + b| + |a - b| = (a + b) + (b - a) = 2b.\]In either case, $|a + b| + |a - b| = 2 \max\{a,b\}.$
Thus, the condition $|x + y + z| + |x + y - z| \le 8$ is equivalent to
\[2 \max \{x + y, z\} \le 8,\... |
We have triangle $\triangle ABC$ where $AB = AC$ and $AD$ is an altitude. Meanwhile, $E$ is a point on $AC$ such that $AB \parallel DE.$ If $BC = 12$ and the area of $\triangle ABC$ is $180,$ what is the area of $ABDE$? | We first draw a diagram: [asy]
pair A, B, C, D, E;
A = (0, 30);
B = (-12, 0);
C = (12, 0);
D = 0.5 * B + 0.5 * C;
E = 0.5 * A + 0.5 * C;
draw(A--B--C--cycle);
draw(A--D);
draw(D--E);
draw(D+(-1, 0)--D+(-1, 1)--D+(0, 1));
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$E$", E, N... |
We have triangle $\triangle ABC$ where $AB = AC$ and $AD$ is an altitude. Meanwhile, $E$ is a point on $AC$ such that $AB \parallel DE.$ If $BC = 12$ and the area of $\triangle ABC$ is $180,$ what is the area of $ABDE$? | We first draw a diagram: [asy]
pair A, B, C, D, E;
A = (0, 30);
B = (-12, 0);
C = (12, 0);
D = 0.5 * B + 0.5 * C;
E = 0.5 * A + 0.5 * C;
draw(A--B--C--cycle);
draw(A--D);
draw(D--E);
draw(D+(-1, 0)--D+(-1, 1)--D+(0, 1));
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$E$", E, N... |
We have triangle $\triangle ABC$ where $AB = AC$ and $AD$ is an altitude. Meanwhile, $E$ is a point on $AC$ such that $AB \parallel DE.$ If $BC = 12$ and the area of $\triangle ABC$ is $180,$ what is the area of $ABDE$? | We first draw a diagram: [asy]
pair A, B, C, D, E;
A = (0, 30);
B = (-12, 0);
C = (12, 0);
D = 0.5 * B + 0.5 * C;
E = 0.5 * A + 0.5 * C;
draw(A--B--C--cycle);
draw(A--D);
draw(D--E);
draw(D+(-1, 0)--D+(-1, 1)--D+(0, 1));
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$E$", E, N... |
How many three-digit positive integers $x$ satisfy $3874x+481\equiv 1205 \pmod{23}$? | We begin by replacing the coefficients and constants in the equation with their residues modulo 23. We find that 3874 divided by 23 gives a remainder of 10, 481 divided by 23 gives a remainder of 21, and 1205 gives a remainder of 9. So the given congruence is equivalent to $$
10x + 21 \equiv 9 \pmod{23}.
$$Now add 2 t... |
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