instruction stringlengths 0 4.41k | output stringlengths 1 5.58k | text stringlengths 13 5.7k |
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The diagonals of a rhombus are $12$ inches and $16$ inches. What is the perimeter of the rhombus, in inches? | The diagonals are perpendicular bisectors of each other, so therefore the side length of the rhombus can be calculated as $\sqrt{6^2+8^2} = 10$. Therefore, the perimeter of the rhombus is $4 \times 10 = \boxed{40}$ inches. The answer is 40. | The diagonals of a rhombus are $12$ inches and $16$ inches. What is the perimeter of the rhombus, in inches?
The diagonals are perpendicular bisectors of each other, so therefore the side length of the rhombus can be calculated as $\sqrt{6^2+8^2} = 10$. Therefore, the perimeter of the rhombus is $4 \times 10 = \boxed... |
$\Delta ABC$ is isosceles with $AC = BC$. If $m\angle C = 50^{\circ}$, what is the number of degrees in $m\angle ABD$? [asy] pair A,B,C,D,E;
C = dir(65); B = C + dir(-65); D = (1,0); E = (2,0);
draw(B--C--A--D); dot(D);
label("$A$",A,S); label("$B$",B,S); label("$D$",D,S); label("$C$",C,N);
[/asy] | Let $x$ be the number of degrees in $\angle ABC$. Since $\triangle ABC$ is isosceles with $AC=BC$, we have $\angle BAC=\angle ABC$.
So, the three interior angles of $\triangle ABC$ measure $x^\circ$, $x^\circ$, and $50^\circ$. The sum of the angles in a triangle is $180^\circ$, so we have $$x+x+50 = 180,$$which we can... | $\Delta ABC$ is isosceles with $AC = BC$. If $m\angle C = 50^{\circ}$, what is the number of degrees in $m\angle ABD$? [asy] pair A,B,C,D,E;
C = dir(65); B = C + dir(-65); D = (1,0); E = (2,0);
draw(B--C--A--D); dot(D);
label("$A$",A,S); label("$B$",B,S); label("$D$",D,S); label("$C$",C,N);
[/asy]
Let $x$ be the numb... |
Find the equation whose graph is a parabola with vertex $(-1,-2)$, horizontal axis of symmetry, and contains the point $(3,0)$. Express your answer in the form "$ay^2+by+c$". | Since the axis of symmetry is horizontal and the vertex is $(-1,-2)$, the parabola may also be written as \[x=a(y+2)^2-1\] for some value of $a$. Plugging the point $(3,0)$ into this expression gives \[3=a(0+2)^2-1=4a-1.\] This tells us $a=1$.
Our equation is \[x=(y+2)^2-1.\] Putting it $x=ay^2+by+c$ form requires... | Find the equation whose graph is a parabola with vertex $(-1,-2)$, horizontal axis of symmetry, and contains the point $(3,0)$. Express your answer in the form "$ay^2+by+c$".
Since the axis of symmetry is horizontal and the vertex is $(-1,-2)$, the parabola may also be written as \[x=a(y+2)^2-1\] for some value of $... |
In triangle $ABC$, $AB=13$, $BC=15$ and $CA=17$. Point $D$ is on $\overline{AB}$, $E$ is on $\overline{BC}$, and $F$ is on $\overline{CA}$. Let $AD=p\cdot AB$, $BE=q\cdot BC$, and $CF=r\cdot CA$, where $p$, $q$, and $r$ are positive and satisfy $p+q+r=1/2$ and $p^2+q^2+r^2=1/3$. The ratio of the area of triangle $DEF$ ... | We let $[\ldots]$ denote area; then the desired value is
$\frac mn = \frac{[DEF]}{[ABC]} = \frac{[ABC] - [ADF] - [BDE] - [CEF]}{[ABC]}$
Using the formula for the area of a triangle $\frac{1}{2}ab\sin C$, we find that
$\frac{[ADF]}{[ABC]} = \frac{\frac 12 \cdot p \cdot AB \cdot (1-r) \cdot AC \cdot \sin \angle CAB}{\fra... | In triangle $ABC$, $AB=13$, $BC=15$ and $CA=17$. Point $D$ is on $\overline{AB}$, $E$ is on $\overline{BC}$, and $F$ is on $\overline{CA}$. Let $AD=p\cdot AB$, $BE=q\cdot BC$, and $CF=r\cdot CA$, where $p$, $q$, and $r$ are positive and satisfy $p+q+r=1/2$ and $p^2+q^2+r^2=1/3$. The ratio of the area of triangle $DEF$ ... |
Let $x=\frac{\sum\limits_{n=1}^{45} \cos n^\circ}{\sum\limits_{n=1}^{45} \sin n^\circ}$. What is the greatest integer that does not exceed $100x$? | Note that $\frac{\sum_{n=1}^{45} \cos n}{\sum_{n=1}^{45} \sin n} = \frac {\cos 1 + \cos 2 + \dots + \cos 45}{\cos 90 + \cos 89 + \dots + \cos 46}$
Now use the sum-product formula $\cos x + \cos y = 2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})$ We want to pair up $[1, 45]$, $[2, 44]$, $[3, 43]$, etc. from the numerator and $... | Let $x=\frac{\sum\limits_{n=1}^{45} \cos n^\circ}{\sum\limits_{n=1}^{45} \sin n^\circ}$. What is the greatest integer that does not exceed $100x$?
Note that $\frac{\sum_{n=1}^{45} \cos n}{\sum_{n=1}^{45} \sin n} = \frac {\cos 1 + \cos 2 + \dots + \cos 45}{\cos 90 + \cos 89 + \dots + \cos 46}$
Now use the sum-product ... |
Consider a sequence $x_1,$ $x_2,$ $x_3,$ $\dots$ defined by
\begin{align*}
x_1 &= \sqrt[3]{2}, \\
x_2 &= (\sqrt[3]{2})^{\sqrt[3]{2}},
\end{align*}and in general,
\[x_n = (x_{n - 1})^{\sqrt[3]{2}}\]for $n > 1.$ What is the smallest value of $n$ for which $x_n$ is an integer? | We have that
\[x_3 = (\sqrt[3]{2})^{\sqrt[3]{2}})^{\sqrt[3]{2}} = (\sqrt[3]{2})^{\sqrt[3]{4}},\]and
\[x_4 = (\sqrt[3]{2})^{\sqrt[4]{2}})^{\sqrt[3]{2}} = (\sqrt[3]{2})^{\sqrt[3]{8}} = (\sqrt[3]{2})^2 = 2,\]so the smallest such $n$ is $\boxed{4}.$ The answer is 4. | Consider a sequence $x_1,$ $x_2,$ $x_3,$ $\dots$ defined by
\begin{align*}
x_1 &= \sqrt[3]{2}, \\
x_2 &= (\sqrt[3]{2})^{\sqrt[3]{2}},
\end{align*}and in general,
\[x_n = (x_{n - 1})^{\sqrt[3]{2}}\]for $n > 1.$ What is the smallest value of $n$ for which $x_n$ is an integer?
We have that
\[x_3 = (\sqrt[3]{2})^{\sqrt... |
A right pyramid has a square base with side length 6 inches. Its apex is 9 inches from each of the other vertices. What is the height of the pyramid from its peak to the center of its square base, in inches? | [asy]
import three;
triple A = (0,0,0);
triple B = (1,0,0);
triple C = (1,1,0);
triple D = (0,1,0);
triple P = (0.5,0.5,1);
draw(B--C--D--P--B);
draw(P--C);
draw(B--A--D,dashed);
draw(P--A,dashed);
label("$A$",A,NW);
label("$B$",B,W);
label("$C$",C,S);
label("$D$",D,E);
label("$P$",P,N);
triple F= (0.5,0.5,0);
label("$... | A right pyramid has a square base with side length 6 inches. Its apex is 9 inches from each of the other vertices. What is the height of the pyramid from its peak to the center of its square base, in inches?
[asy]
import three;
triple A = (0,0,0);
triple B = (1,0,0);
triple C = (1,1,0);
triple D = (0,1,0);
triple P =... |
Find the least positive integer $x$ that satisfies $x+4610 \equiv 2104 \pmod{12}$. | Subtract 4610 from both sides of the congruence to obtain $x\equiv -2506\pmod{12}$. By dividing 2506 by 12, we find that the least integer $k$ for which $-2506+12k>0$ is $k=209$. Adding $12\cdot 209$ to $-2506$, we find that $x\equiv 2\pmod{12}$. Thus $\boxed{2}$ is the least integer satisfying the given congruence. Th... | Find the least positive integer $x$ that satisfies $x+4610 \equiv 2104 \pmod{12}$.
Subtract 4610 from both sides of the congruence to obtain $x\equiv -2506\pmod{12}$. By dividing 2506 by 12, we find that the least integer $k$ for which $-2506+12k>0$ is $k=209$. Adding $12\cdot 209$ to $-2506$, we find that $x\equiv 2... |
Find the remainder when $6x^4-14x^3-4x^2+2x-26$ is divided by $x - 3.$ | By the Remainder Theorem, we can find the remainder by setting $x = 3.$ Thus, the remainder is
\[6 \cdot 3^4 - 14 \cdot 3^3 - 4 \cdot 3^2 + 2 \cdot 3 - 26 = \boxed{52}.\] The answer is 52. | Find the remainder when $6x^4-14x^3-4x^2+2x-26$ is divided by $x - 3.$
By the Remainder Theorem, we can find the remainder by setting $x = 3.$ Thus, the remainder is
\[6 \cdot 3^4 - 14 \cdot 3^3 - 4 \cdot 3^2 + 2 \cdot 3 - 26 = \boxed{52}.\] The answer is 52. |
Determine the value of $\cos \frac{3\pi}{2}$. | Rotating the point $(1,0)$ about the origin by $\frac{3\pi}{2}$ radians counterclockwise gives us the point $(0,-1)$, so $\cos \frac{3\pi}{2} = \boxed{0}$. The answer is 0. | Determine the value of $\cos \frac{3\pi}{2}$.
Rotating the point $(1,0)$ about the origin by $\frac{3\pi}{2}$ radians counterclockwise gives us the point $(0,-1)$, so $\cos \frac{3\pi}{2} = \boxed{0}$. The answer is 0. |
In trapezoid $ABCD$, the sides $AB$ and $CD$ are equal and the height is 3. If the length of $AD$ is 8 and the length of $BC$ is 16, what is the perimeter of $ABCD$? | When the figure is divided as shown, the unknown sides are the hypotenuses of right triangles with legs of 3 and 4. Using the Pythagorean Theorem yields $AB=CD=5$. The total perimeter is $16+5+8+5=\boxed{34}$. [asy]
/* AMC8 1999 #14 Solution */
pen p = linetype("4 4");
pen r = red;
draw((0,0)--(4,3)--(12,3)--(16,0)--cy... | In trapezoid $ABCD$, the sides $AB$ and $CD$ are equal and the height is 3. If the length of $AD$ is 8 and the length of $BC$ is 16, what is the perimeter of $ABCD$?
When the figure is divided as shown, the unknown sides are the hypotenuses of right triangles with legs of 3 and 4. Using the Pythagorean Theorem yields... |
Let $a$ and $b$ be nonzero real numbers such that
\[(3 - 5i)(a + bi)\]is pure imaginary. Find $\frac{a}{b}.$ | Expanding $(3 - 5i)(a + bi),$ we get
\[3a - 5ai + 3bi - 5bi^2 = 3a - 5ai + 3bi + 5b.\]Since this number is pure imaginary, the real part $3a + 5b$ is equal to 0. Hence, $\frac{a}{b} = \boxed{-\frac{5}{3}}.$ The answer is -\frac{5}{3}. | Let $a$ and $b$ be nonzero real numbers such that
\[(3 - 5i)(a + bi)\]is pure imaginary. Find $\frac{a}{b}.$
Expanding $(3 - 5i)(a + bi),$ we get
\[3a - 5ai + 3bi - 5bi^2 = 3a - 5ai + 3bi + 5b.\]Since this number is pure imaginary, the real part $3a + 5b$ is equal to 0. Hence, $\frac{a}{b} = \boxed{-\frac{5}{3}}.$ ... |
Let $P$ be the plane passing through the point $(1,2,3)$ with normal vector $\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}.$ Find the matrix $\mathbf{P}$ such that for any vector $\mathbf{v},$ $\mathbf{P} \mathbf{v}$ is the projection of $\mathbf{v}$ onto plane $P.$ | Let $\mathbf{v} = \begin{pmatrix} x \\ y \\ z \end{pmatrix},$ and let $\mathbf{p}$ be the projection of $\mathbf{p}$ onto plane $P.$ Then $\mathbf{v} - \mathbf{p}$ is the projection of $\mathbf{v}$ onto the normal vector $\mathbf{n} = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}.$
[asy]
import three;
size(160);
curren... | Let $P$ be the plane passing through the point $(1,2,3)$ with normal vector $\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}.$ Find the matrix $\mathbf{P}$ such that for any vector $\mathbf{v},$ $\mathbf{P} \mathbf{v}$ is the projection of $\mathbf{v}$ onto plane $P.$
Let $\mathbf{v} = \begin{pmatrix} x \\ y \\ z \end{pm... |
In the diagram, $\angle EBD=48^\circ$ and $\angle FBC=60^\circ$. What is the value of $x?$ | In $\triangle BFC$ shown below, \begin{align*}
\angle BCF &= 180^{\circ}-\angle FBC-\angle BFC \\
&= 180^{\circ}-60^{\circ}-90^{\circ} \\
&= 30^{\circ}.
\end{align*} Since $\angle ABD$ is a straight angle, \begin{align*}
\angle ABC &= 180^{\circ}-\angle EBD \\
&= 180^{\circ}-48^{\circ} \\
&= 132^{\circ}.
\end{align*} I... | In the diagram, $\angle EBD=48^\circ$ and $\angle FBC=60^\circ$. What is the value of $x?$
In $\triangle BFC$ shown below, \begin{align*}
\angle BCF &= 180^{\circ}-\angle FBC-\angle BFC \\
&= 180^{\circ}-60^{\circ}-90^{\circ} \\
&= 30^{\circ}.
\end{align*} Since $\angle ABD$ is a straight angle, \begin{align*}
\angle... |
Convert $110101_2$ to a base 10 integer. | $110101_2 = 1 \cdot 2^5 + 1 \cdot 2^4 + 0 \cdot 2^3 + 1 \cdot 2^2 + 0 \cdot 2^1 + 1 \cdot 2^0 = 32 + 16 + 4 + 1 = \boxed{53}$. The answer is 53. | Convert $110101_2$ to a base 10 integer.
$110101_2 = 1 \cdot 2^5 + 1 \cdot 2^4 + 0 \cdot 2^3 + 1 \cdot 2^2 + 0 \cdot 2^1 + 1 \cdot 2^0 = 32 + 16 + 4 + 1 = \boxed{53}$. The answer is 53. |
Let $x$, $y$, and $z$ be positive real numbers such that $(x \cdot y) + 2z = (x + 2z) \cdot (y + 2z)$. What is the maximum possible value of $xyz$? | The condition is equivalent to $4z^2+(x+y-2)z=0$. Since $z$ is positive, $z=2-x-y$, so $x+y+z=2$. By the AM-GM inequality, $$xyz \leq \left(\frac{x+y+z}{3}\right)^3 = \boxed{\frac{8}{27}},$$with equality when $x=y=z=\frac{2}{3}$. The answer is \frac{8}{27}. | Let $x$, $y$, and $z$ be positive real numbers such that $(x \cdot y) + 2z = (x + 2z) \cdot (y + 2z)$. What is the maximum possible value of $xyz$?
The condition is equivalent to $4z^2+(x+y-2)z=0$. Since $z$ is positive, $z=2-x-y$, so $x+y+z=2$. By the AM-GM inequality, $$xyz \leq \left(\frac{x+y+z}{3}\right)^3 = \bo... |
If $0.\overline{75}$ is written as a fraction $\frac{a}{b}$ where $a$ and $b$ are positive integers with a greatest common divisor of 1, what is $a+b$? | Let $x = 0.\overline{75}$. Multiplying both sides by 100, we get $100x = 75.\overline{75}$. Subtracting these two equations gives $99x = 75$, so $x = \frac{75}{99} = \frac{25}{33}$. Thus $a+b = 25+33 = \boxed{58}$. The answer is 58. | If $0.\overline{75}$ is written as a fraction $\frac{a}{b}$ where $a$ and $b$ are positive integers with a greatest common divisor of 1, what is $a+b$?
Let $x = 0.\overline{75}$. Multiplying both sides by 100, we get $100x = 75.\overline{75}$. Subtracting these two equations gives $99x = 75$, so $x = \frac{75}{99} = ... |
Suppose $f(z)$ and $g(z)$ are polynomials in $z$, and the degree of $g(z)$ is less than the degree of $f(z)$. If the degree of $f(z)-g(z)$ is two, what is the degree of $f(z)$? | We have $f(z)-g(z)=a_2 \cdot z^2+a_1 \cdot z+a_0$, where $a_2$ is nonzero. If the degree of $g(z)$ is less than two, then $f(z)$ must be of degree two. Therefore, the degree of $f(z)$ is $\boxed{2}$. The answer is 2. | Suppose $f(z)$ and $g(z)$ are polynomials in $z$, and the degree of $g(z)$ is less than the degree of $f(z)$. If the degree of $f(z)-g(z)$ is two, what is the degree of $f(z)$?
We have $f(z)-g(z)=a_2 \cdot z^2+a_1 \cdot z+a_0$, where $a_2$ is nonzero. If the degree of $g(z)$ is less than two, then $f(z)$ must be of... |
What is the number of odd perfect cube factors of $2^4 \cdot 7^9$? | A positive integer is a factor of $2^4\cdot7^9$ if and only if its prime factorization is of the form $2^a\cdot 7^b$ for exponents $a$ and $b$ satisfying $0\leq a \leq 4$ and $0\leq b\leq 9$. A positive integer is a perfect cube if and only if the exponents in its prime factorization are multiples of 3. A positive in... | What is the number of odd perfect cube factors of $2^4 \cdot 7^9$?
A positive integer is a factor of $2^4\cdot7^9$ if and only if its prime factorization is of the form $2^a\cdot 7^b$ for exponents $a$ and $b$ satisfying $0\leq a \leq 4$ and $0\leq b\leq 9$. A positive integer is a perfect cube if and only if the ex... |
The graph shows the total distance Sam drove from 6 a.m to 11 a.m. Find the average rate of change from 6 a.m. to 8 a.m. | Since he traveled 40 miles in 2 hours, the average rate of change from 6 a.m. to 8 a.m. is $\frac{40}{2} = \boxed{20}$ miles per hour. The answer is 20. | The graph shows the total distance Sam drove from 6 a.m to 11 a.m. Find the average rate of change from 6 a.m. to 8 a.m.
Since he traveled 40 miles in 2 hours, the average rate of change from 6 a.m. to 8 a.m. is $\frac{40}{2} = \boxed{20}$ miles per hour. The answer is 20. |
Let $a$, $b$, and $c$ be distinct real numbers such that $a+b+c=3$. Compute \[ \dfrac {ab+bc+ca}{a^2+b^2+c^2}. \] | We have $a+b+c=3,$ and squaring this equation gives \[(a^2+b^2+c^2) + 2(ab+bc+ca) = 9.\]Thus, $a^2+b^2+c^2=9-2(ab+bc+ca).$ Since $a, b, c$ are distinct, it is not possible that $a^2+b^2+c^2=0,$ so we have \[\frac{ab+bc+ca}{a^2+b^2+c^2} = \boxed{\frac{9}{2}}.\] The answer is \frac{9}{2}. | Let $a$, $b$, and $c$ be distinct real numbers such that $a+b+c=3$. Compute \[ \dfrac {ab+bc+ca}{a^2+b^2+c^2}. \]
We have $a+b+c=3,$ and squaring this equation gives \[(a^2+b^2+c^2) + 2(ab+bc+ca) = 9.\]Thus, $a^2+b^2+c^2=9-2(ab+bc+ca).$ Since $a, b, c$ are distinct, it is not possible that $a^2+b^2+c^2=0,$ so we have... |
Find the distance from the point $(-1, 5)$ to the center of the circle with equation $x^2+y^2=2x-4y-4$. | Moving terms to the LHS, we have $x^2-2x+y^2+4y=-4$. Completing the square on the quadratic in $x$, we add $(-2/2)^2=1$ to both sides. Completing the square on the quadratic in $y$, we add $(4/2)^2=4$ to both sides. We are left with the equation $x^2-2x+1+y^2+4y+4=1 \Rightarrow (x-1)^2+(y+2)^2=1$. Thus, our circle has ... | Find the distance from the point $(-1, 5)$ to the center of the circle with equation $x^2+y^2=2x-4y-4$.
Moving terms to the LHS, we have $x^2-2x+y^2+4y=-4$. Completing the square on the quadratic in $x$, we add $(-2/2)^2=1$ to both sides. Completing the square on the quadratic in $y$, we add $(4/2)^2=4$ to both sides... |
If $\sqrt3 \cos 20^\circ$ can be written as $\cos \theta + \sin\theta$ for some acute angle $\theta,$ what is $\theta?$ (Give your answer in degrees, not radians.) | We have $\cos\theta = \sin(90^\circ - \theta),$ so
$$\cos \theta + \sin\theta = \sin(90^\circ-\theta) + \sin\theta.$$Applying the sum of sines formula gives
\begin{align*}
\sin(90^\circ-\theta) + \sin\theta &= 2\sin\frac{(90^\circ-\theta) + \theta}{2}\cos\frac{(90^\circ-\theta) - \theta}{2} \\
&= 2\sin45^\circ\cos\frac... | If $\sqrt3 \cos 20^\circ$ can be written as $\cos \theta + \sin\theta$ for some acute angle $\theta,$ what is $\theta?$ (Give your answer in degrees, not radians.)
We have $\cos\theta = \sin(90^\circ - \theta),$ so
$$\cos \theta + \sin\theta = \sin(90^\circ-\theta) + \sin\theta.$$Applying the sum of sines formula giv... |
Pirate Pete shares his treasure with Pirate Paul in an interesting way. Pete first says, ``Two for me, one for you,'' giving himself two coins and starting Paul's pile with one coin. Then Pete says, ``Three for me, and two for you,'' giving himself three more coins but making Paul's pile two coins in total. Next Pete s... | At the end of distributing the coins, Paul has $x$ coins, and Pete has three times as many, or $3x$ coins. We can also write the number of coins Pete has as $2+3+4+ \dots +x = (x-1)(x+2)/2$. Therefore, \[\frac{(x-1)(x+2)}{2} = 3x.\] Solving for $x$, we find $x = 5$, so the total number of coins they have is $x+3x=4x=... | Pirate Pete shares his treasure with Pirate Paul in an interesting way. Pete first says, ``Two for me, one for you,'' giving himself two coins and starting Paul's pile with one coin. Then Pete says, ``Three for me, and two for you,'' giving himself three more coins but making Paul's pile two coins in total. Next Pete s... |
The function $f$ has the property that for each real number $x$ in its domain, $2/x$ is also in its domain and \[
f(x) + f\left(\frac{2}{x}\right) = \frac{x}{2}.
\]What is the largest set of real numbers that can be in the domain of $f$?
(a) ${\{x\mid x\ne0\}}$
(b) ${\{x\mid x<0\}}$
(c) ${\{x\mid x>0\}}$
(d) ${\{x\... | The conditions on $f$ imply that both \[
\frac{x}{2} = f(x) + f\displaystyle\left(\frac{2}{x}\displaystyle\right)\]and \[\frac{2}{x} = f\left(\frac{2}{x}\right) +
f\displaystyle\left(\frac{1}{2/x}\displaystyle\right) = f\displaystyle\left(\frac{2}{x}\displaystyle\right) + f(x).
\]Thus if $x$ is in the domain of $f$, th... | The function $f$ has the property that for each real number $x$ in its domain, $2/x$ is also in its domain and \[
f(x) + f\left(\frac{2}{x}\right) = \frac{x}{2}.
\]What is the largest set of real numbers that can be in the domain of $f$?
(a) ${\{x\mid x\ne0\}}$
(b) ${\{x\mid x<0\}}$
(c) ${\{x\mid x>0\}}$
(d) ${\{x\... |
For what values of $x$ is $x^2-50x+600\le 11$? Express your answer in interval notation. | We can simplify the inequality to $x^2-50x+589\le 0$. We could solve for the roots using the quadratic formula, but there is an easier solution by factoring: $x^2-50x+589=(x-23)(x-27)$. Thus the parabola $x^2-50x+589$ changes sign at $x=23$ and $x=27$. The solution is either the interval $(-\infty,23]\cup[27,\infty)$ o... | For what values of $x$ is $x^2-50x+600\le 11$? Express your answer in interval notation.
We can simplify the inequality to $x^2-50x+589\le 0$. We could solve for the roots using the quadratic formula, but there is an easier solution by factoring: $x^2-50x+589=(x-23)(x-27)$. Thus the parabola $x^2-50x+589$ changes sig... |
Evaluate $\log_{4}{64}-\log_{4}{\frac{1}{16}}$. | Let $\log_{4}{64}=a$. Then $4^a=64=4^3$, so $a=3$. Let $\log_{4}{\frac{1}{16}}=b$. Then $\frac{1}{16}=4^b$. Express $\frac{1}{16}$ as a power of $4$: $\frac{1}{16}=\frac{1}{4^2}=4^{-2}$. Thus $4^b=4^{-2}$ and $b=-2$. We want to find $\log_{4}{64}-\log_{4}{\frac{1}{16}}=a-b=(3)-(-2)=\boxed{5}$. The answer is 5. | Evaluate $\log_{4}{64}-\log_{4}{\frac{1}{16}}$.
Let $\log_{4}{64}=a$. Then $4^a=64=4^3$, so $a=3$. Let $\log_{4}{\frac{1}{16}}=b$. Then $\frac{1}{16}=4^b$. Express $\frac{1}{16}$ as a power of $4$: $\frac{1}{16}=\frac{1}{4^2}=4^{-2}$. Thus $4^b=4^{-2}$ and $b=-2$. We want to find $\log_{4}{64}-\log_{4}{\frac{1}{16}}=... |
There is a committee composed of eight women and two men. When they meet, they sit in a row---the women in indistinguishable rocking chairs and the men on distinguishable stools. How many distinct ways are there for me to arrange the eight chairs and two stools for a meeting? | Because the rocking chairs are indistinguishable from each other but the stools are distinguishable, we can think of first placing the two stools somewhere in the ten slots and then filling the rest with rocking chairs. The first stool has $10$ slots in which it can go, and the second has $9$. Thus, there are $10 \cdot... | There is a committee composed of eight women and two men. When they meet, they sit in a row---the women in indistinguishable rocking chairs and the men on distinguishable stools. How many distinct ways are there for me to arrange the eight chairs and two stools for a meeting?
Because the rocking chairs are indisting... |
Evaluate
\[i^{14762} + i^{14763} + i^{14764} + i^{14766}.\] | We can take out a factor of $i^{14762}$ to get
\[i^{14762} + i^{14763} + i^{14764} + i^{14766} = i^{14762} (1 + i + i^2 + i^4).\]Since $i^2 = -1$ and $i^4 = 1,$
\[1 + i + i^2 + i^4 = 1 + i - 1 + 1 = 1 + i.\]Therefore, the expression is equal to $\boxed{i^{14762}(1 + i)}.$ The answer is i^{14762}(1 + i). | Evaluate
\[i^{14762} + i^{14763} + i^{14764} + i^{14766}.\]
We can take out a factor of $i^{14762}$ to get
\[i^{14762} + i^{14763} + i^{14764} + i^{14766} = i^{14762} (1 + i + i^2 + i^4).\]Since $i^2 = -1$ and $i^4 = 1,$
\[1 + i + i^2 + i^4 = 1 + i - 1 + 1 = 1 + i.\]Therefore, the expression is equal to $\boxed{i^{14... |
How many different triangles can be formed having a perimeter of 8 units if each side must have integral length? | Let $a,b,$ and $c$ represent the three side lengths of the triangle. The perimeter is $a+b+c=8,$ so $b+c=8-a$. We know by the Triangle Inequality that the sum of two side lengths of a triangle must be greater than the third side length. If we focus on the variable $a$, we have
\[b+c>a\quad\Rightarrow \quad 8-a>a\quad\R... | How many different triangles can be formed having a perimeter of 8 units if each side must have integral length?
Let $a,b,$ and $c$ represent the three side lengths of the triangle. The perimeter is $a+b+c=8,$ so $b+c=8-a$. We know by the Triangle Inequality that the sum of two side lengths of a triangle must be grea... |
Each of the numbers $a_1,$ $a_2,$ $\dots,$ $a_{100}$ is $\pm 1.$ Find the smallest possible positive value of
\[\sum_{1 \le i < j \le 100} a_i a_j.\] | Let $m$ and $n$ denote the number of 1's and $-1$'s among the $a_i,$ respectively. Then $m + n = 100$ and
\[a_1^2 + a_2^2 + \dots + a_{100}^2 = 100.\]Let
\[S = \sum_{1 \le i < j \le 100} a_i a_j.\]Then
\[2S + 100 = (a_1 + a_2 + \dots + a_{100})^2 = (m - n)^2.\]Note that $m - n = m + n - 2n = 100 - 2n$ is even, so $(m ... | Each of the numbers $a_1,$ $a_2,$ $\dots,$ $a_{100}$ is $\pm 1.$ Find the smallest possible positive value of
\[\sum_{1 \le i < j \le 100} a_i a_j.\]
Let $m$ and $n$ denote the number of 1's and $-1$'s among the $a_i,$ respectively. Then $m + n = 100$ and
\[a_1^2 + a_2^2 + \dots + a_{100}^2 = 100.\]Let
\[S = \sum_{... |
Solve for $x$ in the following equation: $\frac{1}{6}+\frac{6}{x}=\frac{13}{x}+\frac{1}{13}$ | Subtracting $\frac{6}{x}$ and $\frac{1}{13}$ from both sides of the equation, we get \[
\frac{7}{78}=\frac{7}{x}.
\] Solving for $x$, we find $x=\boxed{78}$. The answer is 78. | Solve for $x$ in the following equation: $\frac{1}{6}+\frac{6}{x}=\frac{13}{x}+\frac{1}{13}$
Subtracting $\frac{6}{x}$ and $\frac{1}{13}$ from both sides of the equation, we get \[
\frac{7}{78}=\frac{7}{x}.
\] Solving for $x$, we find $x=\boxed{78}$. The answer is 78. |
For how many real values of $c$ do we have $\left|\frac13-ci\right| = \frac45$? | We have $\left|\frac13-ci\right| = \sqrt{{\frac13}^2 + (-c)^2} = \sqrt{c^2 + \frac19}$, so $\left|\frac13-ci\right| = \frac45$ gives us $\sqrt{c^2 + \frac19} = \frac45$. Squaring both sides gives $c^2 + \frac19 = \frac{16}{25}$, so $c^2=\frac{7}{225}$. Taking the square root of both sides gives $c = \frac{\sqrt7}{15}... | For how many real values of $c$ do we have $\left|\frac13-ci\right| = \frac45$?
We have $\left|\frac13-ci\right| = \sqrt{{\frac13}^2 + (-c)^2} = \sqrt{c^2 + \frac19}$, so $\left|\frac13-ci\right| = \frac45$ gives us $\sqrt{c^2 + \frac19} = \frac45$. Squaring both sides gives $c^2 + \frac19 = \frac{16}{25}$, so $c^2=... |
At the grocery store, I bought $5$ different items. I brought $3$ identical bags, and handed them to the cashier. How many ways are there for the cashier to put the items I bought in the $3$ identical bags, assuming he might leave some of the bags empty? | We can divide this into cases.
$\bullet$ Case 1: All $5$ items go in the same bag. There is one possible way to do this.
$\bullet$ Case 2: Four items go in one bag, and the last item goes in another bag. There are $\binom{5}{1}=5$ ways to choose which item goes in a bag by itself.
$\bullet$ Case 3: Three items go... | At the grocery store, I bought $5$ different items. I brought $3$ identical bags, and handed them to the cashier. How many ways are there for the cashier to put the items I bought in the $3$ identical bags, assuming he might leave some of the bags empty?
We can divide this into cases.
$\bullet$ Case 1: All $5$ item... |
The line $y = \frac{1}{2} x + 4$ is paramterized by the form
\[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2 \\ r \end{pmatrix} + t \begin{pmatrix} k \\ -5 \end{pmatrix}.\]Enter the ordered pair $(r,k).$ | Taking $t = 0,$ we find $\begin{pmatrix} 2 \\ r \end{pmatrix}$ lies on the line. Then
\[r = \frac{1}{2} (2) + 4 = 5.\]Taking $t = 1,$ we get
\[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2 \\ 5 \end{pmatrix} + \begin{pmatrix} k \\ -5 \end{pmatrix} = \begin{pmatrix} 2 + k \\ 0 \end{pmatrix}.\]Then
\[0 = \fra... | The line $y = \frac{1}{2} x + 4$ is paramterized by the form
\[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2 \\ r \end{pmatrix} + t \begin{pmatrix} k \\ -5 \end{pmatrix}.\]Enter the ordered pair $(r,k).$
Taking $t = 0,$ we find $\begin{pmatrix} 2 \\ r \end{pmatrix}$ lies on the line. Then
\[r = \frac{1}{2... |
Let $P(x)$ be a polynomial such that when $P(x)$ is divided by $x-15$, the remainder is $10$, and when $P(x)$ is divided by $x-11$, the remainder is $2$. What is the remainder when $P(x)$ is divided by $(x-11)(x-15)$? | Since we are dividing by a quadratic, our remainder will have degree at most $1$. Hence the remainder is of the form $ax+b$ for some constants $a$ and $b$. We have
$$P(x) = (x-11)(x-15)Q(x) + ax+b$$where $Q(x)$ is the quotient when $P(x)$ is divided by $(x-11)(x-15)$. We can eliminate the $Q(x)$ term by plugging in $x=... | Let $P(x)$ be a polynomial such that when $P(x)$ is divided by $x-15$, the remainder is $10$, and when $P(x)$ is divided by $x-11$, the remainder is $2$. What is the remainder when $P(x)$ is divided by $(x-11)(x-15)$?
Since we are dividing by a quadratic, our remainder will have degree at most $1$. Hence the remainde... |
A board game spinner is divided into three regions labeled $A$, $B$ and $C$. The probability of the arrow stopping on region $A$ is $\frac{1}{3}$ and on region $B$ is $\frac{1}{4}$. What is the probability of the arrow stopping on region $C$? Express your answer as a common fraction. | Since the sum of the three probabilities is 1, the probability of stopping on region $C$ is $1 - \frac{1}{3} -
\frac{1}{4} = \frac{12}{12} - \frac{4}{12} - \frac{3}{12} = \boxed{\frac{5}{12}}$. The answer is \frac{5}{12}. | A board game spinner is divided into three regions labeled $A$, $B$ and $C$. The probability of the arrow stopping on region $A$ is $\frac{1}{3}$ and on region $B$ is $\frac{1}{4}$. What is the probability of the arrow stopping on region $C$? Express your answer as a common fraction.
Since the sum of the three pro... |
Calculate: $\left(\frac{1}{2} \right)^{4} \cdot \left(\frac{1}{7} \right)^{2}$. | We have $\left(\frac{1}{2} \right)^{4}=\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}=\frac{1}{16}$ and $\left(\frac{1}{7} \right)^{2}=\frac{1}{7} \cdot \frac{1}{7}=\frac{1}{49}$, so \[\left(\frac{1}{2} \right)^{4} \cdot \left(\frac{1}{7} \right)^{2} = \frac{1}{16} \cdot \frac{1}{49} = \boxed{\frac{1... | Calculate: $\left(\frac{1}{2} \right)^{4} \cdot \left(\frac{1}{7} \right)^{2}$.
We have $\left(\frac{1}{2} \right)^{4}=\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}=\frac{1}{16}$ and $\left(\frac{1}{7} \right)^{2}=\frac{1}{7} \cdot \frac{1}{7}=\frac{1}{49}$, so \[\left(\frac{1}{2} \right)^{4} \cdo... |
Express $\frac{0.\overline{727}}{1.\overline{454}}$ as a common fraction. | We might recognize the top as $\frac{8}{11}$, and the bottom as $\frac{5}{11}$, thereby giving you a value of $\frac{8}{5}$. If not, call the numerator $x$. Multiplying by 100, and subtracting $x$, you get 99x = 72, and thus, $x = \frac{8}{11}$. We then notice that the denominator is $1 + \frac{x}{2}$, thereby givin... | Express $\frac{0.\overline{727}}{1.\overline{454}}$ as a common fraction.
We might recognize the top as $\frac{8}{11}$, and the bottom as $\frac{5}{11}$, thereby giving you a value of $\frac{8}{5}$. If not, call the numerator $x$. Multiplying by 100, and subtracting $x$, you get 99x = 72, and thus, $x = \frac{8}{11... |
How many four-digit numbers $N = \underline{a}\,\underline{b}\,\underline{c}\,\underline{d}$ satisfy all of the following conditions?
$5000 \le N < 7000.$
$N$ is a multiple of $10.$
$2 \le b < c \le 5.$ | The first condition is equivalent to the statement that $a = 5$ or $a = 6$. The second condition is equivalent to the statement that $d = 0$. Finally, the third condition is equivalent to the statement that the ordered pair $(b, c)$ is one of the pairs \[(2,3), (2,4), (2,5), (3,4), (3,5),(4,5).\]In total, $2 \cdot 1 \c... | How many four-digit numbers $N = \underline{a}\,\underline{b}\,\underline{c}\,\underline{d}$ satisfy all of the following conditions?
$5000 \le N < 7000.$
$N$ is a multiple of $10.$
$2 \le b < c \le 5.$
The first condition is equivalent to the statement that $a = 5$ or $a = 6$. The second condition is equivalent to ... |
A car travels 32 mph for 12 miles, 40 mph for 15 miles, 48 mph for 30 minutes and 36 mph for 10 minutes. What is the average speed of the car, in mph? | To find the average speed for the entire trip, we need to divide the total distance by the total time. Remembering that $d=r\cdot t$, and looking at each of the four parts of the trip, these pieces can be determined.
First, a car traveling at 32 mph for 12 miles will be traveling for $12/32=.375$ hours. Next, a car ... | A car travels 32 mph for 12 miles, 40 mph for 15 miles, 48 mph for 30 minutes and 36 mph for 10 minutes. What is the average speed of the car, in mph?
To find the average speed for the entire trip, we need to divide the total distance by the total time. Remembering that $d=r\cdot t$, and looking at each of the four ... |
If $\tan x-\tan y=25$ and $\cot x + \cot y=30$, what is $\tan(x-y)$? | As before, we have $\frac{\tan x + \tan y}{\tan x \tan y} = 30.$ Since $\tan x - \tan y = 25,$ we have $\frac{25}{\tan x \tan y} = 30,$ so $\tan x \tan y = \frac{25}{30} = \frac{5}{6}.$ Then from the angle difference formula,
\[\tan(x-y) = \frac{\tan x - \tan y}{1 + \tan x \tan y} = \frac{25}{1 + \frac{5}{6}} = \boxed{... | If $\tan x-\tan y=25$ and $\cot x + \cot y=30$, what is $\tan(x-y)$?
As before, we have $\frac{\tan x + \tan y}{\tan x \tan y} = 30.$ Since $\tan x - \tan y = 25,$ we have $\frac{25}{\tan x \tan y} = 30,$ so $\tan x \tan y = \frac{25}{30} = \frac{5}{6}.$ Then from the angle difference formula,
\[\tan(x-y) = \frac{\ta... |
Three positive integers have an arithmetic mean of 28 and median of 29. If the median is 3 more than the smallest number, what is the largest number? | The median of three integers is the middle integer. So the middle integer is $29$ and the smallest integer is $29-3=26$. We also know that if the mean is $28$, then the sum of the three numbers is $28\times3=84$. We subtract the other two numbers to find that the largest number is $84-26-29=\boxed{29}$. The answer is 2... | Three positive integers have an arithmetic mean of 28 and median of 29. If the median is 3 more than the smallest number, what is the largest number?
The median of three integers is the middle integer. So the middle integer is $29$ and the smallest integer is $29-3=26$. We also know that if the mean is $28$, then the... |
Evaluate $$\lfloor\sqrt{5}\rfloor + \lfloor\sqrt{6}\rfloor + \lfloor\sqrt{7}\rfloor + \cdots + \lfloor\sqrt{29}\rfloor$$Note: For a real number $x,$ $\lfloor x \rfloor$ denotes the greatest integer that is less than or equal to $x.$ | We note that if $a^2 \leq n < (a+1)^2$ for some integer $a$, then $a \leq \sqrt{x} < a+1$, so $a$ is the greatest integer less than or equal to $x$. Consequently, we break up our sum into the blocks of integers between consecutive perfect squares:
For $5\leq n \leq 9$, $\lfloor\sqrt{n}\rfloor=2$. There are $5$ values ... | Evaluate $$\lfloor\sqrt{5}\rfloor + \lfloor\sqrt{6}\rfloor + \lfloor\sqrt{7}\rfloor + \cdots + \lfloor\sqrt{29}\rfloor$$Note: For a real number $x,$ $\lfloor x \rfloor$ denotes the greatest integer that is less than or equal to $x.$
We note that if $a^2 \leq n < (a+1)^2$ for some integer $a$, then $a \leq \sqrt{x} < ... |
Let $f(x) = \frac{3}{9^x + 3}.$ Find
\[f \left( \frac{1}{1000} \right) + f \left( \frac{2}{1000} \right) + f \left( \frac{3}{1000} \right) + \dots + f \left( \frac{999}{1000} \right).\] | Note that
\begin{align*}
f(x) + f(1 - x) &= \frac{3}{9^x + 3} + \frac{3}{9^{1 - x} + 3} \\
&= \frac{3}{9^x + 3} + \frac{3 \cdot 9^x}{9 + 3 \cdot 9^x} \\
&= \frac{3}{9^x + 3} + \frac{9^x}{3 + 9^x} \\
&= \frac{3 + 9^x}{9^x + 3} \\
&= 1.
\end{align*}Thus, we can pair the 999 terms in the sum into 499 pairs, such that the ... | Let $f(x) = \frac{3}{9^x + 3}.$ Find
\[f \left( \frac{1}{1000} \right) + f \left( \frac{2}{1000} \right) + f \left( \frac{3}{1000} \right) + \dots + f \left( \frac{999}{1000} \right).\]
Note that
\begin{align*}
f(x) + f(1 - x) &= \frac{3}{9^x + 3} + \frac{3}{9^{1 - x} + 3} \\
&= \frac{3}{9^x + 3} + \frac{3 \cdot 9^x... |
What is the minimum value of $7x^2-14x+2022$? | Let $y=7x^2 -14x + 2022$. First, complete the square as follows: $y=7x^2-14x+2022=7(x^2-2x)+2022$. To complete the square, we need to add $\left(\dfrac{2}{2}\right)^2=1$ after the $-2x$. So we have $y+7=7\left(x^2-2x+1\right)+2022$. This gives $y=7\left(x-1\right)^2+2015$.
Now, since $\left(x-1\right)^2\ge0$, the min... | What is the minimum value of $7x^2-14x+2022$?
Let $y=7x^2 -14x + 2022$. First, complete the square as follows: $y=7x^2-14x+2022=7(x^2-2x)+2022$. To complete the square, we need to add $\left(\dfrac{2}{2}\right)^2=1$ after the $-2x$. So we have $y+7=7\left(x^2-2x+1\right)+2022$. This gives $y=7\left(x-1\right)^2+2015... |
A right rectangular prism $P$ (i.e., a rectangular parallelpiped) has sides of integral length $a, b, c,$ with $a\le b\le c.$ A plane parallel to one of the faces of $P$ cuts $P$ into two prisms, one of which is similar to $P,$ and both of which have nonzero volume. Given that $c=1995,$ for how many ordered triples $(a... | Let $P'$ be the prism similar to $P$, and let the sides of $P'$ be of length $x,y,z$, such that $x \le y \le z$. Then
\[\frac{x}{a} = \frac{y}{b} = \frac zc = \frac{z}{1995} < 1.\]
Note that if the ratio of similarity was equal to $1$, we would have a prism with zero volume. As one face of $P'$ is a face of $P$, it fol... | A right rectangular prism $P$ (i.e., a rectangular parallelpiped) has sides of integral length $a, b, c,$ with $a\le b\le c.$ A plane parallel to one of the faces of $P$ cuts $P$ into two prisms, one of which is similar to $P,$ and both of which have nonzero volume. Given that $c=1995,$ for how many ordered triples $(a... |
Find the last three digits of $7^{203}.$ | We can write $7^{203} = (10 - 3)^{203}.$ Then by the Binomial Theorem,
\[(10 - 3)^{203} = 10^{203} - \binom{203}{1} 10^{202} + \binom{203}{2} 10^{201} - \dots + \binom{203}{200} 10^3 - \binom{203}{201} 10^2 + \binom{203}{202} 10 - 1.\]All the terms up to $\binom{203}{200} 10^3$ are divisible by $10^3,$ so for the purp... | Find the last three digits of $7^{203}.$
We can write $7^{203} = (10 - 3)^{203}.$ Then by the Binomial Theorem,
\[(10 - 3)^{203} = 10^{203} - \binom{203}{1} 10^{202} + \binom{203}{2} 10^{201} - \dots + \binom{203}{200} 10^3 - \binom{203}{201} 10^2 + \binom{203}{202} 10 - 1.\]All the terms up to $\binom{203}{200} 10^... |
Find the value of $\tan \left (\operatorname{arccot} \frac{2}{5} \right).$ | Consider a right triangle where the adjacent side is 2 and the opposite side is 5.
[asy]
unitsize (0.5 cm);
draw((0,0)--(2,0)--(2,5)--cycle);
label("$2$", (1,0), S);
label("$5$", (2,5/2), E);
label("$\theta$", (0.4,0.5));
[/asy]
Then $\cot \theta = \frac{2}{5},$ so $\theta = \operatorname{arccot} \frac{2}{5}.$ Hen... | Find the value of $\tan \left (\operatorname{arccot} \frac{2}{5} \right).$
Consider a right triangle where the adjacent side is 2 and the opposite side is 5.
[asy]
unitsize (0.5 cm);
draw((0,0)--(2,0)--(2,5)--cycle);
label("$2$", (1,0), S);
label("$5$", (2,5/2), E);
label("$\theta$", (0.4,0.5));
[/asy]
Then $\cot... |
Find the units digit of $n$ given that $mn = 6^{10}$ and $m$ has a units digit of 3. | The units digit of $mn$ is $6^{10} = 6$. Searching for a units digit for $n$ (which is clearly even), we find that $3 \cdot 2 = 6$, so $\boxed{2}$ is the units digit of $n$. The answer is 2. | Find the units digit of $n$ given that $mn = 6^{10}$ and $m$ has a units digit of 3.
The units digit of $mn$ is $6^{10} = 6$. Searching for a units digit for $n$ (which is clearly even), we find that $3 \cdot 2 = 6$, so $\boxed{2}$ is the units digit of $n$. The answer is 2. |
Let $f : \mathbb{R} \to \mathbb{R}$ be a function such that
\[f(f(x) - y) = f(x) + f(f(y) + f(-x)) + x\]for all real numbers $x$ and $y.$
Let $n$ be the number of possible values of $f(2),$ and let $s$ be the sum of all possible values of $f(2).$ Find $n \times s.$ | Setting $x = y = 0,$ we get
\[f(f(0)) = 2f(0).\]Let $c = f(0),$ so $f(c) = 2c.$
Setting $x = 0$ and $y = c,$ we get
\[f(0) = f(0) + f(f(c) + c).\]Then $f(3c) = 0.$
Setting $x = 0,$ we get
\[f(-y) = f(f(y))\]for all $y.$
Setting $y = f(x),$ we get
\[0 = f(x) + f(f(f(x)) + f(-x)) + x.\]Since $f(f(x)) = f(-x),$ this be... | Let $f : \mathbb{R} \to \mathbb{R}$ be a function such that
\[f(f(x) - y) = f(x) + f(f(y) + f(-x)) + x\]for all real numbers $x$ and $y.$
Let $n$ be the number of possible values of $f(2),$ and let $s$ be the sum of all possible values of $f(2).$ Find $n \times s.$
Setting $x = y = 0,$ we get
\[f(f(0)) = 2f(0).\]Le... |
Find the least common multiple of 28 and 105. | $28 = 2^2 \cdot 7^1$, $105 = 3^1 \cdot 5^1 \cdot 7^1$, so lcm$[28, 105] = 2^2 \cdot 3^1 \cdot 5^1 \cdot 7^1 = \boxed{420}$. The answer is 420. | Find the least common multiple of 28 and 105.
$28 = 2^2 \cdot 7^1$, $105 = 3^1 \cdot 5^1 \cdot 7^1$, so lcm$[28, 105] = 2^2 \cdot 3^1 \cdot 5^1 \cdot 7^1 = \boxed{420}$. The answer is 420. |
What is the area, in square units, of a regular hexagon inscribed in a circle whose circumference is $36\pi$ units? Express your answer in simplest radical form. | Note that since the circumference is $2\pi r = 36 \pi$, where $r$ is the radius, we must have $r=\frac{36 \pi}{2 \pi}=18$. Thus the distance from the center of the hexagon to a vertex is $18$, and we can break up the hexagon into $6$ equilateral triangles, each of which has side length $18$. The area of an equilateral ... | What is the area, in square units, of a regular hexagon inscribed in a circle whose circumference is $36\pi$ units? Express your answer in simplest radical form.
Note that since the circumference is $2\pi r = 36 \pi$, where $r$ is the radius, we must have $r=\frac{36 \pi}{2 \pi}=18$. Thus the distance from the center... |
If $x^2+bx+9/4$ has two non-real roots, find all real possible values of $b$. Express your answer in interval notation. | Using the quadratic formula, we have that the discriminant must be negative for non-real roots: \begin{align*} b^2-4ac&<0
\\\Rightarrow\qquad b^2-4(1)(9/4)&<0
\\\Rightarrow\qquad b^2-9&<0
\\\Rightarrow\qquad (b+3)(b-3)&<0.
\end{align*} Therefore, we find that $ b\in\boxed{(-3, 3)} $. The answer is (-3, 3). | If $x^2+bx+9/4$ has two non-real roots, find all real possible values of $b$. Express your answer in interval notation.
Using the quadratic formula, we have that the discriminant must be negative for non-real roots: \begin{align*} b^2-4ac&<0
\\\Rightarrow\qquad b^2-4(1)(9/4)&<0
\\\Rightarrow\qquad b^2-9&<0
\\\Rightar... |
If $\mathbf{a},$ $\mathbf{b},$ and $\mathbf{c}$ are vectors such that $\mathbf{a} \cdot \mathbf{b} = -3,$ $\mathbf{a} \cdot \mathbf{c} = 4,$ and $\mathbf{b} \cdot \mathbf{c} = 6,$ then find
\[\mathbf{c} \cdot (3 \mathbf{a} + 5 \mathbf{b}).\] | Expanding the dot product, we get
\begin{align*}
\mathbf{c} \cdot (3 \mathbf{a} + 5 \mathbf{b}) &= 3 \mathbf{a} \cdot \mathbf{c} + 5 \mathbf{b} \cdot \mathbf{c} \\
&= 3 \cdot 4 + 5 \cdot 6 = \boxed{42}.
\end{align*} The answer is 42. | If $\mathbf{a},$ $\mathbf{b},$ and $\mathbf{c}$ are vectors such that $\mathbf{a} \cdot \mathbf{b} = -3,$ $\mathbf{a} \cdot \mathbf{c} = 4,$ and $\mathbf{b} \cdot \mathbf{c} = 6,$ then find
\[\mathbf{c} \cdot (3 \mathbf{a} + 5 \mathbf{b}).\]
Expanding the dot product, we get
\begin{align*}
\mathbf{c} \cdot (3 \mathbf... |
Given the three digits 2, 4 and 7, how many different positive two-digit integers can be formed using these digits if a digit may be repeated in an integer? | There are 3 choices for each digit, for a total of $3\cdot3=\boxed{9}$ integers possible. The answer is 9. | Given the three digits 2, 4 and 7, how many different positive two-digit integers can be formed using these digits if a digit may be repeated in an integer?
There are 3 choices for each digit, for a total of $3\cdot3=\boxed{9}$ integers possible. The answer is 9. |
Find the number of real solutions to
\[(x^{2006} + 1)(x^{2004} + x^{2002} + x^{2000} + \dots + x^2 + 1) = 2007x^{2005}.\] | Note that $x = 0$ is not a solution. Also, if $x < 0,$ then the left-hand side is positive and the right-hand side is negative, so $x$ cannot be a solution. Thus, any real roots must be positive. Assume $x > 0.$
Dividing both sides by $x^{2005},$ we get
\[\frac{(x^{2006} + 1)(x^{2004} + x^{2002} + x^{2000} + \dots ... | Find the number of real solutions to
\[(x^{2006} + 1)(x^{2004} + x^{2002} + x^{2000} + \dots + x^2 + 1) = 2007x^{2005}.\]
Note that $x = 0$ is not a solution. Also, if $x < 0,$ then the left-hand side is positive and the right-hand side is negative, so $x$ cannot be a solution. Thus, any real roots must be positive... |
Solve the equation $|y-4| + 3y = 10$ for $y$. | We consider two cases, $y\ge 4$ and $y < 4$.
Case 1: $y \ge 4:$ If $y \ge 4$, then $|y-4| = y-4$ and our equation is $y-4+3y=10$. So, we have $4y = 14$, or $y=\frac{7}{2}$. However, $y=\frac{7}{2}$ does not satisfy $y\ge 4$. Testing $y=\frac{7}{2}$, we have $\left|\frac{7}{2}-4\right| + 3\cdot \frac{7}{2} =\frac{11... | Solve the equation $|y-4| + 3y = 10$ for $y$.
We consider two cases, $y\ge 4$ and $y < 4$.
Case 1: $y \ge 4:$ If $y \ge 4$, then $|y-4| = y-4$ and our equation is $y-4+3y=10$. So, we have $4y = 14$, or $y=\frac{7}{2}$. However, $y=\frac{7}{2}$ does not satisfy $y\ge 4$. Testing $y=\frac{7}{2}$, we have $\left|\fr... |
What is the minimum value of $8x^2-16x+1003$? | Let $y=8x^2 -16x + 1003$. First, complete the square as follows: $y=8x^2-16x+1003=8(x^2-2x)+1003$. To complete the square, we need to add $\left(\dfrac{2}{2}\right)^2=1$ after the $-2x$. So we have $y+8=8\left(x^2-2x+1\right)+1003$. This gives $y=8\left(x-1\right)^2+995$.
Now, since $\left(x-1\right)^2\ge0$, the mini... | What is the minimum value of $8x^2-16x+1003$?
Let $y=8x^2 -16x + 1003$. First, complete the square as follows: $y=8x^2-16x+1003=8(x^2-2x)+1003$. To complete the square, we need to add $\left(\dfrac{2}{2}\right)^2=1$ after the $-2x$. So we have $y+8=8\left(x^2-2x+1\right)+1003$. This gives $y=8\left(x-1\right)^2+995$... |
Five green balls and five red balls are in a bag. A ball is taken from the bag, its color recorded, then placed back in the bag. A second ball is taken and its color recorded. What is the probability the two balls are the same color? | We could have either two greens or two reds. The probability of drawing two greens is $\left(\dfrac{5}{10}\right)^{\!2}=\dfrac{1}{4}$. The probability of drawing two reds is $\left(\dfrac{5}{10}\right)^{\!2}=\dfrac{1}{4}$. So the answer is $\dfrac{1}{4} + \dfrac{1}{4} = \boxed{\dfrac{1}{2}}$. The answer is \dfrac{1}{2... | Five green balls and five red balls are in a bag. A ball is taken from the bag, its color recorded, then placed back in the bag. A second ball is taken and its color recorded. What is the probability the two balls are the same color?
We could have either two greens or two reds. The probability of drawing two green... |
Find the remainder when $x^4$ is divided by $x^2 + 5x + 1.$ | \[
\begin{array}{c|cc ccc}
\multicolumn{3}{r}{x^2} & -5x & +24 \\
\cline{3-6}
x^2 + 5x + 1 & x^4& & & & \\
\multicolumn{3}{r}{x^4} & +5x^3 & +x^2 \\
\cline{3-5}
\multicolumn{3}{r}{} & -5x^3 & -x^2 & \\
\multicolumn{3}{r}{} & -5x^3 & -25x^2 & -5x \\
\cline{4-6}
\multicolumn{3}{r}{} & & 24x^2 & +5x \\
\multicolumn{3}{r... | Find the remainder when $x^4$ is divided by $x^2 + 5x + 1.$
\[
\begin{array}{c|cc ccc}
\multicolumn{3}{r}{x^2} & -5x & +24 \\
\cline{3-6}
x^2 + 5x + 1 & x^4& & & & \\
\multicolumn{3}{r}{x^4} & +5x^3 & +x^2 \\
\cline{3-5}
\multicolumn{3}{r}{} & -5x^3 & -x^2 & \\
\multicolumn{3}{r}{} & -5x^3 & -25x^2 & -5x \\
\cline{... |
What is the number of centimeters in the length of $EF$ if $AB\parallel CD\parallel EF$ and $AB = 120$ cm?
[asy]
size(4cm,4cm);
pair A,B,C,D,E,F,X;
A=(0,1);
B=(1,1);
C=(1,0);
X=(0,0);
D=(1/3)*C+(2/3)*X;
draw (A--B--C--D);
draw(D--B);
draw(A--C);
E=(0.6,0.4);
F=(1,0.4);
draw(E--F);
label("$A$",A,NW);
label("$B$",... | Since $AB\parallel EF,$ we know that $\angle BAC = \angle FEC$ and $\angle ABC = \angle EFC.$ Therefore, we see that $\triangle ABC \sim \triangle EFC$ by AA Similarity. Likewise, $\triangle BDC \sim \triangle BEF.$
From our similarities, we can come up with two equations: $\dfrac{BF}{BC} = \dfrac{EF}{DC}$ and $\dfrac... | What is the number of centimeters in the length of $EF$ if $AB\parallel CD\parallel EF$ and $AB = 120$ cm?
[asy]
size(4cm,4cm);
pair A,B,C,D,E,F,X;
A=(0,1);
B=(1,1);
C=(1,0);
X=(0,0);
D=(1/3)*C+(2/3)*X;
draw (A--B--C--D);
draw(D--B);
draw(A--C);
E=(0.6,0.4);
F=(1,0.4);
draw(E--F);
label("$A$",A,NW);
label("$B$",... |
What is the greatest integer less than 50 for which the greatest common factor of that integer and 24 is 3? | The prime factors of 24 are 2, 2, 2, and 3. If the greatest common factor with 24 is 3, that means the other number is a multiple of 3 but not 2, 4, 8, or 16. Since the other number cannot be an even number (multiple of 2), we start with 49 and look at decreasing odd numbers. 49 is not a multiple of 3, 47 and 45 are mu... | What is the greatest integer less than 50 for which the greatest common factor of that integer and 24 is 3?
The prime factors of 24 are 2, 2, 2, and 3. If the greatest common factor with 24 is 3, that means the other number is a multiple of 3 but not 2, 4, 8, or 16. Since the other number cannot be an even number (mu... |
The side lengths of a triangle are 12 cm, 35 cm and 37 cm. What is the area of the triangle in square centimeters? | Shrinking the triangle by dividing every side length by 1, we recognize the resulting set $$\{12,35,37\}$$ of side lengths as a Pythagorean triple. Therefore, the original triangle is also a right triangle, and its legs measure 12 cm and 35 cm. The area of the triangle is $\frac{1}{2}(12\text{ cm})(35\text{ cm})=\box... | The side lengths of a triangle are 12 cm, 35 cm and 37 cm. What is the area of the triangle in square centimeters?
Shrinking the triangle by dividing every side length by 1, we recognize the resulting set $$\{12,35,37\}$$ of side lengths as a Pythagorean triple. Therefore, the original triangle is also a right trian... |
What is the value of the sum $\frac{1}{2}+\frac{1^2}{2^2}+\frac{1^3}{2^3}+ \ldots +\frac{1^{12}}{2^{12}}$? Express your answer as a common fraction. | This is the sum of the series $a_1 + a_2 + \ldots + a_{12}$ with $a_1 = \frac{1}{2}$ and $r = \frac{1}{2}$.
Thus, \begin{align*}
S &= \frac{a(1-r^{n})}{1-r}= \frac{1}{2} \cdot \frac{1-\left(\frac{1}{2}\right)^{12}}{1-\frac{1}{2}}\\
& = \frac{1}{2}\cdot\frac{1-\frac{1}{4096}}{\frac{1}{2}}=\frac{1}{2}\cdot\frac{2}{1}\c... | What is the value of the sum $\frac{1}{2}+\frac{1^2}{2^2}+\frac{1^3}{2^3}+ \ldots +\frac{1^{12}}{2^{12}}$? Express your answer as a common fraction.
This is the sum of the series $a_1 + a_2 + \ldots + a_{12}$ with $a_1 = \frac{1}{2}$ and $r = \frac{1}{2}$.
Thus, \begin{align*}
S &= \frac{a(1-r^{n})}{1-r}= \frac{1}{... |
Let $P$ be a point inside triangle $ABC$ such that
\[\overrightarrow{PA} + 3 \overrightarrow{PB} + 5 \overrightarrow{PC} = \mathbf{0}.\]Find the ratio of the area of triangle $ABC$ to the area of triangle $APC.$ | We let $\mathbf{a} = \overrightarrow{A},$ etc. Then the equation $\overrightarrow{PA} + 3 \overrightarrow{PB} + 5 \overrightarrow{PC} = \mathbf{0}$ becomes
\[\mathbf{a} - \mathbf{p} + 3 (\mathbf{b} - \mathbf{p}) + 5 (\mathbf{c} - \mathbf{p}) = \mathbf{0}.\]Solving for $\mathbf{p},$ we find
\[\mathbf{p} = \frac{\mathbf... | Let $P$ be a point inside triangle $ABC$ such that
\[\overrightarrow{PA} + 3 \overrightarrow{PB} + 5 \overrightarrow{PC} = \mathbf{0}.\]Find the ratio of the area of triangle $ABC$ to the area of triangle $APC.$
We let $\mathbf{a} = \overrightarrow{A},$ etc. Then the equation $\overrightarrow{PA} + 3 \overrightarrow... |
A sports conference has 16 teams in two divisions of 8. How many games are in a complete season for the conference if each team must play every other team in its own division twice and every team in the other division once? | Each team plays 7 other teams in its division twice, and the 8 teams in the other division once, for a total of $7 \times 2 + 8 = 22$ games for each team. There are 16 teams total, which gives a preliminary count of $22 \times 16 = 352$ games, but we must divide by two because we have counted each game twice (once for... | A sports conference has 16 teams in two divisions of 8. How many games are in a complete season for the conference if each team must play every other team in its own division twice and every team in the other division once?
Each team plays 7 other teams in its division twice, and the 8 teams in the other division on... |
$x = {1+\frac{\sqrt{3}}{1+\frac{\sqrt{3}}{1+...}}}$. Evaluate $\frac{1}{(x+2)(x-1)}$. When your answer is in the form $\frac{A+\sqrt{B}}{C}$, where $A$, $B$, and $C$ are integers, and $B$ is not divisible by the square of a prime, what is $|A|+|B|+|C|$? | We see that $x-1=\frac{\sqrt{3}}{1+\frac{\sqrt{3}}{1+...}}$, so $\frac{\sqrt{3}}{x-1}=1+\frac{\sqrt{3}}{1+\frac{\sqrt{3}}{1+...}}=x$. Solving for $x$, we find $\sqrt{3}=x(x-1)$, which means $x^{2}-x=\sqrt{3}$. Simplify the denominator of $\frac{1}{(x+2)(x-1)}$ to obtain $\frac{1}{x^2+x-2}$. Substituting for $x^2-x$, we... | $x = {1+\frac{\sqrt{3}}{1+\frac{\sqrt{3}}{1+...}}}$. Evaluate $\frac{1}{(x+2)(x-1)}$. When your answer is in the form $\frac{A+\sqrt{B}}{C}$, where $A$, $B$, and $C$ are integers, and $B$ is not divisible by the square of a prime, what is $|A|+|B|+|C|$?
We see that $x-1=\frac{\sqrt{3}}{1+\frac{\sqrt{3}}{1+...}}$, so ... |
The sides of an isosceles triangle are $\sin x,$ $\sin x,$ and $\sin 7x,$ and its vertex angle is $2x.$ (All angle measurements are in degrees.) Enter all possible values of $x,$ separated by commas. | Note that angle $x$ must be acute.
If we drop an altitude from the vertex of the isosceles triangle, then we obtain two right triangles, where one of the angles is $x,$ the opposite side is $\frac{\sin 7x}{2},$ and the hypotenuse is $\sin x.$ Hence,
\[\cos x = \frac{\frac{\sin 7x}{2}}{\sin x} = \frac{\sin 7x}{2 \sin ... | The sides of an isosceles triangle are $\sin x,$ $\sin x,$ and $\sin 7x,$ and its vertex angle is $2x.$ (All angle measurements are in degrees.) Enter all possible values of $x,$ separated by commas.
Note that angle $x$ must be acute.
If we drop an altitude from the vertex of the isosceles triangle, then we obtain... |
Let $\mathbf{a} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}.$ Find the area of the triangle with vertices $\mathbf{0},$ $\mathbf{a},$ and $\mathbf{b}.$ | The area of the triangle formed by $\mathbf{0},$ $\mathbf{a},$ and $\mathbf{b}$ is half the area of the parallelogram formed by $\mathbf{0},$ $\mathbf{a},$ $\mathbf{b},$ and $\mathbf{a} + \mathbf{b}.$
[asy]
unitsize(0.8 cm);
pair A, B, O;
A = (3,-2);
B = (-1,3);
O = (0,0);
draw(O--A,Arrow(6));
draw(O--B,Arrow(6));
... | Let $\mathbf{a} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}.$ Find the area of the triangle with vertices $\mathbf{0},$ $\mathbf{a},$ and $\mathbf{b}.$
The area of the triangle formed by $\mathbf{0},$ $\mathbf{a},$ and $\mathbf{b}$ is half the area of the parallel... |
The line with equation $y = -x$ is an axis of symmetry of the curve with equation
\[y = \frac{px + q}{rx + s},\]where $p,$ $q,$ $r,$ $s$ are all nonzero. Which of the following statements must hold?
(A) $p + q = 0$
(B) $p - r = 0$
(C) $p - s = 0$
(D) $q - r = 0$
(E) $q - s = 0$
(F) $r - s = 0$ | Since $y = -x$ is an axis of symmetry, if point $(a,b)$ lies on the graph, then so does $(-b,-a).$ Thus, the equation of the graph can also be written as
\[-x = \frac{py + q}{ry + s}.\]Substituting $y = \frac{px + q}{rx + s},$ we get
\[-x = \frac{p \cdot \frac{px + q}{rx + s} + q}{r \cdot \frac{px + q}{rx + s} + s} = ... | The line with equation $y = -x$ is an axis of symmetry of the curve with equation
\[y = \frac{px + q}{rx + s},\]where $p,$ $q,$ $r,$ $s$ are all nonzero. Which of the following statements must hold?
(A) $p + q = 0$
(B) $p - r = 0$
(C) $p - s = 0$
(D) $q - r = 0$
(E) $q - s = 0$
(F) $r - s = 0$
Since $y = -x$ i... |
Let $z$ be a complex number such that
\[|z - 10| + |z - 6i| = 15.\]Find the smallest possible value of $|z|.$ | By the Triangle Inequality,
\[|z - 10| + |z - 6i| = |z - 10| + |6i - z| \ge |(z - 10) + (6i - z)| = |-10 + 6i| = 15.\]But we are told that $|z - 10| + |z - 6i| = 15.$ The only way that equality can occur is if $z$ lies on the line segment connecting 10 and $6i$ in the complex plane.
[asy]
unitsize(0.4 cm);
pair Z = ... | Let $z$ be a complex number such that
\[|z - 10| + |z - 6i| = 15.\]Find the smallest possible value of $|z|.$
By the Triangle Inequality,
\[|z - 10| + |z - 6i| = |z - 10| + |6i - z| \ge |(z - 10) + (6i - z)| = |-10 + 6i| = 15.\]But we are told that $|z - 10| + |z - 6i| = 15.$ The only way that equality can occur is ... |
If
\[\frac{\cos^4 \alpha}{\cos^2 \beta} + \frac{\sin^4 \alpha}{\sin^2 \beta} = 1,\]and $\alpha + \beta = \frac{\pi}{4}$, then find the sum of all possible values of
\[\frac{\sin^4 \beta}{\sin^2 \alpha} + \frac{\cos^4 \beta}{\cos^2 \alpha}.\] | As before, we have that $\cos^2 \alpha = \cos^2 \beta.$ Since $\alpha + \beta = \frac{\pi}{4}$, we have that
\[\cos^2 \alpha = \cos^2 \beta = \cos^2 \left(\frac{\pi}{4} - \alpha\right) = \sin^2 \alpha.\]Therefore,
\[\frac{\sin^4 \beta}{\sin^2 \alpha} + \frac{\cos^4 \beta}{\cos^2 \alpha} = \frac{\sin^4 \beta}{\cos^2 \b... | If
\[\frac{\cos^4 \alpha}{\cos^2 \beta} + \frac{\sin^4 \alpha}{\sin^2 \beta} = 1,\]and $\alpha + \beta = \frac{\pi}{4}$, then find the sum of all possible values of
\[\frac{\sin^4 \beta}{\sin^2 \alpha} + \frac{\cos^4 \beta}{\cos^2 \alpha}.\]
As before, we have that $\cos^2 \alpha = \cos^2 \beta.$ Since $\alpha + \be... |
For what value of $x$ does $5^{x^{2}-3x+2} = 5^{x^{2}+5x-6}$? Express your answer as a common fraction. | If $5^{x^{2}-3x+2} = 5^{x^{2}+5x-6}$, then $x^{2}-3x+2 = x^{2}+5x-6$. We can eliminate the $x^2$ term from each side and solve $-3x+2=5x-6$ for $x$ to get $x=\boxed{1}$. The answer is 1. | For what value of $x$ does $5^{x^{2}-3x+2} = 5^{x^{2}+5x-6}$? Express your answer as a common fraction.
If $5^{x^{2}-3x+2} = 5^{x^{2}+5x-6}$, then $x^{2}-3x+2 = x^{2}+5x-6$. We can eliminate the $x^2$ term from each side and solve $-3x+2=5x-6$ for $x$ to get $x=\boxed{1}$. The answer is 1. |
Given that $a$ is a multiple of $380$, find the greatest common divisor of $3a^3+a^2+4a+57$ and $a$. | We use the Euclidean Algorithm. \begin{align*}
\text{gcd}\,(3a^3+a^2+4a+57,a)
&=\text{gcd}\,(3a^3+a^2+4a+57-(3a^2+a+4)a,a)\\
&=\text{gcd}\,(57,a).
\end{align*}Since $57$ is not a divisor of $380$, but $19$ is a divisor of both $57$ and $380$, and $a$ is a multiple of $380$, the greatest common divisor is $\boxed{19}$.... | Given that $a$ is a multiple of $380$, find the greatest common divisor of $3a^3+a^2+4a+57$ and $a$.
We use the Euclidean Algorithm. \begin{align*}
\text{gcd}\,(3a^3+a^2+4a+57,a)
&=\text{gcd}\,(3a^3+a^2+4a+57-(3a^2+a+4)a,a)\\
&=\text{gcd}\,(57,a).
\end{align*}Since $57$ is not a divisor of $380$, but $19$ is a divis... |
What is the smallest four-digit positive integer congruent to $2 \pmod{23}?$ | An integer that is congruent to $2 \pmod{23}$ is of the form $23n+2$.
Therefore, we form the inequality $23n+2>1000$, and find the smallest possible integer $n$. We get \begin{align*}
23n+2&>1000 \\
23n&>998\\
n&>\frac{998}{23} \approx 43.39.
\end{align*} The smallest possible positive integer $n$ is $44$. We plug it ... | What is the smallest four-digit positive integer congruent to $2 \pmod{23}?$
An integer that is congruent to $2 \pmod{23}$ is of the form $23n+2$.
Therefore, we form the inequality $23n+2>1000$, and find the smallest possible integer $n$. We get \begin{align*}
23n+2&>1000 \\
23n&>998\\
n&>\frac{998}{23} \approx 43.3... |
Let $x$ and $y$ be complex numbers such that
\[\frac{x + iy}{x - iy} + \frac{x - iy}{x + iy} = 1.\]Find
\[\frac{x^4 + y^4}{x^4 - y^4} + \frac{x^4 - y^4}{x^4 + y^4}.\] | Putting $\frac{x + iy}{x - iy} + \frac{x - iy}{x + iy}$ over a common denominator, we get
\[\frac{2x^2 + 2y^2}{x^2 + y^2} = 1.\]Then $2x^2 + 2y^2 = x^2 + y^2,$ so $x^2 = y^2.$
Then
\begin{align*}
\frac{x^4 + y^4}{x^4 - y^4} + \frac{x^4 - y^4}{x^4 + y^4} &= \frac{2x^4}{0} + \frac{0}{2x^4} \\
&= \boxed{\text{undefined}}... | Let $x$ and $y$ be complex numbers such that
\[\frac{x + iy}{x - iy} + \frac{x - iy}{x + iy} = 1.\]Find
\[\frac{x^4 + y^4}{x^4 - y^4} + \frac{x^4 - y^4}{x^4 + y^4}.\]
Putting $\frac{x + iy}{x - iy} + \frac{x - iy}{x + iy}$ over a common denominator, we get
\[\frac{2x^2 + 2y^2}{x^2 + y^2} = 1.\]Then $2x^2 + 2y^2 = x^2... |
Xanthia buys hot dogs that come in packages of six, and she buys hot dog buns that come in packages of ten. What is the smallest number of hot dog packages she can buy in order to be able to buy an equal number of hot dogs and hot dog buns? | To solve this problem, we first find the least common multiple (LCM) of 6 and 10. $6=2\cdot3$ and $10=2\cdot5$, so their LCM is $2\cdot3\cdot5=30$. Therefore, Xanthia can buy $30\div6=\boxed{5}$ hot dog packages and $30\div10=3$ hot dog bun packages to have an equal number of hot dogs and hot dog buns. The answer is 5. | Xanthia buys hot dogs that come in packages of six, and she buys hot dog buns that come in packages of ten. What is the smallest number of hot dog packages she can buy in order to be able to buy an equal number of hot dogs and hot dog buns?
To solve this problem, we first find the least common multiple (LCM) of 6 and... |
Jeremy's friend Steven will pay him $\dfrac{13}{4}$ dollars for every room he cleans. Yesterday, Jeremy cleaned $\dfrac{11}{5}$ rooms. How many dollars does Steven owe Jeremy? Leave your answer in fraction form. | Since Steven owes $\frac{13}{4}$ for each of the $\frac{11}{5}$ rooms, he owes $\frac{13}{4}\cdot \frac{11}{5} = \frac{13\cdot 11}{4\cdot 5} = \boxed{\frac{143}{20}}$ dollars. The answer is \frac{143}{20}. | Jeremy's friend Steven will pay him $\dfrac{13}{4}$ dollars for every room he cleans. Yesterday, Jeremy cleaned $\dfrac{11}{5}$ rooms. How many dollars does Steven owe Jeremy? Leave your answer in fraction form.
Since Steven owes $\frac{13}{4}$ for each of the $\frac{11}{5}$ rooms, he owes $\frac{13}{4}\cdot \frac{11... |
Subtract $123.45$ from $456.78.$ Express the result as a decimal to the nearest thousandth. | We can organize the subtraction concisely using columns as follows: \[
\begin{array}{@{}c@{}c@{}c@{}c@{}c@{}c}
& 4 & 5 & 6. & 7 & 8 \\
- & 1 & 2 & 3. & 4 & 5
\\ \cline{1-6}
& 3 & 3 & 3. & 3 & 3 \\
\end{array}
\] The answer is $\boxed{333.333}$. The answer is 333.333. | Subtract $123.45$ from $456.78.$ Express the result as a decimal to the nearest thousandth.
We can organize the subtraction concisely using columns as follows: \[
\begin{array}{@{}c@{}c@{}c@{}c@{}c@{}c}
& 4 & 5 & 6. & 7 & 8 \\
- & 1 & 2 & 3. & 4 & 5
\\ \cline{1-6}
& 3 & 3 & 3. & 3 & 3 \\
\end{array}
\] The answer is... |
What is $(-1)^1+(-1)^2+\cdots+(-1)^{2007}$ ? | Because $(-1)^k$ equals 1 if $k$ is even and $-1$ if $k$ is odd, the sum can be written as \[
(-1+1)+(-1+1)+\cdots+(-1+1)-1 =0+0+\cdots+0-1=\boxed{-1}.
\] The answer is -1. | What is $(-1)^1+(-1)^2+\cdots+(-1)^{2007}$ ?
Because $(-1)^k$ equals 1 if $k$ is even and $-1$ if $k$ is odd, the sum can be written as \[
(-1+1)+(-1+1)+\cdots+(-1+1)-1 =0+0+\cdots+0-1=\boxed{-1}.
\] The answer is -1. |
Three tiles are marked X and three other tiles are marked O. The six tiles are randomly arranged in a row. What is the probability that the arrangement reads XOXOXO? | There are three X's and three O's, and the tiles are selected without replacement, so the probability is \[
\frac{3}{6}\cdot\frac{3}{5}\cdot\frac{2}{4}\cdot\frac{2}{3}\cdot\frac{1}{2}\cdot\frac{1}{1}= \frac{1}{20}.
\]OR
The three tiles marked X are equally likely to lie in any of $\binom{6}{3}=20$ positions, so the pr... | Three tiles are marked X and three other tiles are marked O. The six tiles are randomly arranged in a row. What is the probability that the arrangement reads XOXOXO?
There are three X's and three O's, and the tiles are selected without replacement, so the probability is \[
\frac{3}{6}\cdot\frac{3}{5}\cdot\frac{2}{4}\... |
What is the smallest four-digit positive integer that gives a remainder of 13 when divided by 47? | We divide 1000 by 47 and get a remainder of 13. Therefore, the smallest four-digit integer that gives a remainder of 13 when divided by 47 is $\boxed{1000}$. The answer is 1000. | What is the smallest four-digit positive integer that gives a remainder of 13 when divided by 47?
We divide 1000 by 47 and get a remainder of 13. Therefore, the smallest four-digit integer that gives a remainder of 13 when divided by 47 is $\boxed{1000}$. The answer is 1000. |
Let $ABCDEF$ be a regular hexagon, and let $G,H,I$ be the midpoints of sides $AB,CD,EF$ respectively. If the area of $\triangle GHI$ is $100$, what is the area of the hexagon $ABCDEF$? | We begin with a diagram of the given information: [asy]
size(4cm);
real x=sqrt(3);
pair d=(2,0); pair c=(1,x); pair b=(-1,x); pair a=-d; pair f=-c; pair e=-b;
pair g=(a+b)/2; pair h=(c+d)/2; pair i=(e+f)/2;
draw(a--b--c--d--e--f--a);
dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); dot(g); dot(h); dot(i);
draw(g--h--i--... | Let $ABCDEF$ be a regular hexagon, and let $G,H,I$ be the midpoints of sides $AB,CD,EF$ respectively. If the area of $\triangle GHI$ is $100$, what is the area of the hexagon $ABCDEF$?
We begin with a diagram of the given information: [asy]
size(4cm);
real x=sqrt(3);
pair d=(2,0); pair c=(1,x); pair b=(-1,x); pair a=... |
Mary can mow a lawn in four hours and Tom can mow the lawn in 5 hours. If Tom works for 2 hours alone, and then Mary joins him, how many additional minutes will it take them to finish mowing the lawn? | As calculated before, Tom mows $\frac{2}{5}$ of the lawn in 2 hours. Mary can mow the whole lawn in 4 hours, so in one hour, she can mow $\frac{1}{4}$ of the lawn. Together, they can mow $\frac{2}{5} + \frac{1}{4} = \frac{13}{20}$ of the lawn in one hour. Since $\frac{3}{5}$ of the lawn remains, it will take them $\fra... | Mary can mow a lawn in four hours and Tom can mow the lawn in 5 hours. If Tom works for 2 hours alone, and then Mary joins him, how many additional minutes will it take them to finish mowing the lawn?
As calculated before, Tom mows $\frac{2}{5}$ of the lawn in 2 hours. Mary can mow the whole lawn in 4 hours, so in on... |
Gage skated 1 hr 25 min each day for 6 days and 1 hr 40 min each day for 4 days. How many minutes would he have to skate the eleventh day in order to average 90 minutes of skating each day for the entire time? | In 6 days, Gage skated for $6 \times 85 =510$ minutes, and in 4 days he skated for $4 \times 100 = 400$ minutes. So, in 10 days he skated for $510 + 400 = 910$ minutes. To average 90 minutes per day for 11 days he must skate $11 \times 90 = 990$ minutes, so he must skate $990-910=\boxed{80}$ minutes the eleventh day.... | Gage skated 1 hr 25 min each day for 6 days and 1 hr 40 min each day for 4 days. How many minutes would he have to skate the eleventh day in order to average 90 minutes of skating each day for the entire time?
In 6 days, Gage skated for $6 \times 85 =510$ minutes, and in 4 days he skated for $4 \times 100 = 400$ min... |
If \[f(x) =
\begin{cases}
x^2-4 &\quad \text{if } x \ge 4, \\
x + 3 &\quad \text{otherwise},
\end{cases}
\]then for how many values of $x$ is $f(f(x)) = 5$? | Let $y = f(x)$. Then, $f(f(x)) = f(y) = 5$, so either $x^2 - 4 = 5$ or $x + 3 = 5$. Solving the first equations yields that $y = f(x) = \pm 3$, both of which are less than $4$. The second equation yields that $y = 2$, but we discard this solution because $y < 4$.
Hence $f(x) = \pm 3$, so $x^2 - 4 = \pm 3$ or $x + 3 = ... | If \[f(x) =
\begin{cases}
x^2-4 &\quad \text{if } x \ge 4, \\
x + 3 &\quad \text{otherwise},
\end{cases}
\]then for how many values of $x$ is $f(f(x)) = 5$?
Let $y = f(x)$. Then, $f(f(x)) = f(y) = 5$, so either $x^2 - 4 = 5$ or $x + 3 = 5$. Solving the first equations yields that $y = f(x) = \pm 3$, both of which are... |
A line is expressed in the form
\[\begin{pmatrix} 3 \\ 2 \end{pmatrix} \cdot \left( \begin{pmatrix} x \\ y \end{pmatrix} - \begin{pmatrix} -2 \\ 5 \end{pmatrix} \right) = 0.\]The equation of the line can be expressed in the form $y = mx + b.$ Enter the ordered pair $(m,b).$ | Expanding, we get
\[\begin{pmatrix} 3 \\ 2 \end{pmatrix} \cdot \left( \begin{pmatrix} x \\ y \end{pmatrix} - \begin{pmatrix} -2 \\ 5 \end{pmatrix} \right) = \begin{pmatrix} 3 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} x + 2 \\ y - 5 \end{pmatrix} = (3)(x + 2) + (2)(y - 5) = 0.\]Solving for $y,$ we find
\[y = -\frac{3}{2}... | A line is expressed in the form
\[\begin{pmatrix} 3 \\ 2 \end{pmatrix} \cdot \left( \begin{pmatrix} x \\ y \end{pmatrix} - \begin{pmatrix} -2 \\ 5 \end{pmatrix} \right) = 0.\]The equation of the line can be expressed in the form $y = mx + b.$ Enter the ordered pair $(m,b).$
Expanding, we get
\[\begin{pmatrix} 3 \\ 2... |
The graph of \[(y-2)^4 - 4(x+1)^4 = 2(y-2)^2 - 1\]is the union of the graphs of two different conic sections. Which two types of conic sections are they? | We can rewrite the given equation as \[(y-2)^4 - 2(y-2)^2 + 1 = 4(x+1)^4.\]The left-hand side is the perfect square of a binomial: \[((y-2)^2-1)^2 = 4(x+1)^4.\]Therefore, either $(y-2)^2-1=2(x+1)^2$ or $(y-2)^2-1=-2(x+1)^2.$ That is, either $(y-2)^2-2(x+1)^2=1$ or $(y-2)^2+2(x+1)^2=1.$ These are the equations for a hyp... | The graph of \[(y-2)^4 - 4(x+1)^4 = 2(y-2)^2 - 1\]is the union of the graphs of two different conic sections. Which two types of conic sections are they?
We can rewrite the given equation as \[(y-2)^4 - 2(y-2)^2 + 1 = 4(x+1)^4.\]The left-hand side is the perfect square of a binomial: \[((y-2)^2-1)^2 = 4(x+1)^4.\]Ther... |
We have a triangle $\triangle ABC$ such that $AB = 8,$ $AC = 10,$ and $BC = 12.$ What is the length of the median $AM$? | Let's draw a sketch first. Since $\triangle ABC$ is not isosceles, we cannot conclude that $AM$ is perpendicular to $BC.$ [asy]
pair A, B, C, M;
A = (0, 6);
B = (-6, 0);
C = (6, 0);
M = 0.5 * B + 0.5 * C;
draw(A--B--C--cycle);
draw(A--M);
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$M$", M, S);... | We have a triangle $\triangle ABC$ such that $AB = 8,$ $AC = 10,$ and $BC = 12.$ What is the length of the median $AM$?
Let's draw a sketch first. Since $\triangle ABC$ is not isosceles, we cannot conclude that $AM$ is perpendicular to $BC.$ [asy]
pair A, B, C, M;
A = (0, 6);
B = (-6, 0);
C = (6, 0);
M = 0.5 * B + 0.... |
Marina solved the quadratic equation $16x^2+24x-864=0$ by completing the square. In the process, she came up with the equivalent equation $$(x+r)^2 = s,$$where $r$ and $s$ are constants.
What is $s$? | Dividing both sides of the equation $16x^2+24x-864=0$ by $16$, we have $$x^2+\frac{3}{2}x-\frac{27}{2}=0.$$The square which agrees with $x^2+\frac{3}{2}x-\frac{27}{2}$ except for the constant term is $\left(x+\frac{3}{4}\right)^2$, which is equal to $x^2+\frac{3}{2}x+\frac{9}{16}$ and thus to $\left(x^2+\frac{3}{2}x-\f... | Marina solved the quadratic equation $16x^2+24x-864=0$ by completing the square. In the process, she came up with the equivalent equation $$(x+r)^2 = s,$$where $r$ and $s$ are constants.
What is $s$?
Dividing both sides of the equation $16x^2+24x-864=0$ by $16$, we have $$x^2+\frac{3}{2}x-\frac{27}{2}=0.$$The square... |
Each of the boxes in the figure below is a square. How many different squares can be traced using the lines in the figure, excluding the outermost square? | The number of squares that can be traced excluding the outermost square is the same as the total number of squares minus the number of squares that form the outermost square. The outermost square is a $4 \times 4$ square, which contains 9 $1 \times 1$ squares, 4 $2 \times 2$ squares, and 1 $3 \times 3$ square. Therefor... | Each of the boxes in the figure below is a square. How many different squares can be traced using the lines in the figure, excluding the outermost square?
The number of squares that can be traced excluding the outermost square is the same as the total number of squares minus the number of squares that form the outerm... |
What is the largest positive multiple of $12$ that is less than $360?$ | Dividing $360$ by $12$ gives a quotient $30$ with a remainder of $0$. In other words, \[360=12\cdot30+0.\]Thus, $30\cdot12=\boxed{360}$ is the largest multiple of $12$ which is less than $360.$ The answer is 360. | What is the largest positive multiple of $12$ that is less than $360?$
Dividing $360$ by $12$ gives a quotient $30$ with a remainder of $0$. In other words, \[360=12\cdot30+0.\]Thus, $30\cdot12=\boxed{360}$ is the largest multiple of $12$ which is less than $360.$ The answer is 360. |
John has 5 green marbles and 3 purple marbles. He chooses a marble at random, writes down its color, and then puts the marble back. He performs this process 4 times. What is the probability that he chooses exactly one green marble? | There are $\binom{4}{1}=4$ ways to choose which of the four marbles John chose was green. For any of those 4 choices, there is a $\left( \frac{5}{8} \right) \left( \frac{3}{8} \right) ^3 = \frac{135}{4096}$ chance for that choice to happen. The total probability is then $4 \cdot \frac{135}{4096}= \boxed{\frac{135}{1024... | John has 5 green marbles and 3 purple marbles. He chooses a marble at random, writes down its color, and then puts the marble back. He performs this process 4 times. What is the probability that he chooses exactly one green marble?
There are $\binom{4}{1}=4$ ways to choose which of the four marbles John chose was gr... |
As $n$ ranges over the positive integers, what is the sum of all possible values of the greatest common divisor of $2n+3$ and $n$? | Using the Euclidean algorithm, we have:
\begin{align*}
\gcd(2n+3, n) &= \gcd(n, 2n+3 - 2n) \\
&= \gcd(n, 3).
\end{align*}There are two cases to consider:
Case 1: $n$ is not a multiple of 3. Therefore, $n$ and 3 are relatively prime and have a greatest common divisor of 1.
Case 2: $n$ is a multiple of 3. In this case,... | As $n$ ranges over the positive integers, what is the sum of all possible values of the greatest common divisor of $2n+3$ and $n$?
Using the Euclidean algorithm, we have:
\begin{align*}
\gcd(2n+3, n) &= \gcd(n, 2n+3 - 2n) \\
&= \gcd(n, 3).
\end{align*}There are two cases to consider:
Case 1: $n$ is not a multiple of... |
If $\mathbf{a}$ and $\mathbf{b}$ are vectors such that $\|\mathbf{a}\| = 7$, $\|\mathbf{b}\| = 11$, and the angle between them is $60^\circ$, then find the value of $\mathbf{a} \cdot \mathbf{b}$. | We know that $\mathbf{a}\cdot\mathbf{b}=\|\mathbf{a}\|\cdot\|\mathbf{b}\|\cdot\cos \theta =7\cdot 11\cdot\cos 60^\circ =7\cdot 11\cdot\frac{1}{2}=\boxed{\frac{77}{2}}$. The answer is \frac{77}{2}. | If $\mathbf{a}$ and $\mathbf{b}$ are vectors such that $\|\mathbf{a}\| = 7$, $\|\mathbf{b}\| = 11$, and the angle between them is $60^\circ$, then find the value of $\mathbf{a} \cdot \mathbf{b}$.
We know that $\mathbf{a}\cdot\mathbf{b}=\|\mathbf{a}\|\cdot\|\mathbf{b}\|\cdot\cos \theta =7\cdot 11\cdot\cos 60^\circ =7\... |
Machiavelli subtracted $2-4i$ from $1+3i$. What number did he get? | Subtracting the real part and imaginary parts separately, we have $(1-2)+(3-(-4))i=\boxed{-1+7i}$. The answer is -1+7i. | Machiavelli subtracted $2-4i$ from $1+3i$. What number did he get?
Subtracting the real part and imaginary parts separately, we have $(1-2)+(3-(-4))i=\boxed{-1+7i}$. The answer is -1+7i. |
In how many ways can 7 people be seated in a row of chairs if two of the people, Wilma and Paul, refuse to sit next to each other or at the ends of the row? | The number of acceptable arrangements from the previous question is 3600. The number of seating arrangements in which Wilma and Paul sit at the ends is $5!\times 2!$. Thus the number of acceptable arrangements is $3600-5!\times 2!=\boxed{3360}$. The answer is 3360. | In how many ways can 7 people be seated in a row of chairs if two of the people, Wilma and Paul, refuse to sit next to each other or at the ends of the row?
The number of acceptable arrangements from the previous question is 3600. The number of seating arrangements in which Wilma and Paul sit at the ends is $5!\times... |
Define a regular $n$-pointed star to be the union of $n$ line segments $P_1P_2, P_2P_3,\ldots, P_nP_1$ such that
the points $P_1, P_2,\ldots, P_n$ are coplanar and no three of them are collinear,
each of the $n$ line segments intersects at least one of the other line segments at a point other than an endpoint,
all of t... | We use the Principle of Inclusion-Exclusion (PIE).
If we join the adjacent vertices of the regular $n$-star, we get a regular $n$-gon. We number the vertices of this $n$-gon in a counterclockwise direction: $0, 1, 2, 3, \ldots, n-1.$
A regular $n$-star will be formed if we choose a vertex number $m$, where $0 \le m \le... | Define a regular $n$-pointed star to be the union of $n$ line segments $P_1P_2, P_2P_3,\ldots, P_nP_1$ such that
the points $P_1, P_2,\ldots, P_n$ are coplanar and no three of them are collinear,
each of the $n$ line segments intersects at least one of the other line segments at a point other than an endpoint,
all of t... |
Given that $16^{-1} \equiv 43 \pmod{97}$, find $64^{-1} \pmod{97}$, as a residue modulo 97. (Give an answer between 0 and 96, inclusive.) | Since $16^{-1} \equiv 43 \pmod{97}$, $64^{-1} \equiv (16^2)^{-1} \equiv (16^{-1})^2 \equiv 43^2 \equiv \boxed{52} \pmod{97}$. The answer is 52. | Given that $16^{-1} \equiv 43 \pmod{97}$, find $64^{-1} \pmod{97}$, as a residue modulo 97. (Give an answer between 0 and 96, inclusive.)
Since $16^{-1} \equiv 43 \pmod{97}$, $64^{-1} \equiv (16^2)^{-1} \equiv (16^{-1})^2 \equiv 43^2 \equiv \boxed{52} \pmod{97}$. The answer is 52. |
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