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The science club has 25 members: 10 boys and 15 girls. A 5-person committee is chosen at random. What is the probability that the committee has at least 1 boy and at least 1 girl?

Level 5

Counting & Probability

We can use the idea of complementary probability to solve this problem without too much nasty casework. The probability that the committee has at least 1 boy and 1 girl is equal to 1 minus the probability that the committee is either all boys or all girls. The number of ways to choose a committee of all boys is $\binom{10}{5}=252$, the number of ways to choose a committee of all girls is $\binom{15}{5}=3,\!003$, and the total number of committees is $\binom{25}{5}=53,\!130$, so the probability of selecting a committee of all boys or all girls is $\dfrac{252+3003}{53,\!130}=\dfrac{31}{506}$. Thus the probability that the committee contains at least one boy and one girl is $1-\dfrac{31}{506} = \boxed{\dfrac{475}{506}}$.

In a rectangular array of points, with 5 rows and $N$ columns, the points are numbered consecutively from left to right beginning with the top row. Thus the top row is numbered 1 through $N,$ the second row is numbered $N + 1$ through $2N,$ and so forth. Five points, $P_1, P_2, P_3, P_4,$ and $P_5,$ are selected so that each $P_i$ is in row $i.$ Let $x_i$ be the number associated with $P_i.$ Now renumber the array consecutively from top to bottom, beginning with the first column. Let $y_i$ be the number associated with $P_i$ after the renumbering. It is found that $x_1 = y_2,$ $x_2 = y_1,$ $x_3 = y_4,$ $x_4 = y_5,$ and $x_5 = y_3.$ Find the smallest possible value of $N.$

Level 5

Number Theory

Let each point $P_i$ be in column $c_i$. The numberings for $P_i$ can now be defined as follows.\begin{align*}x_i &= (i - 1)N + c_i\\ y_i &= (c_i - 1)5 + i \end{align*} We can now convert the five given equalities.\begin{align}x_1&=y_2 & \Longrightarrow & & c_1 &= 5 c_2-3\\ x_2&=y_1 & \Longrightarrow & & N+c_2 &= 5 c_1-4\\ x_3&=y_4 & \Longrightarrow & & 2 N+c_3 &= 5 c_4-1\\ x_4&=y_5 & \Longrightarrow & & 3 N+c_4 &= 5 c_5\\ x_5&=y_3 & \Longrightarrow & & 4 N+c_5 &= 5 c_3-2 \end{align}Equations $(1)$ and $(2)$ combine to form\[N = 24c_2 - 19\]Similarly equations $(3)$, $(4)$, and $(5)$ combine to form\[117N +51 = 124c_3\]Take this equation modulo 31\[24N+20\equiv 0 \pmod{31}\]And substitute for N\[24 \cdot 24 c_2 - 24 \cdot 19 +20\equiv 0 \pmod{31}\]\[18 c_2 \equiv 2 \pmod{31}\] Thus the smallest $c_2$ might be is $7$ and by substitution $N = 24 \cdot 7 - 19 = 149$ The column values can also easily be found by substitution\begin{align*}c_1&=32\\ c_2&=7\\ c_3&=141\\ c_4&=88\\ c_5&=107 \end{align*}As these are all positive and less than $N$, $\boxed{149}$ is the solution.

The diagram shows a rectangle that has been dissected into nine non-overlapping squares. Given that the width and the height of the rectangle are relatively prime positive integers, find the perimeter of the rectangle. [asy]draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36)); draw((36,33)--(69,33));draw((41,33)--(41,61));draw((25,36)--(25,61)); draw((34,36)--(34,45)--(25,45)); draw((36,36)--(36,38)--(34,38)); draw((36,38)--(41,38)); draw((34,45)--(41,45));[/asy]

Level 5

Geometry

Call the squares' side lengths from smallest to largest $a_1,\ldots,a_9$, and let $l,w$ represent the dimensions of the rectangle. The picture shows that\begin{align*} a_1+a_2 &= a_3\\ a_1 + a_3 &= a_4\\ a_3 + a_4 &= a_5\\ a_4 + a_5 &= a_6\\ a_2 + a_3 + a_5 &= a_7\\ a_2 + a_7 &= a_8\\ a_1 + a_4 + a_6 &= a_9\\ a_6 + a_9 &= a_7 + a_8.\end{align*} Expressing all terms 3 to 9 in terms of $a_1$ and $a_2$ and substituting their expanded forms into the previous equation will give the expression $5a_1 = 2a_2$. We can guess that $a_1 = 2$. (If we started with $a_1$ odd, the resulting sides would not be integers and we would need to scale up by a factor of $2$ to make them integers; if we started with $a_1 > 2$ even, the resulting dimensions would not be relatively prime and we would need to scale down.) Then solving gives $a_9 = 36$, $a_6=25$, $a_8 = 33$, which gives us $l=61,w=69$. These numbers are relatively prime, as desired. The perimeter is $2(61)+2(69)=\boxed{260}$.

Real numbers $a$ and $b$ satisfy the equations $3^a=81^{b+2}$ and $125^b=5^{a-3}$. What is $ab$?

Level 5

Algebra

The given equations are equivalent, respectively, to \[ 3^a=3^{4(b+2)}\quad\text{and}\quad 5^{3b}=5^{a-3}. \] Therefore $a=4(b+2)$ and $3b=a-3$. The solution of this system is $a=-12$ and $b=-5$, so $ab=\boxed{60}$.

Let $n$ be the number of ordered quadruples $(x_1,x_2,x_3,x_4)$ of positive odd integers that satisfy $\sum_{i = 1}^4 x_i = 98.$ Find $\frac n{100}.$

Level 5

Counting & Probability

Define $x_i = 2y_i - 1$. Then $2\left(\sum_{i = 1}^4 y_i\right) - 4 = 98$, so $\sum_{i = 1}^4 y_i = 51$. So we want to find four natural numbers that sum up to 51; we can imagine this as trying to split up 51 on the number line into 4 ranges. This is equivalent to trying to place 3 markers on the numbers 1 through 50; thus the answer is $n = {50\choose3} = \frac{50 * 49 * 48}{3 * 2} = 19600$, and $\frac n{100} = \boxed{196}$.

In the configuration below, $\theta$ is measured in radians, $C$ is the center of the circle, $BCD$ and $ACE$ are line segments and $AB$ is tangent to the circle at $A$. [asy] defaultpen(fontsize(10pt)+linewidth(.8pt)); pair A=(0,-1), E=(0,1), C=(0,0), D=dir(10), F=dir(190), B=(-1/sin(10*pi/180))*dir(10); fill(Arc((0,0),1,10,90)--C--D--cycle,mediumgray); fill(Arc((0,0),1,190,270)--B--F--cycle,mediumgray); draw(unitcircle); draw(A--B--D^^A--E); label("$A$",A,S); label("$B$",B,W); label("$C$",C,SE); label("$\theta$",C,SW); label("$D$",D,NE); label("$E$",E,N); [/asy] A necessary and sufficient condition for the equality of the two shaded areas, given $0 < \theta < \frac{\pi}{2}$, is $\textbf{(A)}\ \tan \theta = \theta\qquad \textbf{(B)}\ \tan \theta = 2\theta\qquad \textbf{(C)}\ \tan\theta = 4\theta\qquad \textbf{(D)}\ \tan 2\theta =\theta\qquad\\ \textbf{(E)}\ \tan\frac{\theta}{2}=\theta$

Level 5

Geometry

Well, the shaded sector's area is basically $\text{(ratio of } \theta \text{ to total angle of circle)} \times \text{(total area)} = \frac{\theta}{2\pi} \cdot (\pi r^2) = \frac{\theta}{2} \cdot (AC)^2$. In addition, if you let $\angle{ACB} = \theta$, then\[\tan \theta = \frac{AB}{AC}\]\[AB = AC\tan \theta = r\tan \theta\]\[[ABC] = \frac{AB \cdot AC}{2} = \frac{r^2\tan \theta}{2}\]Then the area of that shaded thing on the left becomes\[\frac{r^2\tan \theta}{2} - \frac{\theta \cdot r^2}{2}\]We want this to be equal to the sector area so\[\frac{r^2\tan \theta}{2} - \frac{\theta \cdot r^2}{2} = \frac{\theta \cdot r^2}{2}\]\[\frac{r^2\tan \theta}{2} = \theta \cdot r^2\]\[\boxed{\tan \theta = 2\theta}\]

Determine the largest positive integer $n$ such that there exist positive integers $x, y, z$ so that \[ n^2 = x^2+y^2+z^2+2xy+2yz+2zx+3x+3y+3z-6 \]

Level 5

Intermediate Algebra

The given equation rewrites as $n^2 = (x+y+z+1)^2+(x+y+z+1)-8$. Writing $r = x+y+z+1$, we have $n^2 = r^2+r-8$. Clearly, one possibility is $n=r=\boxed{8}$, which is realized by $x=y=1, z=6$. On the other hand, for $r > 8$, we have $r^2 < r^2+r-8 < (r+1)^2.$

Let \[f(x) = \begin{cases} 2x^2 - 3&\text{if } x\le 2, \\ ax + 4 &\text{if } x>2. \end{cases} \]Find $a$ if the graph of $y=f(x)$ is continuous (which means the graph can be drawn without lifting your pencil from the paper).

Level 5

Algebra

If the graph of $f$ is continuous, then the graphs of the two cases must meet when $x=2,$ which (loosely speaking) is the dividing point between the two cases. Therefore, we must have $2\cdot 2^2 -3 = 2a + 4.$ Solving this equation gives $a = \boxed{\frac{1}{2}}.$

If $k$ and $\ell$ are positive 4-digit integers such that $\gcd(k,\ell)=3$, what is the smallest possible value for $\mathop{\text{lcm}}[k,\ell]$?

Level 5

Number Theory

The identity $\gcd(k,\ell)\cdot\mathop{\text{lcm}}[k,\ell] = k\ell$ holds for all positive integers $k$ and $\ell$. Thus, we have $$\mathop{\text{lcm}}[k,\ell] = \frac{k\ell}{3}.$$Also, $k$ and $\ell$ must be 4-digit multiples of $3$, so our choices for each are $$1002,1005,1008,1011,1014,\ldots,$$and by minimizing the product $k\ell$, we minimize the least common multiple of $k$ and $\ell$. However, $k$ and $\ell$ cannot both be $1002$, since their greatest common divisor would then be $1002$ (not $3$). Setting $k=1002$ and $\ell=1005$, we obtain $\gcd(k,\ell)=3$ as desired, and we obtain the smallest possible value for the least common multiple: \begin{align*} \mathop{\text{lcm}}[1002,1005] &= \frac{1002\cdot 1005}{3} \\ &= 1002\cdot 335 \\ &= (1000\cdot 335)+(2\cdot 335)\\ &= \boxed{335{,}670}. \end{align*}

Find all real numbers $k$ for which there exists a nonzero, 2-dimensional vector $\mathbf{v}$ such that \[\begin{pmatrix} 1 & 8 \\ 2 & 1 \end{pmatrix} \mathbf{v} = k \mathbf{v}.\]Enter all the solutions, separated by commas.

Level 5

Precalculus

Let $\mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix}$. Then \[\begin{pmatrix} 1 & 8 \\ 2 & 1 \end{pmatrix} \mathbf{v} = \begin{pmatrix} 1 & 8 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x + 8y \\ 2x + y \end{pmatrix},\]and \[k \mathbf{v} = k \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} kx \\ ky \end{pmatrix}.\]Thus, we want $k$, $x$, and $y$ to satisfy \begin{align*} x + 8y &= kx, \\ 2x + y &= ky. \end{align*}From the first equation, $(k - 1) x = 8y$. If $x = 0$, then this equation implies $y = 0$. But the vector $\mathbf{v}$ is nonzero, so $x$ is nonzero. From the second equation, $2x = (k - 1) y$. Similarly, if $y = 0$, then this equation implies $x = 0$, so $y$ is nonzero. We also see that $k \neq 1$, because if $k = 1$, then $y = 0$, which again implies $x = 0$. Hence, we can write \[\frac{x}{y} = \frac{8}{k - 1} = \frac{k - 1}{2}.\]Cross-multiplying, we get $(k - 1)^2 = 16$. Then $k - 1 = \pm 4.$ Therefore, $k = \boxed{5}$ or $k = \boxed{-3}$. To make sure that these values of $k$ work, we should check if the corresponding vector $\mathbf{v}$ exists. For $k = 5$, we can take $\mathbf{v} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$, and for $k = -3$, we can take $\mathbf{v} = \begin{pmatrix} -2 \\ 1 \end{pmatrix}$, so both values of $k$ are possible.