Answer: 2.0

How many solutions does the equation $\sin \left( \frac{\pi}2 \cos x\right)=\cos \left( \frac{\pi}2 \sin x\right)$ have in the closed interval $[0,\pi]$
$\textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3\qquad \textbf{(E) }4$

Solution: The ranges of $\frac{\pi}2 \sin x$ and $\frac{\pi}2 \cos x$ are both $\left[-\frac{\pi}2, \frac{\pi}2 \right],$ which is included in the range of $\arcsin,$ so we can use it with no issues. \begin{align*} \frac{\pi}2 \cos x &= \arcsin \left( \cos \left( \frac{\pi}2 \sin x\right)\right) \\ \frac{\pi}2 \cos x &= \arcsin \left( \sin \left( \frac{\pi}2 - \frac{\pi}2 \sin x\right)\right) \\ \frac{\pi}2 \cos x &= \frac{\pi}2 - \frac{\pi}2 \sin x \\ \cos x &= 1 - \sin x \\ \cos x + \sin x &= 1. \end{align*} This only happens at $x = 0, \frac{\pi}2$ on the interval $[0,\pi],$ because one of $\sin$ and $\cos$ must be $1$ and the other $0.$ Therefore, the answer is $\boxed{2}.$

Solution:
By the Cofunction Identity $\cos\theta=\sin\left(\frac{\pi}{2}-\theta\right),$ we rewrite the given equation: \[\sin \left(\frac{\pi}2 \cos x\right) = \sin \left(\frac{\pi}2 - \frac{\pi}2 \sin x\right).\] Recall that if $\sin\theta=\sin\phi,$ then $\theta=\phi+2n\pi$ or $\theta=\pi-\phi+2n\pi$ for some integer $n.$ Therefore, we have two cases:
Together, we obtain $\boxed{2}.$

Solution:
This problem is equivalent to counting the intersections of the graphs of $y=\sin\left(\frac{\pi}{2}\cos x\right)$ and $y=\cos\left(\frac{\pi}{2}\sin x\right)$ in the closed interval $[0,\pi].$ We construct a table of values, as shown below: \[\begin{array}{c|ccc} & & & \\ [-2ex] & \boldsymbol{x=0} & \boldsymbol{x=\frac{\pi}{2}} & \boldsymbol{x=\pi} \\ [1.5ex] \hline & & & \\ [-1ex] \boldsymbol{\cos x} & 1 & 0 & -1 \\ [1.5ex] \boldsymbol{\frac{\pi}{2}\cos x} & \frac{\pi}{2} & 0 & -\frac{\pi}{2} \\ [1.5ex] \boldsymbol{\sin\left(\frac{\pi}{2}\cos x\right)} & 1 & 0 & -1 \\ [1.5ex] \hline  & & & \\ [-1ex] \boldsymbol{\sin x} & 0 & 1 & 0 \\ [1.5ex] \boldsymbol{\frac{\pi}{2}\sin x} & 0 & \frac{\pi}{2} & 0 \\ [1.5ex] \boldsymbol{\cos\left(\frac{\pi}{2}\sin x\right)} & 1 & 0 & 1 \\ [1ex] \end{array}\] For $x\in[0,\pi],$ note that:
For the graphs to intersect, we need $\sin\left(\frac{\pi}{2}\cos x\right)\in[0,1].$ This occurs when $\frac{\pi}{2}\cos x\in\left[0,\frac{\pi}{2}\right].$
By the Cofunction Identity $\cos\theta=\sin\left(\frac{\pi}{2}-\theta\right),$ we rewrite the given equation: \[\sin\left(\frac{\pi}{2}\cos x\right) = \sin\left(\frac{\pi}{2}-\frac{\pi}{2}\sin x\right).\] Since $\frac{\pi}{2}\cos x\in\left[0,\frac{\pi}{2}\right]$ and $\frac{\pi}{2}\sin x\in\left[0,\frac{\pi}{2}\right],$ it follows that $x\in\left[0,\frac{\pi}{2}\right]$ and $\frac{\pi}{2}-\frac{\pi}{2}\sin x\in\left[0,\frac{\pi}{2}\right].$
We can apply the arcsine function to both sides, then rearrange and simplify: \begin{align*} \frac{\pi}{2}\cos x &= \frac{\pi}{2}-\frac{\pi}{2}\sin x \\ \sin x + \cos x &= 1. \end{align*} From Case 1 in Solution 2, we conclude that $(0,1)$ and $\left(\frac{\pi}{2},0\right)$ are the only points of intersection, as shown below:  Therefore, the answer is $\boxed{2}.$


Answer: 5.0

How many solutions does the equation $\tan(2x)=\cos(\tfrac{x}{2})$ have on the interval $[0,2\pi]?$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution:
We count the intersections of the graphs of $y=\tan(2x)$ and $y=\cos\left(\frac x2\right):$
The graphs of $y=\tan(2x)$ and $y=\cos\left(\frac x2\right)$ intersect once on each of the five branches of $y=\tan(2x),$ as shown below:  Therefore, the answer is $\boxed{5}.$


Answer: 50.0

How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line $y=\pi x$ , the line $y=-0.1$ and the line $x=5.1?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 41 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 57$

Solution:
The region is a right triangle which contains the following lattice points: $(0,0); (1,0)\rightarrow(1,3); (2,0)\rightarrow(2,6); (3,0)\rightarrow(3,9); (4,0)\rightarrow(4,12); (5,0)\rightarrow(5,15)$

Squares $1\times 1$ :
Suppose that the top-right corner is $(x,y)$ , with $2\le x\le 5$ . Then to include all other corners, we need $1\le y\le 3(x-1)$ .
This produces $3+6+9+12=30$ squares.
Squares $2\times 2$ :
Here $3\le x\le 5$ . To include all other corners, we need $2\le y\le 3(x-2)$ .
This produces $2+5+8=15$ squares.
Squares $3\times 3$ :
Similarly, this produces $5$ squares.
No other squares will fit in the region. Therefore the answer is $\boxed{50}$

Solution:
The vertical line is just to the right of $x = 5$ , the horizontal line is just under $y = 0$ , and the sloped line will always be above the $y$ value of $3x$ .
This means they will always miss being on a coordinate with integer coordinates so you just have to count the number of squares to the left, above, and under these lines. After counting the number of $1\cdot1$ $2\cdot2$ , and $3\cdot3$ squares and getting $30$ $15$ , and $5$ respectively, and we end up with $\boxed{50}$

Solution:
The endpoint lattice points are $(1,3), (2,6), (3,9), (4,12), (5,15).$ Now we split this problem into cases.
Case 1: Square has length $\bf1$
The $x$ coordinates must be $(1,2)$ or $(2,3)$ and so on to $(4,5).$ The idea is that you start at $y=1$ and add at the endpoint, namely $y=3.$ The number ends up being $3+6+9+12 = 30$ squares for this case.
Case 2: Square has length $\bf2$
The $x$ coordinates must be $(1,3)$ or $(2,4)$ or $(3,15)$ and so now it starts at $y=2.$ It ends up being $2+5+8 = 15.$
Case 3: Square has length $\bf3$
The $x$ coordinates must be $(1,4)$ or $(2,5)$ so there is $1+4 = 5$ squares for this case.
Therefore, $30+15+5 = \boxed{50}$


Answer: 50.0

Solution:
Solution by e_power_pi_times_i
Revised by Kinglogic and RJ5303707
 (red shows lattice points within the triangle)
If we draw a picture showing the triangle, we see that it would be easier to count the squares vertically and not horizontally. The upper bound is $16$ squares $(y=5.1*\pi)$ , and the limit for the $x$ -value is $5$ squares. First we count the $1*1$ squares. In the rightmost column, there are $12$ squares with length $1$ because $y=4*\pi$ generates squares from $(4,0)$ to $(4,4\pi)$ , and continuing on we have $9$ $6$ , and $3$ for $x$ -values for $1$ $2$ , and $3$ in the equation $y=\pi x$ . So there are $12+9+6+3 = 30$ squares with length $1$ in the figure.
For $2*2$ squares, each square takes up $2$ units left and $2$ units up. Squares can also overlap. For $2*2$ squares, the rightmost column stretches from $(3,0)$ to $(3,3\pi)$ , so there are $8$ squares with length $2$ in a $2$ by $9$ box. Repeating the process, the next column stretches from $(2,0)$ to $(2,2\pi)$ , so there are $5$ squares. Continuing and adding up in the end, there are $8+5+2=15$ squares with length $2$ in the figure.
Squares with length $3$ in the rightmost column start at $(2,0)$ and end at $(2,2\pi)$ , so there are $4$ such squares in the right column. As the left row starts at $(1,0)$ and ends at $(1,\pi)$ there are $4+1=5$ squares with length $3$ . As squares with length $4$ would not fit in the triangle, the answer is $30+15+5$ which is $\boxed{50}$


Answer: 1296.0

How many strings of length $5$ formed from the digits $0$ $1$ $2$ $3$ $4$ are there such that for each $j \in \{1,2,3,4\}$ , at least $j$ of the digits are less than $j$ ? (For example, $02214$ satisfies this condition
because it contains at least $1$ digit less than $1$ , at least $2$ digits less than $2$ , at least $3$ digits less
than $3$ , and at least $4$ digits less than $4$ . The string $23404$ does not satisfy the condition because it
does not contain at least $2$ digits less than $2$ .)
$\textbf{(A) }500\qquad\textbf{(B) }625\qquad\textbf{(C) }1089\qquad\textbf{(D) }1199\qquad\textbf{(E) }1296$

Solution:
For some $n$ , let there be $n+1$ parking spaces counterclockwise in a circle. Consider a string of $n$ integers $c_1c_2 \ldots c_n$ each between $0$ and $n$ , and let $n$ cars come into this circle so that the $i$ th car tries to park at spot $c_i$ , but if it is already taken then it instead keeps going counterclockwise and takes the next available spot. After this process, exactly one spot will remain empty.
Then the strings of $n$ numbers between $0$ and $n-1$ that contain at least $k$ integers $<k$ for $1 \leq k \leq n$ are exactly the set of strings that leave spot $n$ empty. Also note for any string $c_1c_2 \ldots c_n$ , we can add $1$ to each $c_i$ (mod $n+1$ ) to shift the empty spot counterclockwise, meaning for each string there exists exactly one $j$ with $0 \leq j \leq n$ so that $(c_1+j)(c_2+j) \ldots (c_n+j)$ leaves spot $n$ empty. This gives there are $\frac{(n+1)^{n}}{n+1} = (n+1)^{n-1}$ such strings.
Plugging in $n = 5$ gives $\boxed{1296}$ such strings.

Solution:
Note that a valid string must have at least one $0.$
We perform casework on the number of different digits such strings can have. For each string, we list the digits in ascending order, then consider permutations:
Together, the answer is $1+75+500+600+120=\boxed{1296}.$

Solution:
Denote by $N \left( p, q \right)$ the number of $p$ -digit strings formed by using numbers $0, 1, \cdots, q$ , where for each $j \in \{1,2, \cdots , q\}$ , at least $j$ of the digits are less than $j$
We have the following recursive equation: \[N \left( p, q \right) = \sum_{i = 0}^{p - q} \binom{p}{i} N \left( p - i, q - 1 \right) , \ \forall \ p \geq q \mbox{ and } q \geq 1\] and the boundary condition $N \left( p, 0 \right) = 1$ for any $p \geq 0$
By solving this recursive equation, for $q = 1$ and $p \geq q$ , we get \begin{align*} N \left( p , 1 \right) & = \sum_{i = 0}^{p - 1} \binom{p}{i} N \left( p - i, 0 \right) \\ & = \sum_{i = 0}^{p - 1} \binom{p}{i} \\ & = \sum_{i = 0}^p \binom{p}{i} - \binom{p}{p} \\ & = 2^p - 1 . \end{align*}
For $q = 2$ and $p \geq q$ , we get \begin{align*} N \left( p , 2 \right) & = \sum_{i = 0}^{p - 2} \binom{p}{i} N \left( p - i, 1 \right) \\ & = \sum_{i = 0}^{p - 2} \binom{p}{i} \left( 2^{p - i} - 1 \right) \\ & = \sum_{i = 0}^p \binom{p}{i} \left( 2^{p - i} - 1 \right) - \sum_{i = p - 1}^p \binom{p}{i} \left( 2^{p - i} - 1 \right) \\ & = \sum_{i = 0}^p \left( \binom{p}{i} 1^i 2^{p - i} - \binom{p}{i} 1^i 1^{p - i} \right) - p \\ & = \left( 1 + 2 \right)^p - \left( 1 + 1 \right)^p - p \\ & = 3^p - 2^p - p . \end{align*}
For $q = 3$ and $p \geq q$ , we get \begin{align*} N \left( p , 3 \right) & = \sum_{i = 0}^{p - 3} \binom{p}{i} N \left( p - i, 2 \right) \\ & = \sum_{i = 0}^{p - 3} \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) \\ & = \sum_{i = 0}^p \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) - \sum_{i = p - 2}^p \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) \\ & = \sum_{i = 0}^p \left( \binom{p}{i} 1^i 3^{p - i} - \binom{p}{i} 1^i 2^{p - i} - \binom{p}{i} \left( p - i \right) \right) - \frac{3}{2} p \left( p - 1 \right) \\ & = \left( 1 + 3 \right)^p - \left( 1 + 2 \right)^p - \frac{d \left( 1 + x \right)^p}{dx} \bigg|_{x = 1} - \frac{3}{2} p \left( p - 1 \right) \\ & = 4^p - 3^p - 2^{p-1} p - \frac{3}{2} p \left( p - 1 \right) . \end{align*}
For $q = 4$ and $p = 5$ , we get \begin{align*} N \left( 5 , 4 \right) & = \sum_{i = 0}^1 \binom{5}{i} N \left( 5 - i , 3 \right) \\ & = N \left( 5 , 3 \right) + 5 N \left( 4 , 3 \right) \\ & = \boxed{1296}

Solution: The number of strings is $(n+1)^{(n-1)}$ as shown by Solution 1 (Parking Function), which is always equivalent to 1 (mod n). Thus you can choose $\boxed{1296}$

Solution:
Solution 4 tried to observe the answer modulo $5$ to easily solve the problem, but apparently had faulty logic. This solution is still completely viable though:
Notice that for any valid set $\{a_1, a_2, a_3, a_4, a_5\}$ , if there is at least one element in the set that is unique (i.e. there is at least one digit in the set that is found nowhere else in the set), then the number of distinct permutations of the set is clearly divisible by $5$ . Therefore to evaluate the answer modulo $5$ , we only need to look at sets where each element has a multiplicity of at least $2$ (i.e. appears twice or more in the set).
These sets are of the form $\{a,a,b,b,b\}$ and $\{a,a,a,a,a\}$ . The first set can be permuted in $\binom{5}{2,3}=10 \equiv 0\pmod{5}$ , and the second set can be permuted one way, and the only set of the form $\{a,a,a,a,a\}$ is $\{0,0,0,0,0\}$ . Therefore the answer is congruent to $1\pmod{5}$ and you CAN choose $\boxed{1296}$

Solution:
We list out all possibilities for the string, sorted in increasing order. Note that 1 cannot appear before the 2nd position, 2 cannot appear before the 3rd position, and so on. We also make a list of the number of possible permutations of each sorted string.
00000 - 1
00001 - 5
00002 - 5
00003 - 5
00004 - 5
00011 - 10
00012 - 20
00013 - 20
00014 - 20
00022 - 10
00023 - 20
00024 - 20
00033 - 10
00034 - 20
00111 - 10
00112 - 30
00113 - 30
00114 - 30
00122 - 30
00123 - 60
00124 - 60
00133 - 30
00134 - 60
00222 - 10
00223 - 30
00224 - 30
00233 - 30
00234 - 60
01111 - 5
01112 - 20
01113 - 20
01114 - 20
01122 - 30
01123 - 60
01124 - 60
01133 - 30
01134 - 60
01222 - 20
01223 - 60
01224 - 60
01233 - 60
01234 - 120
Finally, adding all of these up, we obtain $1\cdot 1+5\cdot 5+5\cdot 10+10\cdot 20+10\cdot 30 + 10\cdot 60 + 1\cdot 120 = \boxed{1296}$ . (Or alternatively, we could notice as in solution 5 that the only possible string with only 1 permutation is 00000, and thus the answer must be congruent to 1 modulo 5)


Answer: 1296.0

Solution:
For some $n$ , let there be $n+1$ parking spaces counterclockwise in a circle. Consider a string of $n$ integers $c_1c_2 \ldots c_n$ each between $0$ and $n$ , and let $n$ cars come into this circle so that the $i$ th car tries to park at spot $c_i$ , but if it is already taken then it instead keeps going counterclockwise and takes the next available spot. After this process, exactly one spot will remain empty.
Then the strings of $n$ numbers between $0$ and $n-1$ that contain at least $k$ integers $<k$ for $1 \leq k \leq n$ are exactly the set of strings that leave spot $n$ empty. Also note for any string $c_1c_2 \ldots c_n$ , we can add $1$ to each $c_i$ (mod $n+1$ ) to shift the empty spot counterclockwise, meaning for each string there exists exactly one $j$ with $0 \leq j \leq n$ so that $(c_1+j)(c_2+j) \ldots (c_n+j)$ leaves spot $n$ empty. This gives there are $\frac{(n+1)^{n}}{n+1} = (n+1)^{n-1}$ such strings.
Plugging in $n = 5$ gives $\boxed{1296}$ such strings.

Solution:
Note that a valid string must have at least one $0.$
We perform casework on the number of different digits such strings can have. For each string, we list the digits in ascending order, then consider permutations:
Together, the answer is $1+75+500+600+120=\boxed{1296}.$

Solution:
Denote by $N \left( p, q \right)$ the number of $p$ -digit strings formed by using numbers $0, 1, \cdots, q$ , where for each $j \in \{1,2, \cdots , q\}$ , at least $j$ of the digits are less than $j$
We have the following recursive equation: \[N \left( p, q \right) = \sum_{i = 0}^{p - q} \binom{p}{i} N \left( p - i, q - 1 \right) , \ \forall \ p \geq q \mbox{ and } q \geq 1\] and the boundary condition $N \left( p, 0 \right) = 1$ for any $p \geq 0$
By solving this recursive equation, for $q = 1$ and $p \geq q$ , we get \begin{align*} N \left( p , 1 \right) & = \sum_{i = 0}^{p - 1} \binom{p}{i} N \left( p - i, 0 \right) \\ & = \sum_{i = 0}^{p - 1} \binom{p}{i} \\ & = \sum_{i = 0}^p \binom{p}{i} - \binom{p}{p} \\ & = 2^p - 1 . \end{align*}
For $q = 2$ and $p \geq q$ , we get \begin{align*} N \left( p , 2 \right) & = \sum_{i = 0}^{p - 2} \binom{p}{i} N \left( p - i, 1 \right) \\ & = \sum_{i = 0}^{p - 2} \binom{p}{i} \left( 2^{p - i} - 1 \right) \\ & = \sum_{i = 0}^p \binom{p}{i} \left( 2^{p - i} - 1 \right) - \sum_{i = p - 1}^p \binom{p}{i} \left( 2^{p - i} - 1 \right) \\ & = \sum_{i = 0}^p \left( \binom{p}{i} 1^i 2^{p - i} - \binom{p}{i} 1^i 1^{p - i} \right) - p \\ & = \left( 1 + 2 \right)^p - \left( 1 + 1 \right)^p - p \\ & = 3^p - 2^p - p . \end{align*}
For $q = 3$ and $p \geq q$ , we get \begin{align*} N \left( p , 3 \right) & = \sum_{i = 0}^{p - 3} \binom{p}{i} N \left( p - i, 2 \right) \\ & = \sum_{i = 0}^{p - 3} \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) \\ & = \sum_{i = 0}^p \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) - \sum_{i = p - 2}^p \binom{p}{i} \left( 3^{p - i} - 2^{p - i} - \left( p - i \right) \right) \\ & = \sum_{i = 0}^p \left( \binom{p}{i} 1^i 3^{p - i} - \binom{p}{i} 1^i 2^{p - i} - \binom{p}{i} \left( p - i \right) \right) - \frac{3}{2} p \left( p - 1 \right) \\ & = \left( 1 + 3 \right)^p - \left( 1 + 2 \right)^p - \frac{d \left( 1 + x \right)^p}{dx} \bigg|_{x = 1} - \frac{3}{2} p \left( p - 1 \right) \\ & = 4^p - 3^p - 2^{p-1} p - \frac{3}{2} p \left( p - 1 \right) . \end{align*}
For $q = 4$ and $p = 5$ , we get \begin{align*} N \left( 5 , 4 \right) & = \sum_{i = 0}^1 \binom{5}{i} N \left( 5 - i , 3 \right) \\ & = N \left( 5 , 3 \right) + 5 N \left( 4 , 3 \right) \\ & = \boxed{1296}

Solution: The number of strings is $(n+1)^{(n-1)}$ as shown by Solution 1 (Parking Function), which is always equivalent to 1 (mod n). Thus you can choose $\boxed{1296}$

Solution:
Solution 4 tried to observe the answer modulo $5$ to easily solve the problem, but apparently had faulty logic. This solution is still completely viable though:
Notice that for any valid set $\{a_1, a_2, a_3, a_4, a_5\}$ , if there is at least one element in the set that is unique (i.e. there is at least one digit in the set that is found nowhere else in the set), then the number of distinct permutations of the set is clearly divisible by $5$ . Therefore to evaluate the answer modulo $5$ , we only need to look at sets where each element has a multiplicity of at least $2$ (i.e. appears twice or more in the set).
These sets are of the form $\{a,a,b,b,b\}$ and $\{a,a,a,a,a\}$ . The first set can be permuted in $\binom{5}{2,3}=10 \equiv 0\pmod{5}$ , and the second set can be permuted one way, and the only set of the form $\{a,a,a,a,a\}$ is $\{0,0,0,0,0\}$ . Therefore the answer is congruent to $1\pmod{5}$ and you CAN choose $\boxed{1296}$

Solution:
We list out all possibilities for the string, sorted in increasing order. Note that 1 cannot appear before the 2nd position, 2 cannot appear before the 3rd position, and so on. We also make a list of the number of possible permutations of each sorted string.
00000 - 1
00001 - 5
00002 - 5
00003 - 5
00004 - 5
00011 - 10
00012 - 20
00013 - 20
00014 - 20
00022 - 10
00023 - 20
00024 - 20
00033 - 10
00034 - 20
00111 - 10
00112 - 30
00113 - 30
00114 - 30
00122 - 30
00123 - 60
00124 - 60
00133 - 30
00134 - 60
00222 - 10
00223 - 30
00224 - 30
00233 - 30
00234 - 60
01111 - 5
01112 - 20
01113 - 20
01114 - 20
01122 - 30
01123 - 60
01124 - 60
01133 - 30
01134 - 60
01222 - 20
01223 - 60
01224 - 60
01233 - 60
01234 - 120
Finally, adding all of these up, we obtain $1\cdot 1+5\cdot 5+5\cdot 10+10\cdot 20+10\cdot 30 + 10\cdot 60 + 1\cdot 120 = \boxed{1296}$ . (Or alternatively, we could notice as in solution 5 that the only possible string with only 1 permutation is 00000, and thus the answer must be congruent to 1 modulo 5)


Answer: 12.0

How many subsets containing three different numbers can be selected from the set \[\{ 89,95,99,132, 166,173 \}\] so that the sum of the three numbers is even?
$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 24$

Solution:
To have an even sum with three numbers, we must add either $E+O+O$ , or $E + E + E$ , where $O$ represents an odd number, and $E$ represents an even number.
Since there are not three even numbers in the given set, $E+E+E$ is impossible.  Thus, we must choose two odd numbers, and one even number.
There are $2$ choices for the even number.
There are $4$ choices for the first odd number.  There are $3$ choices for the last odd number.  But the order in which we pick these 2 numbers doesn't matter, so this overcounts the pairs of odd numbers by a factor of $2$ .   Thus, we have $\frac{4\cdot 3}{2} = 6$ choices for a pair of odd numbers.
In total, there are $2$ choices for an even number, and $6$ choices for the odd numbers, giving a total of $2\cdot 6 = 12$ possible choices for a 3-element set that has an even sum.  This is option $\boxed{12}$


Answer: 240.0

How many subsets of $\{2,3,4,5,6,7,8,9\}$ contain at least one prime number?
$\textbf{(A) } 128 \qquad \textbf{(B) } 192 \qquad \textbf{(C) } 224 \qquad \textbf{(D) } 240 \qquad \textbf{(E) } 256$

Solution: Since an element of a subset is either in or out, the total number of subsets of the $8$ -element set is $2^8 = 256$ . However, since we are only concerned about the subsets with at least $1$ prime in it, we can use complementary counting to count the subsets without a prime and subtract that from the total. Because there are $4$ non-primes, there are $2^8 -2^4 = \boxed{240}$ subsets with at least $1$ prime.

Solution: We can construct our subset by choosing which primes are included and which composites are included. There are $2^4-1$ ways to select the primes (total subsets minus the empty set) and $2^4$ ways to select the composites. Thus, there are $15\cdot16=\boxed{240}$ ways to choose a subset of the eight numbers.

Solution: We complement count. We know that we have $2^8$ subsets in all, including the empty subset. We subtract $\dbinom{4}{0}+\dbinom{4}{1}+\dbinom{4}{2}+\dbinom{4}{3}+\dbinom{4}{4} = 2^4 = 16$ We have $2^8-16 = 256-16 = \boxed{240}$ ways to choose a subset of the eight numbers that include a prime number. ~hh99754539


Answer: 240.0

How many subsets of $\{2,3,4,5,6,7,8,9\}$ contain at least one prime number?
$\textbf{(A)} \text{ 128} \qquad \textbf{(B)} \text{ 192} \qquad \textbf{(C)} \text{ 224} \qquad \textbf{(D)} \text{ 240} \qquad \textbf{(E)} \text{ 256}$

Solution:
Well, there are 4 composite numbers, and you can list them in a 1 number format, a 2 number, 3 number, and a 4 number format. Now, we can use combinations.
$\binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4} = 15$ . Using the answer choices, the only multiple of 15 is $\boxed{240}$

Solution:
Subsets of $\{2,3,4,5,6,7,8,9\}$ include a single digit up to all eight numbers. Therefore, we must add the combinations of all possible subsets and subtract from each of the subsets formed by the composite numbers.
Hence:
$\binom{8}{1} - \binom{4}{1} + \binom{8}{2} - \binom{4}{2} + \binom{8}{3} - \binom{4}{3} + \binom{8}{4} - 1 + \binom{8}{5} + \binom{8}{6} + \binom{8}{7} + 1 = \boxed{240}$

Solution:
Total subsets is $(2^8) = 256$ Using complementary counting and finding the sets with composite numbers:
only 4,6,8 and 9 are composite. Each one can be either in the set or out: $2^4$ = 16 $256-16=240$ $\boxed{240}$

Solution:
We multiply the number of possibilities of the set having prime numbers and the set having composites.
The possibilities of primes are $2^4-1=15$ (As there is one solution not containing any primes)
The possibilities of the set containing composites are $2^4=16$ (There can be a set with no composites)
Multiplying this we get $15 \cdot 16 = \boxed{240}$


Answer: 5.0

How many subsets of two elements can be removed from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$ so that the mean (average) of the remaining numbers is 6?
$\textbf{(A)}\text{ 1}\qquad\textbf{(B)}\text{ 2}\qquad\textbf{(C)}\text{ 3}\qquad\textbf{(D)}\text{ 5}\qquad\textbf{(E)}\text{ 6}$

Solution: Since there will be $9$ elements after removal, and their mean is $6$ , we know their sum is $54$ . We also know that the sum of the set pre-removal is $66$ . Thus, the sum of the $2$ elements removed is $66-54=12$ . There are only $\boxed{5}$

Solution: We can simply remove $5$ subsets of $2$ numbers while leaving only $6$ behind. The average of this one-number set is still $6$ , so the answer is $\boxed{5}$


Answer: 21.0

How many terms are in the arithmetic sequence $13$ $16$ $19$ $\dotsc$ $70$ $73$
$\textbf{(A)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61$

Solution:
$73-13 = 60$ , so the amount of terms in the sequence $13$ $16$ $19$ $\dotsc$ $70$ $73$ is the same as in the sequence $0$ $3$ $6$ $\dotsc$ $57$ $60$
In this sequence, the terms are the multiples of $3$ going up to $60$ , and there are $20$ multiples of $3$ in $60$
However, the number 0 must also be included, adding another multiple. So, the answer is $\boxed{21}$

Solution: Using the formula for arithmetic sequence's nth term, we see that $a + (n-1)d \Longrightarrow13 + (n-1)3 =73, \Longrightarrow n = 21$ $\boxed{21}$

Solution:
Minus each of the terms by $12$ to make the the sequence $1 , 4 , 7,..., 61$ $\frac{61-1}{3}=20, 20 + 1 = 21$
$\boxed{21}$

Solution: Subtract each of the terms by $10$ to make the sequence $3 , 6 , 9,..., 60, 63$ . Then divide the each term in the sequence by $3$ to get $1, 2, 3,..., 20, 21$ . Now it is clear to see that there are $21$ terms in the sequence. $\boxed{21}$


Answer: 69.0

How many three-digit numbers are divisible by 13?
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 67\qquad\textbf{(C)}\ 69\qquad\textbf{(D)}\ 76\qquad\textbf{(E)}\ 77$

Solution: Let $k$ be any positive integer so that $13k$ is a multiple of $13$ . For the smallest three-digit number, $13k>100$ and $k>\frac{100}{13} \approx 7.7$ . For the greatest three-digit number, $13k<999$ and $k<\frac{999}{13} \approx 76.8$ . The number $k$ can range from $8$ to $76$ so there are $\boxed{69}$ three-digit numbers.


Answer: 60.0

How many three-digit numbers are not divisible by $5$ , have digits that sum to less than $20$ , and have the first digit equal to the third digit?
$\textbf{(A)}\ 52 \qquad\textbf{(B)}\ 60  \qquad\textbf{(C)}\ 66 \qquad\textbf{(D)}\ 68 \qquad\textbf{(E)}\ 70$

Solution:
We use a casework approach to solve the problem. These three digit numbers are of the form $\overline{xyx}$ .( $\overline{abc}$ denotes the number $100a+10b+c$ ).  We see that $x\neq 0$ and $x\neq 5$ , as $x=0$ does not yield a three-digit integer and $x=5$ yields a number divisible by 5.
The second condition is that the sum $2x+y<20$ .  When $x$ is $1$ $2$ $3$ , or $4$ $y$ can be any digit from $0$ to $9$ , as $2x<10$ .  This yields $10(4) = 40$ numbers.
When $x=6$ , we see that $12+y<20$ so $y<8$ .  This yields $8$ more numbers.
When $x=7$ $14+y<20$ so $y<6$ .  This yields $6$ more numbers.
When $x=8$ $16+y<20$ so $y<4$ .  This yields $4$ more numbers.
When $x=9$ $18+y<20$ so $y<2$ .  This yields $2$ more numbers.
Summing, we get $40 + 8 + 6 + 4 + 2 = \boxed{60}$


Answer: 52.0

How many three-digit numbers have at least one $2$ and at least one $3$
$\text{(A) }52 \qquad \text{(B) }54  \qquad \text{(C) }56 \qquad \text{(D) }58 \qquad \text{(E) }60$

Solution:
We can do this problem with some simple case work.
Case 1: The hundreds place is not $2$ or $3.$ This means that the tens place and ones place must be $2$ and $3$ respectively or $3$ and $2$ respectively. This case covers $1, 4, 5, 6, 7, 8,$ and $9,$ so it gives us $2 \cdot 7 = 14$ cases.
Case 2: The hundreds place is $2.$ This means that $3$ must be in the tens place or ones place. Starting with cases in which the tens place is not $3$ , we get $203, 213, 223, 243, 253, 263, 273, 283,$ and $293.$ With cases in which the tens place is $3$ , we have $230-239$ , or $10$ more cases. This gives us $9 + 10=19$ cases.
Case 3: The hundreds place is $3.$ This case is almost identical to the second case, just swap the $2$ s with $3$ s and $3$ s with $2$ s in the reasoning and its the same, giving us an additional $19$ cases.
Addition up all of these cases gives $14+19+19=52$ cases, or $\boxed{52}.$


Answer: 45.0

How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?
$\textbf{(A) } 41\qquad \textbf{(B) } 42\qquad \textbf{(C) } 43\qquad \textbf{(D) } 44\qquad \textbf{(E) } 45$

Solution:
If the middle digit is the average of the first and last digits, twice the middle digit must be equal to the sum of the first and last digits.
Doing some casework
If the middle digit is $1$ , possible numbers range from $111$ to $210$ . So there are $2$ numbers in this case.
If the middle digit is $2$ , possible numbers range from $123$ to $420$ . So there are $4$ numbers in this case.
If the middle digit is $3$ , possible numbers range from $135$ to $630$ . So there are $6$ numbers in this case.
If the middle digit is $4$ , possible numbers range from $147$ to $840$ . So there are $8$ numbers in this case.
If the middle digit is $5$ , possible numbers range from $159$ to $951$ . So there are $9$ numbers in this case.
If the middle digit is $6$ , possible numbers range from $369$ to $963$ . So there are $7$ numbers in this case.
If the middle digit is $7$ , possible numbers range from $579$ to $975$ . So there are $5$ numbers in this case.
If the middle digit is $8$ , possible numbers range from $789$ to $987$ . So there are $3$ numbers in this case.
If the middle digit is $9$ , the only possible number is $999$ . So there is $1$ number in this case.
So the total number of three-digit numbers that satisfy the property is $2+4+6+8+9+7+5+3+1=\boxed{45}$

Solution:
Alternatively, we could note that the middle digit is uniquely defined by the first and third digits since it is half of their sum. This also means that the sum of the first and third digits must be even. Since even numbers are formed either by adding two odd numbers or two even numbers, we can split our problem into $2$ cases:
If both the first digit and the last digit are odd, then we have $1, 3, 5, 7,$ or $9$ as choices for each of these digits, and there are $5\cdot5=25$ numbers in this case.
If both the first and last digits are even, then we have $2, 4, 6, 8$ as our choices for the first digit and $0, 2, 4, 6, 8$ for the third digit. There are $4\cdot5=20$ numbers here.
The total number, then, is $20+25=\boxed{45}$

Solution: As we noted in Solution 2, we note that the sum of the first and third digits has to be even. The first digit can have $9$ possibilities $(1-9)$ , and the third digit can have $10$ possibilities $(0-9)$ . This means there can be $9\cdot10=90$ possible two-digit numbers in which the first digit and the third digit are digits. Exactly half of these would have their sum be divisible by $2$ (since $90$ is even), so our answer is $\frac{90}{2}=\boxed{45}$

Solution:
If our first digit is odd, the last digit also has to be odd for there to be a whole number middle digit that is the average of the first and the last digits.
If our first digit is even, the last digit has to be even, or $0$ for there to be a whole number middle digit that is the average of the first and the last digits.
As there are $5$ different possible values for each starting digit of our number, we can multiply that by the number of possible first digits $(1-9)$ to achieve our desired solution. $5 \cdot 9 = \boxed{45}$


Answer: 14.0

How many three-digit positive integers $N$ satisfy the following properties?
$\textbf{(A) } 13 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 16 \qquad \textbf{(E) } 17$

Solution:
Multiples of $5$ will always end in $0$ or $5$ , and since the numbers have to be a three-digit numbers (otherwise it would be a two-digit number), it cannot start with 0, narrowing our choices to 3-digit numbers starting with $5$ . Since the numbers must be divisible by 7, all possibilities have to be in the range from $7 \cdot 72$ to $7 \cdot 85$ inclusive.
$85 - 72 + 1 = 14$ $\boxed{14}$

Solution:
Let $N=\overline{cab}=100c+10a+b.$ We know that $\overline{bac}$ is divisible by $5$ , so $c$ is either $0$ or $5$ . However, since $c$ is the first digit of the three-digit number $N$ , it can not be $0$ , so therefore, $c=5$ . Thus, $N=\overline{5ab}=500+10a+b.$ There are no further restrictions on digits $a$ and $b$ aside from $N$ being divisible by $7$
The smallest possible $N$ is $504$ . The next smallest $N$ is $511$ , then $518$ , and so on, all the way up to $595$ . Thus, our set of possible $N$ is $\{504,511,518,\dots,595\}$ . Dividing by $7$ for each of the terms will not affect the cardinality of this set, so we do so and get $\{72,73,74,\dots,85\}$ . We subtract $71$ from each of the terms, again leaving the cardinality unchanged. We end up with $\{1,2,3,\cdots,14\}$ , which has a cardinality of $14$ . Therefore, our answer is $\boxed{14.}$

Solution:
We first proceed as in the above solution, up to $N=500+10a+b$ .
We then use modular arithmetic:
\begin{align*} 0&\equiv N \:(\text{mod }7)\\ &\equiv500+10a+b\:(\text{mod }7)\\ &\equiv3+3a+b\:(\text{mod }7)\\ 3a+b&\equiv-3\:(\text{mod }7)\\ &\equiv4\:(\text{mod }7)\\ \end{align*}
We know that $0\le a,b<10$ . We then look at each possible value of $a$
If $a=0$ , then $b$ must be $4$
If $a=1$ , then $b$ must be $1$ or $8$
If $a=2$ , then $b$ must be $5$
If $a=3$ , then $b$ must be $2$ or $9$
If $a=4$ , then $b$ must be $6$
If $a=5$ , then $b$ must be $3$
If $a=6$ , then $b$ must be $0$ or $7$
If $a=7$ , then $b$ must be $4$
If $a=8$ , then $b$ must be $1$ or $8$
If $a=9$ , then $b$ must be $5$
Each of these cases are unique, so there are a total of $1+2+1+2+1+1+2+1+2+1=\boxed{14.}$

Solution:
The key point is that when reversed, the number must start with a $0$ or a $5$ based on the second restriction. But numbers can't start with a $0$
So the problem is simply counting the number of multiples of $7$ in the $500$ s.
$7 \times 72 = 504$ , so the first multiple is $7 \times 72$
$7 \times 85 = 595$ , so the last multiple is $7 \times 85$
Now, we just have to count $7\times 72, 7\times 73, 7\times 74,\cdots, 7\times 85$
We have a set that numbers $85-71=\boxed{14}$


Answer: 13.0

How many three-digit positive integers $\underline{a} \ \underline{b} \ \underline{c}$ are there whose nonzero digits $a,b,$ and $c$ satisfy \[0.\overline{\underline{a}~\underline{b}~\underline{c}} = \frac{1}{3} (0.\overline{a} + 0.\overline{b} + 0.\overline{c})?\] (The bar indicates repetition, thus $0.\overline{\underline{a}~\underline{b}~\underline{c}}$ is the infinite repeating decimal $0.\underline{a}~\underline{b}~\underline{c}~\underline{a}~\underline{b}~\underline{c}~\cdots$
$\textbf{(A) } 9 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 11 \qquad \textbf{(D) } 13 \qquad \textbf{(E) } 14$

Solution:
We rewrite the given equation, then rearrange: \begin{align*} \frac{100a+10b+c}{999} &= \frac13\left(\frac a9 + \frac b9 + \frac c9\right) \\ 100a+10b+c &= 37a + 37b + 37c \\ 63a &= 27b+36c \\ 7a &= 3b+4c. \end{align*} Now, this problem is equivalent to counting the ordered triples $(a,b,c)$ that satisfies the equation.
Clearly, the $9$ ordered triples $(a,b,c)=(1,1,1),(2,2,2),\ldots,(9,9,9)$ are solutions to this equation.
The expression $3b+4c$ has the same value when:
We find $4$ more solutions from the $9$ solutions above: $(a,b,c)=(4,8,1),(5,1,8),(5,9,2),(6,2,9).$ Note that all solutions are symmetric about $(a,b,c)=(5,5,5).$
Together, we have $9+4=\boxed{13}$ ordered triples $(a,b,c).$


Answer: 450.0

How many three-digit positive integers have an odd number of even digits?
$\textbf{(A) }150\qquad\textbf{(B) }250\qquad\textbf{(C) }350\qquad\textbf{(D) }450\qquad\textbf{(E) }550$

Solution:
We use simple case work to solve this problem.
Case 1: even, even, even = $4 \cdot 5 \cdot 5 = 100$
Case 2: even, odd, odd = $4 \cdot 5 \cdot 5 = 100$
Case 3: odd, even, odd = $5 \cdot 5 \cdot 5 = 125$
Case 4: odd, odd, even = $5 \cdot 5 \cdot 5 = 125$
Simply sum up the cases to get your answer. $100 + 100 + 125 + 125 = \boxed{450}$

Solution: We will show that the answer is $450$ by proving a bijection between the three digit integers that have an even number of even digits and the three digit integers that have an odd number of even digits. For every even number with an odd number of even digits, we increment the number's last digit by $1$ , unless the last digit is $9$ , in which case it becomes $0$ . It is very easy to show that every number with an even number of even digits is mapped to every number with an odd number of even digits, and vice versa. Thus, the answer is half the number of three digit numbers, or $\boxed{450}$


Answer: 5.0

How many triangles are in this figure? (Some triangles may overlap other triangles.)
[asy] draw((0,0)--(42,0)--(14,21)--cycle); draw((14,21)--(18,0)--(30,9)); [/asy]
$\text{(A)}\ 9 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 5$

Solution:
By inspection, we have that there $5$ triangles:  Each of the $3$ small triangles, $1$ medium triangle made of the rightmost two small triangles, and the $1$ large triangle.
$\boxed{5}$


Answer: 4.0

How many triangles have area $10$ and vertices at $(-5,0),(5,0)$ and $(5\cos \theta, 5\sin \theta)$ for some angle $\theta$
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }4 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ } 8$

Solution: The triangle can be seen as having the base on the $x$ axis and height $|5\sin\theta|$ . The length of the base is $10$ , thus the height must be $2$ . The equation $|\sin\theta| = \frac 25$ has $\boxed{4}$ solutions, one in each quadrant.

Solution: Alternatively, we use shoelace to get: \[\frac {1}{2}|(-5*0+5*5\sin (\theta)+5 \cos (\theta) *0)-(0*5+0*5 \cos (\theta)-5*5\sin (\theta))|=10 \implies \frac {1}{2}|50\sin (\theta)|=10\] This means $\sin (\theta)=\pm \frac {2}{5}$ . We see that if it equals $\frac {2}{5}$ , then $\cos (\theta)=\pm \frac {\sqrt {21}}{5}$ . Likewise, we see that if $\sin (\theta)=-\frac {2}{5}$ , then $\cos (\theta)$ has $2$ solutions. Thus, there are $\boxed{4}$ unique points such that the triangle has an area of $10$ , or $C$


Answer: 2148.0

How many triangles with positive area have all their vertices at points $(i,j)$ in the coordinate plane, where $i$ and $j$ are integers between $1$ and $5$ , inclusive?
$\textbf{(A)}\ 2128 \qquad\textbf{(B)}\ 2148 \qquad\textbf{(C)}\ 2160 \qquad\textbf{(D)}\ 2200 \qquad\textbf{(E)}\ 2300$

Solution:
We can solve this by finding all the combinations, then subtracting the ones that are on the same line. There are $25$ points in all, from $(1,1)$ to $(5,5)$ , so $\dbinom{25}3$ is $\frac{25\cdot 24\cdot 23}{3\cdot 2 \cdot 1}$ , which simplifies to $2300$ .
Now we count the ones that are on the same line. We see that any three points chosen from $(1,1)$ and $(1,5)$ would be on the same line, so $\dbinom53$ is $10$ , and there are $5$ rows, $5$ columns, and $2$ long diagonals, so that results in $120$ .
We can also count the ones with $4$ on a diagonal. That is $\dbinom43$ , which is 4, and there are $4$ of those diagonals, so that results in $16$ .
We can count the ones with only $3$ on a diagonal, and there are $4$ diagonals like that, so that results in $4$ .
We can also count the ones with a slope of $\frac12$ $2$ $-\frac12$ , or $-2$ , with $3$ points in each. Note that there are $3$ such lines, for each slope, present in the grid. In total, this results in $12$ .
Finally, we subtract all the ones in a line from $2300$ , so we have $2300-120-16-4-12=\boxed{2148}$

Solution:
There are $5 \times 5 = 25$ total points in all. So, there are $\dbinom{25}3 = 2300$ ways to choose the three vertices for the triangle. However, there are some cases where they 3 points chosen results in a straight line.
There are $10 \times 10 = 100$ cases where the 3 points chosen make up a vertical or horizontal line.
There are $2\left(1+\dbinom{4}3+\dbinom{5}3+\dbinom{4}3+1\right)=40$ cases where the 3 points all land on the diagonals of the square.
There are $3 \times 4=12$ ways where the 3 points make the a slope of $\frac{1}{2}$ $-\frac{1}{2}$ $2$ , and $-2$
Hence, there are $100+40+12=152$ cases where the chosen 3 points make a line. The answer would be $2300-152=\boxed{2148}$


Answer: 17.0

How many two-digit numbers have digits whose sum is a perfect square?
$\textbf{(A)}\ 13\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 17\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 19$

Solution:
There is $1$ integer whose digits sum to $1$ $10$
There are $4$ integers whose digits sum to $4$ $13, 22, 31, \text{and } 40$
There are $9$ integers whose digits sum to $9$ $18, 27, 36, 45, 54, 63, 72, 81, \text{and } 90$
There are $3$ integers whose digits sum to $16$ $79, 88, \text{and } 97$
Two digits cannot sum to $25$ or any greater square since the greatest sum of digits of a two-digit number is $9 + 9 = 18$
Thus, the answer is $1 + 4 + 9 + 3 = \boxed{17}$


Answer: 8.0

How many two-digit positive integers $N$ have the property that the sum of $N$ and the number obtained by reversing the order of the digits of is a perfect square?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$

Solution:
Let $N = 10t + u$ , where $t$ is the tens digit and $u$ is the units digit.
The condition of the problem is that $10t + u + 10u + t$ is a perfect square.
Simplifying and factoring, we want $11(t+u)$ to be a perfect square.
Thus, $t+u$ must at least be a multiple of $11$ , and since $t$ and $u$ are digits, the only multiple of $11$ that works is $11$ itself.
Thus, $(t,u) = (2,9)$ is the first solution, and $(t,u) = (9,2)$ is the last solution.  There are $8$ solutions in total, leading to answer $\boxed{8}$


Answer: 18.0

How many two-digit positive integers have at least one $7$ as a digit?
$\mathrm{(A) \ } 10 \qquad \mathrm{(B) \ } 18\qquad \mathrm{(C) \ } 19 \qquad \mathrm{(D) \ } 20\qquad \mathrm{(E) \ } 30$

Solution:
Ten numbers $(70,71,\dots,79)$ have $7$ as the tens digit. Nine numbers $(17,27,\dots,97)$ have it as the ones digit. Number $77$ is in both sets.
Thus the result is $10+9-1=18 \Rightarrow$ $\boxed{18}$

Solution:
We use complementary counting. The complement of having at least one $7$ as a digit is having no $7$ s as a digit.
We have $9$ digits to choose from for the first digit and $10$ digits for the second. This gives a total of $9 \times 10 = 90$ two-digit numbers.
But since we cannot have $7$ as a digit, we have $8$ first digits and $9$ second digits to choose from.
Thus there are $8 \times 9 = 72$ two-digit numbers without a $7$ as a digit.
$90$ (The total number of two-digit numbers) $-  72$ (The number of two-digit numbers without a $7$ $= 18 \Rightarrow$ $\boxed{18}$


Answer: 42.0

How many unordered pairs of edges of a given cube determine a plane?
$\textbf{(A) } 12 \qquad \textbf{(B) } 28 \qquad \textbf{(C) } 36\qquad \textbf{(D) } 42 \qquad \textbf{(E) } 66$

Solution:
Without loss of generality, choose one of the $12$ edges of the cube to be among the two selected. We now calculate the probability that a randomly-selected second edge makes the pair satisfy the condition in the problem statement.
For two lines in space to determine a common plane, they must either intersect or be parallel (in other words, they cannot be skew lines). If all $12$ line segments are extended to lines, the first (arbitrarily chosen) edge's line intersects $4$ lines and is parallel to another $3$ . Thus $4+3=7$ of the $12-1=11$ remaining line segments (which could be chosen for the second edge) give a pair of lines determining a common plane. To see this, observe that in the diagram below, the red edge is parallel to the $3$ green edges and intersects with the $4$ blue edges.

This means that the probability that a randomly-selected pair of edges determine a plane is $\frac{7}{11}$ , and we calculate that there are ${12 \choose 2} = 66$ total pairs of edges that could be chosen (without the restriction). Thus the answer is $\frac{7}{11} \cdot 66 =\boxed{42}$

Solution:
As in Solution 1, we observe that the two edges must either be parallel or intersect. Clearly the edges will intersect if and only if they are part of the same face. We can thus divide into two cases:
Case 1 : The two edges are part of the same face. There are $6$ faces, and ${4 \choose 2}=6$ ways to choose $2$ of the $4$ edges of the square, giving a total of $6 \cdot 6 = 36$ possibilities.
Case 2 : The two edges are parallel and not part of the same face. Observe that each of the $12$ edges is parallel to exactly $1$ edge that is not part of its face. The edges can thus be paired up, giving $\frac{12}{2} = 6$ possibilities for this case.
Adding the two cases, the answer is hence $36 + 6 = \boxed{42}$

Solution:
As in solution 1, we see that for an arbitrary edge of the cube, only $7$ of the other edges can be chosen to form a plane.
For each of the $12$ edges, this gives $(7)(12) = 84$ total possible pairings. However each pair has been counted twice (e.g edge $A$ with $B$ , and edge $B$ with $A$ ) so we divide by 2 to get $\boxed{42}$


Answer: 6.0

How many values of $\theta$ in the interval $0<\theta\le 2\pi$ satisfy \[1-3\sin\theta+5\cos3\theta = 0?\] $\textbf{(A) }2 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5\qquad \textbf{(D) }6 \qquad \textbf{(E) }8$

Solution:
We rearrange to get \[5\cos3\theta = 3\sin\theta-1.\] We can graph two functions in this case: $y=5\cos{3x}$ and $y=3\sin{x} -1$ .
Using transformation of functions, we know that $5\cos{3x}$ is just a cosine function with amplitude $5$ and period $\frac{2\pi}{3}$ . Similarly, $3\sin{x} -1$ is just a sine function with amplitude $3$ and shifted $1$ unit downward:  So, we have $\boxed{6}$ solutions.


Answer: 432.0

How many ways are there to paint each of the integers $2, 3, \cdots , 9$ either red, green, or blue so that each number has a different color from each of its proper divisors?
$\textbf{(A)}~144\qquad\textbf{(B)}~216\qquad\textbf{(C)}~256\qquad\textbf{(D)}~384\qquad\textbf{(E)}~432$

Solution:
The $5$ and $7$ can be painted with no restrictions because the set of integers does not contain a multiple or proper factor of $5$ or $7$ . There are 3 ways to paint each, giving us $\underline{9}$ ways to paint both. The $2$ is the most restrictive number. There are $\underline{3}$ ways to paint $2$ , but without loss of generality, let it be painted red. $4$ cannot be the same color as $2$ or $8$ , so there are $\underline{2}$ ways to paint $4$ , which automatically determines the color for $8$ $6$ cannot be painted red, so there are $\underline{2}$ ways to paint $6$ , but WLOG, let it be painted blue. There are $\underline{2}$ choices for the color for $3$ , which is either red or green in this case. Lastly, there are $\underline{2}$ ways to choose the color for $9$
$9 \cdot 3 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = \boxed{432}$

Solution:
We note that the primes can be colored any of the $3$ colors since they don't have any proper divisors other than $1$ , which is not in the list. Furthermore, $6$ is the only number in the list that has $2$ distinct prime factors (namely, $2$ and $3$ ), so we do casework on $6$
Case 1 $2$ and $3$ are the same color
In this case, we have $3$ primes to choose the color for ( $2$ $5$ , and $7$ ). Afterwards, $4$ $6$ , and $9$ have two possible colors, which will determine the color of $8$ . Thus, there are $3^3\cdot 2^3=216$ possibilities here.
Case 2 $2$ and $3$ are different colors
In this case, we have $4$ primes to color. Without loss of generality, we'll color the $2$ first, then the $3$ . Then there are $3$ color choices for $2,5,7$ , and $2$ color choices for $3$ . This will determine the color of $6$ as well. After that, we only need to choose the color for $4$ and $9$ , which each have $2$ choices. Thus, there are $3^3\cdot 2^3=216$ possibilities here as well.
Adding up gives $216+216=\boxed{432}$

Solution:
$2,4,8$ require different colors each, so there are $6$ ways to color these.
$5$ and $7$ can be any color, so there are $3\times 3$ ways to color these.
$6$ can have $2$ colors once $2$ is colored, and thus $3$ also has $2$ colors following $6$ , which leaves another $2$ for $9$
All together: $6\times 3 \times 3 \times 2 \times 2 \times 2 = 432 \Rightarrow \boxed{432}$


Answer: 36.0

How many ways are there to place $3$ indistinguishable red chips, $3$ indistinguishable blue chips, and $3$ indistinguishable green chips in the squares of a $3 \times 3$ grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally?
$\textbf{(A)} ~12\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24\qquad\textbf{(D)} ~30\qquad\textbf{(E)} ~36$

Solution:
Call the different colors A,B,C. There are $3!=6$ ways to rearrange these colors to these three letters, so $6$ must be multiplied after the letters are permuted in the grid.
WLOG assume that A is in the center. \[\begin{tabular}{ c c c }  ? & ? & ? \\   ? & A & ? \\    ? & ? & ?     \end{tabular}\] In this configuration, there are two cases, either all the A's lie on the same diagonal: \[\begin{tabular}{ c c c }  ? & ? & A \\   ? & A & ? \\    A & ? & ?     \end{tabular}\] or all the other two A's are on adjacent corners: \[\begin{tabular}{ c c c }  A & ? & A \\   ? & A & ? \\    ? & ? & ?     \end{tabular}\] In the first case there are two ways to order them since there are two diagonals, and in the second case there are four ways to order them since there are four pairs of adjacent corners.
In each case there is only one way to put the three B's and the three C's as shown in the diagrams. \[\begin{tabular}{ c c c }   C & B & A \\    B & A & C \\     A & C & B     \end{tabular}\] \[\begin{tabular}{ c c c }  A & B & A \\   C & A & C \\    B & C & B     \end{tabular}\] This means that there are $4+2=6$ ways to arrange A,B, and C in the grid, and there are $6$ ways to rearrange the colors. Therefore, there are $6\cdot6=36$ ways in total, which is $\boxed{36}$

Solution:
Without the loss of generality, we fix the top-left square with a red chip. We apply casework to its two adjacent chips:
Case (1): The top-center and center-left chips have different colors.  There are three subcases for Case (1):  As there are $3!=6$ permutations of the three colors, each subcase has $6$ ways. So, Case (1) has $3\cdot6=18$ ways in total.
Case (2): The top-center and center-left chips have the same color.  There are three subcases for Case (2):  As there are $3!=6$ permutations of the three colors, each subcase has $6$ ways. So, Case (2) has $3\cdot6=18$ ways in total.
Answer
Together, the answer is $18+18=\boxed{36}.$

Solution: We consider all possible configurations of the red chips for which rotations matter:  As there are $2!=2$ permutations of blue and green for each configuration, the answer is $2\cdot(2+4+4+4+4)=\boxed{36}.$

Solution:
$(1) \quad$ $\begin{tabular}{ c c c }  R & G & ? \\   B & R & ? \\    ? & ? & ?     \end{tabular}$ $\Longrightarrow$ $\begin{tabular}{ c c c }  R & G & B \\   B & R & G \\    R & G & B     \end{tabular}$ $\Longrightarrow$ $\begin{tabular}{ c c c }  R & G & R \\   B & R & B \\    G & B & G     \end{tabular}$ $\quad 3 \cdot 2 \cdot 2 = 12$
There are $3$ choices for $R$ $2$ choices for $G$ $R$ on the down left corner can be switched with $B$ on the upper right corner.
$(2) \quad$ $\begin{tabular}{ c c c }  R & G & ? \\   B & R & ? \\    ? & ? & ?     \end{tabular}$ $\Longrightarrow$ $\begin{tabular}{ c c c }  R & G & B \\   B & R & G \\    G & B & R     \end{tabular}$ $\quad 3 \cdot 2 = 6$
There are $3$ choices for $R$ $2$ choices for $G$
$(3) \quad$ $\begin{tabular}{ c c c }  G & R & ? \\   R & B & ? \\    ? & ? & ?     \end{tabular}$ $\Longrightarrow$ $\begin{tabular}{ c c c }  G & R & B \\   R & B & G \\    G & R & B     \end{tabular}$ $\Longrightarrow$ $\begin{tabular}{ c c c }  G & R & G \\   R & B & R \\    B & G & B     \end{tabular}$ $\quad 3 \cdot 2 \cdot 2 = 12$
Note that $(3)$ is a $180$ ° rotation of $(1)$
$(4) \quad$ $\begin{tabular}{ c c c }  G & R & ? \\   R & B & ? \\    ? & ? & ?     \end{tabular}$ $\Longrightarrow$ $\begin{tabular}{ c c c }  G & R & B \\   R & B & G \\    B & G & R     \end{tabular}$ $\quad 3 \cdot 2 = 6$
Note that $(4)$ is a $90$ ° rotation of $(2)$
Therefore, the answer is $2 \cdot (12 + 6) = \boxed{36}$

Solution:
Case (1) : We have a permutation of R, B, and G as all of the rows. There are $3!$ ways to rearrange these three colors. After finishing the first row, we move onto the second. Notice how the second row must be a derangement of the first one. By the derangement formula, $\frac{3!}{e} \approx 2$ , so there are two possible permutations of the second row. (Note: You could have also found the number of derangements of PIE). Finally, there are $2$ possible permutations for the last row. Thus, there are $3!\cdot2\cdot2=24$ possibilities.
Case (2) : All of the rows have two chips that are the same color and one that is different. There are obviously $3$ possible configurations for the first row, $2$ for the second, and $2$ for the third. Thus, there are $3\cdot2\cdot2=12$ possibilities.
Therefore, our answer is $24+12=\boxed{36}.$


Answer: 144.0

How many ways are there to split the integers $1$ through $14$ into $7$ pairs such that in each pair, the greater number is at least $2$ times the lesser number?
$\textbf{(A) } 108 \qquad \textbf{(B) } 120 \qquad \textbf{(C) } 126 \qquad \textbf{(D) } 132 \qquad \textbf{(E) } 144$

Solution:
Clearly, the integers from $8$ through $14$ must be in different pairs, and $7$ must pair with $14.$
Note that $6$ can pair with either $12$ or $13.$ From here, we consider casework:
Together, the answer is $72+72=\boxed{144}.$

Solution:
As said in Solution 1, clearly, the integers from $8$ through $14$ must be in different pairs.
We know that $8$ or $9$ can pair with any integer from $1$ to $4$ $10$ or $11$ can pair with any integer from $1$ to $5$ , and $12$ or $13$ can pair with any integer from $1$ to $6$ . Thus, $8$ will have $4$ choices to pair with, $9$ will then have $3$ choices to pair with ( $9$ cannot pair with the same number as the one $8$ pairs with). $10$ cannot pair with the numbers $8$ and $9$ has paired with but can also now pair with $5$ , so there are $3$ choices. $11$ cannot pair with $8$ 's, $9$ 's, or $10$ 's paired numbers, so there will be $2$ choices for $11$ $12$ can pair with an integer from $1$ to $5$ that hasn't been paired with already, or it can pair with $6$ $13$ will only have one choice left, and $7$ must pair with $14$
So, the answer is $4\cdot3\cdot3\cdot2\cdot2\cdot1\cdot1=\boxed{144}.$

Solution:
The integers $x \in \{8, \ldots , 14 \}$ must each be the larger elements of a distinct pair.
Assign partners in decreasing order for $x \in \{7, \dots, 1\}$
Note that $7$ must pair with $14$ $\mathbf{1} \textbf{ choice}$
For $5 \leq x \leq 7$ , the choices are $\{2x, \dots, 14\} - \{ \text{previous choices}\}$ . As $x$ decreases by 1, The minuend increases by 2 elements, and the subtrahend increases by 1 element, so the difference increases by 1, yielding $\mathbf{3!} \textbf{ combined choices}$
After assigning a partner to $5$ , there are no invalid pairings for yet-unpaired numbers, so there are $\mathbf{4!} \textbf{ ways}$ to choose partners for $\{1,2,3,4\}$
The answer is $3! \cdot 4! = \boxed{144}$


Answer: 144.0

Solution:
Clearly, the integers from $8$ through $14$ must be in different pairs, and $7$ must pair with $14.$
Note that $6$ can pair with either $12$ or $13.$ From here, we consider casework:
Together, the answer is $72+72=\boxed{144}.$

Solution:
As said in Solution 1, clearly, the integers from $8$ through $14$ must be in different pairs.
We know that $8$ or $9$ can pair with any integer from $1$ to $4$ $10$ or $11$ can pair with any integer from $1$ to $5$ , and $12$ or $13$ can pair with any integer from $1$ to $6$ . Thus, $8$ will have $4$ choices to pair with, $9$ will then have $3$ choices to pair with ( $9$ cannot pair with the same number as the one $8$ pairs with). $10$ cannot pair with the numbers $8$ and $9$ has paired with but can also now pair with $5$ , so there are $3$ choices. $11$ cannot pair with $8$ 's, $9$ 's, or $10$ 's paired numbers, so there will be $2$ choices for $11$ $12$ can pair with an integer from $1$ to $5$ that hasn't been paired with already, or it can pair with $6$ $13$ will only have one choice left, and $7$ must pair with $14$
So, the answer is $4\cdot3\cdot3\cdot2\cdot2\cdot1\cdot1=\boxed{144}.$

Solution:
The integers $x \in \{8, \ldots , 14 \}$ must each be the larger elements of a distinct pair.
Assign partners in decreasing order for $x \in \{7, \dots, 1\}$
Note that $7$ must pair with $14$ $\mathbf{1} \textbf{ choice}$
For $5 \leq x \leq 7$ , the choices are $\{2x, \dots, 14\} - \{ \text{previous choices}\}$ . As $x$ decreases by 1, The minuend increases by 2 elements, and the subtrahend increases by 1 element, so the difference increases by 1, yielding $\mathbf{3!} \textbf{ combined choices}$
After assigning a partner to $5$ , there are no invalid pairings for yet-unpaired numbers, so there are $\mathbf{4!} \textbf{ ways}$ to choose partners for $\{1,2,3,4\}$
The answer is $3! \cdot 4! = \boxed{144}$


Answer: 337.0

How many ways are there to write $2016$ as the sum of twos and threes, ignoring order? (For example, $1008\cdot 2 + 0\cdot 3$ and $402\cdot 2 + 404\cdot 3$ are two such ways.)
$\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672$

Solution:
The amount of twos in our sum ranges from $0$ to $1008$ , with differences of $3$ because $2 \cdot 3 = \operatorname{lcm}(2, 3)$
The possible amount of twos is $\frac{1008 - 0}{3} + 1 \Rightarrow \boxed{337}$

Solution: You can also see that you can rewrite the word problem into an equation $2x + 3y$ $2016$ . Therefore the question is just how many multiples of $3$ subtracted from 2016 will be an even number. We can see that $x = 1008$ $y = 0$ all the way to $x = 0$ , and $y = 672$ works, with $y$ being incremented by $2$ 's.Therefore, between $0$ and $672$ , the number of multiples of $2$ is $\boxed{337}$

Solution:
We can utilize the stars-and-bars distribution technique to solve this problem.
We have 2 "buckets" in which we will distribute parts of our total sum, 2016. By doing this, we know we will have $\binom{2016-1}{2-1}$ "total" answers. We want every third x and second y, so we divide our previous total by 6, which will result in $2015/6$ . We have to round down to the nearest integer, and we have to add 2 because we did not consider the 2 solutions involving x or y being 0.
So, $2015/6\implies335\implies335+2=\boxed{337}$

Solution:
Note that $2016 = 6 \cdot 336$ . In other words, we can write $2016$ as the sum of $336$ sixes.
In turn, we can express each $6$ as either $2 + 2 + 2$ or $3 + 3$
Therefore, we can write $2016$ as $n (2 + 2 + 2) + (336 - n) (3 + 3)$ , where $n$ is an integer between $0$ and $336$ , inclusive. Since each value of $n$ corresponds to a unique way to write the sum, we get $336 + 1 = \boxed{337}$


Answer: 24.0

How many ways can a student schedule $3$ mathematics courses -- algebra, geometry, and number theory -- in a $6$ -period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other $3$ periods is of no concern here.)
$\textbf{(A) }3\qquad\textbf{(B) }6\qquad\textbf{(C) }12\qquad\textbf{(D) }18\qquad\textbf{(E) }24$

Solution:
We must place the classes into the periods such that no two classes are in the same period or in consecutive periods.
Ignoring distinguishability, we can thus list out the ways that three periods can be chosen for the classes when periods cannot be consecutive:
Periods $1, 3, 5$
Periods $1, 3, 6$
Periods $1, 4, 6$
Periods $2, 4, 6$
There are $4$ ways to place $3$ nondistinguishable classes into $6$ periods such that no two classes are in consecutive periods. For each of these ways, there are $3! = 6$ orderings of the classes among themselves.
Therefore, there are $4 \cdot 6 = \boxed{24}$ ways to choose the classes.

Solution:
Counting what we don't want is another slick way to solve this problem. Use PIE (Principle of Inclusion and Exclusion) to count two cases: 1. Two classes consecutive, 2. Three classes consecutive.
Case 1: Consider two consecutive periods as a "block" of which there are 5 places to put in(1,2; 2,3; 3,4; 4,5; 5,6). Then we simply need to place two classes within the block, $3 \cdot 2$ . Finally we have 4 periods remaining to place the final math class. Thus there are $5 \cdot 3 \cdot 2 \cdot 4$ ways to place two consecutive math classes with disregard to the third.
Case 2: Now consider three consecutive periods as a "block" of which there are now 4 places to put in(1,2,3; 2,3,4; 3,4,5; 4,5,6). We now need to arrange the math classes in the block, $3 \cdot 2 \cdot 1$ . Thus there are $4 \cdot 3 \cdot 2 \cdot 1$ ways to place all three consecutive math classes.
By PIE we subtract Case 1 by Case 2 in order to not overcount: $120-24$ . Then we subtract that answer from the total ways to place the classes with no restrictions: $(6 \cdot 5 \cdot 4) - 96= \boxed{24}$

Solution:
We can tackle this problem with a stars-and-bars-ish approach. First, letting math class be 1 and non-math-class be 0, place 0s in between 3 1s: \[10101\] Now we need to place 1 additional 0. There are 4 places to put it: \[\underline{\hspace{0.3cm}}1\underline{\hspace{0.3cm}}01\underline{\hspace{0.3cm}}01\underline{\hspace{0.3cm}}\] It can be placed in any 1 of the underscores.
Since there are $3!=6$ ways to order the math classes, the answer is $6\cdot 4=\boxed{24}$

Solution:
Let $M$ represent a math class and $N$ represent a non-math class. We have _N_N_N_, where the spaces represent the possible spots the $M’s$ could go in. There are $4\choose 3$ ways to choose the spots for the math classes and $3!=6$ ways to order the classes. Hence, the answer is $4 \cdot 6 = \boxed{24}$ .
~azc1027


Answer: 24.0

Solution:
We must place the classes into the periods such that no two classes are in the same period or in consecutive periods.
Ignoring distinguishability, we can thus list out the ways that three periods can be chosen for the classes when periods cannot be consecutive:
Periods $1, 3, 5$
Periods $1, 3, 6$
Periods $1, 4, 6$
Periods $2, 4, 6$
There are $4$ ways to place $3$ nondistinguishable classes into $6$ periods such that no two classes are in consecutive periods. For each of these ways, there are $3! = 6$ orderings of the classes among themselves.
Therefore, there are $4 \cdot 6 = \boxed{24}$ ways to choose the classes.

Solution:
Counting what we don't want is another slick way to solve this problem. Use PIE (Principle of Inclusion and Exclusion) to count two cases: 1. Two classes consecutive, 2. Three classes consecutive.
Case 1: Consider two consecutive periods as a "block" of which there are 5 places to put in(1,2; 2,3; 3,4; 4,5; 5,6). Then we simply need to place two classes within the block, $3 \cdot 2$ . Finally we have 4 periods remaining to place the final math class. Thus there are $5 \cdot 3 \cdot 2 \cdot 4$ ways to place two consecutive math classes with disregard to the third.
Case 2: Now consider three consecutive periods as a "block" of which there are now 4 places to put in(1,2,3; 2,3,4; 3,4,5; 4,5,6). We now need to arrange the math classes in the block, $3 \cdot 2 \cdot 1$ . Thus there are $4 \cdot 3 \cdot 2 \cdot 1$ ways to place all three consecutive math classes.
By PIE we subtract Case 1 by Case 2 in order to not overcount: $120-24$ . Then we subtract that answer from the total ways to place the classes with no restrictions: $(6 \cdot 5 \cdot 4) - 96= \boxed{24}$

Solution:
We can tackle this problem with a stars-and-bars-ish approach. First, letting math class be 1 and non-math-class be 0, place 0s in between 3 1s: \[10101\] Now we need to place 1 additional 0. There are 4 places to put it: \[\underline{\hspace{0.3cm}}1\underline{\hspace{0.3cm}}01\underline{\hspace{0.3cm}}01\underline{\hspace{0.3cm}}\] It can be placed in any 1 of the underscores.
Since there are $3!=6$ ways to order the math classes, the answer is $6\cdot 4=\boxed{24}$

Solution:
Let $M$ represent a math class and $N$ represent a non-math class. We have _N_N_N_, where the spaces represent the possible spots the $M’s$ could go in. There are $4\choose 3$ ways to choose the spots for the math classes and $3!=6$ ways to order the classes. Hence, the answer is $4 \cdot 6 = \boxed{24}$ .
~azc1027


Answer: 6.0

How many whole numbers are between $\sqrt{8}$ and $\sqrt{80}$
$\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9$

Solution:
No... of course you're not supposed to know what the square root of 8 is, or the square root of 80. There aren't any formulas, either. Approximation seems like the best strategy.
Clearly it must be true that for any positive integers $a$ $b$ , and $c$ with $a>b>c$ \[\sqrt{a}>\sqrt{b}>\sqrt{c}\]
If we let $a=9$ $b=8$ , and $c=4$ , then we get \[\sqrt{9}>\sqrt{8}>\sqrt{4}\] \[3>\sqrt{8}>2\]
Therefore, the smallest whole number between $\sqrt{8}$ and $\sqrt{80}$ is $3$
Similarly, if we let $a=81$ $b=80$ , and $c=64$ , we get \[\sqrt{81}>\sqrt{80}>\sqrt{64}\] \[9>\sqrt{80}>8\]
So $8$ is the largest whole number between $\sqrt{8}$ and $\sqrt{80}$
So we know that we just have to find the number of integers from 3 to 8 inclusive. If we subtract 2 from every number in this set (which doesn't change the number of integers in the set at all), we find that now all we need to do is find the number of integers there are from 1 to 6, which is obviously 6.
$\boxed{6}$


Answer: 138.0

How many whole numbers between $100$ and $400$ contain the digit $2$
$\text{(A)}\ 100 \qquad \text{(B)}\ 120 \qquad \text{(C)}\ 138 \qquad \text{(D)}\ 140 \qquad \text{(E)}\ 148$

Solution:
This is a very common type of counting problem that you'll see quite often. Doing this the simple way would take too long, and might even make lots of mistakes.
If you ever learned about complementary counting , this would be the best time to utilize it. Instead of counting how many DO have 2's, why don't we count how many that DON'T?
So let's find the number of numbers. Obviously, we'd start by subtracting 100 from 400, getting us 300, but we're not done. Since just subtracting includes the number 400, we must subtract one (because 400 isn't allowed - it says between), getting us 299.
So how many numbers are there that DON'T have a 2? Well, we have 2 possibilities for the hundreds digit (1, 3, note that 2 is not allowed), 9 possibilities for the tens digit (1, 3, 4, 5, ... , 9, 0), and 9 possibilities for the ones digit. $2 \times 9 \times 9 = 162$ . However, one of the numbers we counted is $100$ , which isn't allowed, so there are $162-1=161$ numbers without a 2.
Since there are 299 numbers in total and 161 that DON'T have any 2's, $299 - 161 = 138$ numbers WILL have at least one two.
$\boxed{138}$

Solution:
As in the previous solution, get rid of the $400$ to make things easier. This gives us the numbers from $100$ to $399$
Let $A$ be the event that the first digit is $2$ $B$ be the event that the second digit is $2$ , and $C$ be the event that the third digit is $2$ . Then, PIE says that our answer will be $|A|+|B|+|C|-|A\cup B|-|A\cup C|-|B\cup C|+|A\cup B\cup C|$
We have that $|A|$ is just $1\times10\times10=100$ $|B|=3\times1\times10=30$ , and $|C|=3\times10\times1=30$
Next, $|A\cup B|$ is just having something in the form $22\_$ , so there are $10$ ways. Similarly, $|A\cup C|$ means that we have $2\_2$ , so there are again $10$ ways. Finally, $|B\cup C|$ means that our number is like $\_22$ , so there are $3$ ways.
Finally, $|A\cup B\cup C|$ counts the number of three digit numbers with all three digits $2$ , which there is only $1$ of: $222$
Putting this together, our answer will be $100+30+30-10-10-3+1=\boxed{138}$


Answer: 728.0

How many whole numbers between 1 and 1000 do not contain the digit 1?
$\textbf{(A)}\ 512 \qquad \textbf{(B)}\ 648 \qquad \textbf{(C)}\ 720 \qquad \textbf{(D)}\ 728 \qquad \textbf{(E)}\ 800$

Solution:
Note that this is the same as finding how many numbers with up to three digits do not contain 1.
Since there are 10 total possible digits, and only one of them is not allowed (1), each place value has its choice of 9 digits, for a total of $9*9*9=729$ such numbers. However, we overcounted by one; 0 is not between 1 and 1000, so there are $\boxed{728}$ numbers.


Answer: 162.0

How many whole numbers between 99 and 999 contain exactly one 0?
$\text{(A)}\ 72\qquad\text{(B)}\ 90\qquad\text{(C)}\ 144\qquad\text{(D)}\ 162\qquad\text{(E)}\ 180$

Solution: Numbers with exactly one zero have the form $\overline{a0b}$ or $\overline{ab0}$ , where the $a,b \neq 0$ . There are $(9\cdot1\cdot9)+(9\cdot9\cdot1) = 81+81 = \boxed{162}$


Answer: 21.0

How many whole numbers from $1$ through $46$ are divisible by either $3$ or $5$ or both?
$\text{(A)}\ 18 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 24 \qquad \text{(D)}\ 25 \qquad \text{(E)}\ 27$

Solution:
There are $\left\lfloor \frac{46}{3}\right\rfloor =15$ numbers divisible by $3$ $\left\lfloor\frac{46}{5}\right\rfloor =9$ numbers divisible by $5$ , so at first we have $15+9=24$ numbers that are divisible by $3$ or $5$ , except we counted the multiples of $\text{LCM}(3,5)=15$ twice, once for $3$ and once for $5$
There are $\left\lfloor \frac{46}{15}\right\rfloor =3$ numbers divisible by $15$ , so there are $24-3=21$ numbers divisible by $3$ or $5$ $\rightarrow \boxed{21}$


Answer: 5.0

How many whole numbers lie in the interval between $\frac{5}{3}$ and $2\pi$
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ \text{infinitely many}$

Solution: The smallest whole number in the interval is $2$ because $5/3$ is more than $1$ but less than $2$ . The largest whole number in the interval is $6$ because $2\pi$ is more than $6$ but less than $7$ . There are five whole numbers in the interval. They are $2$ $3$ $4$ $5$ , and $6$ , so the answer is $\boxed{5}$


Answer: 9.0

How many zeros are at the end of the product \[25\times 25\times 25\times 25\times 25\times 25\times 25\times 8\times 8\times 8?\]
$\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 12$

Solution: \begin{align*} (5^2)^7\times (2^3)^3 &=  5^{14}\times 2^{9} \\ &= 5^5\times 10^9  \end{align*} Since $5^5$ doesn't end in a $0$ , we can conclude that the product ends in $9$ zeroes $\rightarrow \boxed{9}$


Answer: 240.0

Hui is an avid reader. She bought a copy of the best seller Math is Beautiful . On the first day, Hui read $1/5$ of the pages plus $12$ more, and on the second day she read $1/4$ of the remaining pages plus $15$ pages. On the third day she read $1/3$ of the remaining pages plus $18$ pages. She then realized that there were only $62$ pages left to read, which she read the next day. How many pages are in this book?
$\textbf{(A)}\ 120 \qquad\textbf{(B)}\ 180\qquad\textbf{(C)}\ 240\qquad\textbf{(D)}\ 300\qquad\textbf{(E)}\ 360$

Solution:
Let $x$ be the number of pages in the book. After the first day, Hui had $\frac{4x}{5}-12$ pages left to read. After the second, she had $\left(\frac{3}{4}\right)\left(\frac{4x}{5}-12\right)-15 = \frac{3x}{5}-24$ left. After the third, she had $\left(\frac{2}{3}\right)\left(\frac{3x}{5}-24\right)-18=\frac{2x}{5}-34$ left. This is equivalent to $62.$
\begin{align*} \frac{2x}{5}-34&=62\\ 2x - 170 &= 310\\ 2x &= 480\\ x &= \boxed{240}

Solution:
If Hui has $62$ pages left to read at the end of Day 3, then on Day 3, before she read those extra 18 pages, she had $80$ pages left to read. $80$ pages is $\frac{2}{3}$ of the pages she had left at the end of Day 2. So at the end of Day 2, she had $80 \cdot \frac{3}{2} = 120$ pages left to read.
Similarly, we can work backwards on Day 2 to find that at the end of Day 1 Hui had $180$ pages left to read. Again working backwards on Day 1 we find that Hui originally had $\boxed{240}$ pages to read.


Answer: 8.0

I'm thinking of two whole numbers. Their product is 24 and their sum is 11. What is the larger number?
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 12$

Solution: Let the numbers be $x$ and $y$ . Then we have $x+y=11$ and $xy=24$ . Solving for $x$ in the first equation yields $x=11-y$ , and substituting this into the second equation gives $(11-y)(y)=24$ . Simplifying this gives $-y^2+11y=24$ , or $y^2-11y+24=0$ . This factors as $(y-3)(y-8)=0$ , so $y=3$ or $y=8$ , and the corresponding $x$ values are $x=8$ and $x=3$ . These are essentially the same answer: one number is $3$ and one number is $8$ , so the largest number is $8, \boxed{8}$

Solution:
Use the answers to attempt to "reverse engineer" an appropriate pair of numbers.  Looking at option $A$ , guess that one of the numbers is $3$ .  If the sum of two numbers is $11$ and one is $3$ , then other must be $11 - 3 = 8$ .  The product of those numbers is $3\cdot 8 = 24$ , which is the second condition of the problem, so our number are $3$ and $8$
However, $3$ is the smaller of the two numbers, so the answer is $8$ or $\boxed{8}$


Answer: 128.0

Solution:
Solution by e_power_pi_times_i
Notice that if $x=1$ , then $a_7x^7 + a_6x^6 + \cdots + a_0 =  a_7 + a_6 + \cdots + a_0$ . Therefore the answer is $(3(1)-1)^7) = \boxed{128}$


Answer: 69.0

If $(x, y)$ is a solution to the system $xy=6$ and $x^2y+xy^2+x+y=63$ ,
find $x^2+y^2$
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ \frac{1173}{32} \qquad \textbf{(C)}\ 55 \qquad \textbf{(D)}\ 69 \qquad \textbf{(E)}\ 81$

Solution: First note that $x^2y+xy^2+x+y= (xy+1)(x+y)$ . Substituting $6$ for $xy$ gives $7(x+y)= 63$ , giving a result of $x+y=9$ . Squaring this equation and subtracting by $12$ , gives us $x^2+y^2= \boxed{69}$


Answer: 1.0

If $1-\frac{4}{x}+\frac{4}{x^2}=0$ , then $\frac{2}{x}$ equals
$\textbf{(A) }-1\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }-1\text{ or }2\qquad  \textbf{(E) }-1\text{ or }-2$

Solution: By guessing and checking, 2 works. $\frac{2}{x} = \boxed{1}$ ~awin

Solution: Multiplying each side by $x^2$ , we get $x^2-4x+4 = 0$ . Factoring, we get $(x-2)(x-2) = 0$ . Therefore, $x = 2$ $\frac{2}{x} = \boxed{1}$ ~awin

Solution: Directly factoring, we get $(1-\frac{2}{x})^2 = 0$ . Thus $\frac{2}{x}$ must equal $\boxed{1}$


Answer: 17.0

If $1998$ is written as a product of two positive integers whose difference is as small as possible, then the difference is
$\mathrm{(A) \ }8 \qquad \mathrm{(B) \ }15 \qquad \mathrm{(C) \ }17 \qquad \mathrm{(D) \ }47 \qquad \mathrm{(E) \ } 93$

Solution:
If we want the difference of the two factors to be as small as possible, then the two numbers must be as close to $\sqrt{1998}$ as possible.
Since $45^2 = 2025$ , the factors should be as close to $44$ or $45$ as possible.
Breaking down $1998$ into its prime factors gives $1998 = 2\cdot 3^3 \cdot 37$
$37$ is relatively close to $44$ , and no numbers between $38$ and $44$ are factors of $1998$ .  Thus, the two factors are $37$ and $2\cdot 3^3 = 54$ , and the difference is $54 - 37 = 17$ , and the answer is $\boxed{17}$


Answer: 9.0

If $2$ is a solution (root) of $x^3+hx+10=0$ , then $h$ equals:
$\textbf{(A) }10\qquad \textbf{(B) }9 \qquad \textbf{(C) }2\qquad \textbf{(D) }-2\qquad \textbf{(E) }-9$

Solution: Substitute $2$ for $x$ . We are given that this equation is true. Solving for $h$ gives $h=-9$ . The answer is $\boxed{9}$


Answer: 18.0

If $20\%$ of a number is $12$ , what is $30\%$ of the same number?
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 30$

Solution:
$20\%$ of a number is equal to $\frac{1}{5}$ of that number. Let $n$ =the number
$\frac{1}{5}n = 12$ Multiply both sides by 5
$n = 60$
$30\%$ of $n$ is equal to $\frac{3}{10}n = \frac{3}{10}\cdot60 = 3\cdot6 = \boxed{18}$


Answer: 7.0

If $2137^{753}$ is multiplied out, the units' digit in the final product is:
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 9$

Solution:
$7^1$ has a unit digit of $7$ $7^2$ has a unit digit of $9$ $7^3$ has a unit digit of $3$ $7^4$ has a unit digit of $1$ $7^5$ has a unit digit of $7$
Notice that the unit digit eventually cycles to itself when the exponent is increased by $4$ .  It also does not matter what the other digits are in the base because the units digit is found by multiplying by only the units digit.  Since $753$ leaves a remainder of $1$ after being divided by $4$ , the units digit of $2137^{753}$ is $7$ , which is answer choice $\boxed{7}$


Answer: 0.0

If $2^a+2^b=3^c+3^d$ , the number of integers $a,b,c,d$ which can possibly be negative, is, at most:
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ 0$

Solution:
Assume $c,d \ge 0$ , and WLOG, assume $a<0$ and $a \le b$ .  This also takes into account when $b$ is negative.  That means \[\frac{1}{2^{-a}} + 2^b = 3^c + 3^d\] Multiply both sides by $2^{-a}$ to get \[1 + 2^{-a+b} = 2^{-a} (3^c + 3^d)\] Note that both sides are integers.  If $a \ne b$ , then the right side is even while the left side is odd, so equality can not happen.  If $a = b$ , then $2^{-a} (3^c + 3^d) = 2$ , and since $a<0$ $a = -1$ and $3^c + 3^d = 1$ .  No nonnegative value of $c$ and $d$ works, so equality can not happen.  Thus, $a$ and $b$ can not be negative when $c,d \ge 0$
Assume $a,b \ge 0$ , and WLOG, assume $c < 0$ and $c \le d$ .  This also takes into account when $d$ is negative.  That means \[2^a + 2^b = \frac{1}{3^{-c}} + 3^d\] Multiply both sides by $3^{-c}$ to get \[3^{-c} (2^a + 2^b) = 1 + 3^{d-c}\] That makes both sides integers.  The left side is congruent to $0$ modulo $3$ while the right side is congruent to $1$ or $2$ modulo $3$ , so equality can not happen.  Thus, $c$ and $d$ can not be negative when $a,b \ge 0$
Assume $a,c < 0$ , and WLOG, let $a \le b$ and $c \le d$ .  This also takes into account when $b$ or $d$ is negative.  That means \[\frac{1}{2^{-a}} + 2^b = \frac{1}{3^{-c}} + 3^d\] Multiply both sides by $2^{-a} \cdot 3^{-c}$ to get \[3^{-c} (1 + 2^{b-a}) = 2^{-a} (1 + 3^{d-c})\] That makes both sides integers.  The left side is congruent to $0$ modulo $3$ while the right side is congruent to $1$ or $2$ modulo $3$ , so equality can not happen.  Thus, $a$ and $c$ can not be negative.
Putting all the information together, none of $a,b,c,d$ can be negative, so the answer is $\boxed{0}$


Answer: 3.0

Solution:
Divide both sides of the original equation by $2^{1995}$ , giving:
$2^3 - 2^2 - 2^1 + 1 = k$
$k = 3$ , and the answer is $\boxed{3}$


Answer: 15.0

If $2x+1=8$ , then $4x+1=$
$\mathrm{(A)\ } 15 \qquad \mathrm{(B) \ }16 \qquad \mathrm{(C) \  } 17 \qquad \mathrm{(D) \  } 18 \qquad \mathrm{(E) \  }19$

Solution: We have \begin{align*}2x+1 = 8 &\iff 2x = 7 \\ &\iff x = \frac{7}{2},\end{align*} so \begin{align*}4x+1 &= 4\left(\frac{7}{2}\right)+1 \\ &= 2(7)+1 \\ &= \boxed{15}

Solution: From $2x = 7$ (as above), we can directly compute \begin{align*}4x &= 2(2x) \\ &= 2(7) \\ &= 14,\end{align*} so $4x+1 = 14+1 = \boxed{15}$


Answer: 2.0

If $60^a=3$ and $60^b=5$ , then $12^{(1-a-b)/\left(2\left(1-b\right)\right)}$ is
$\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 2\sqrt{3}\qquad$

Solution: We have that $12=\frac{60}{5}$ . We can substitute our value for 5, to get \[12=\frac{60}{60^b}=60\cdot 60^{-b}=60^{1-b}.\] Hence \[12^{(1-a-b)/\left(2\left(1-b\right)\right)}=60^{(1-b)(1-a-b)/\left(2\left(1-b\right)\right)}=60^{(1-a-b)/2}.\] Since $4=\frac{60}{3\cdot 5}$ , we have \[4=\frac{60}{60^a 60^b}=60^{1-a-b}.\] Therefore, we have \[60^{(1-a-b)/2}=4^{1/2}=\boxed{2}\]


Answer: 25.0

Solution: \begin{align*} 991+993+995+997+999=5000-N \\ &\Rightarrow (1000-9)+(1000-7)+(1000-5)+(1000-3)+(1000-1) = 5000-N \\ &\Rightarrow 5\times 1000-(1+3+5+7+9) = 5000 -N \\ &\Rightarrow 5000-25=5000-N \\ &\Rightarrow N=25\rightarrow \boxed{25}


Answer: 0.5

If $9^{x + 2} = 240 + 9^x$ , then the value of $x$ is:
$\textbf{(A)}\ 0.1 \qquad \textbf{(B)}\ 0.2\qquad \textbf{(C)}\ 0.3\qquad \textbf{(D)}\ 0.4\qquad \textbf{(E)}\ 0.5$

Solution: $9^{x+2}$ can be rewritten as $9^x*9^2=9^x*81$ , which means the equation can be rewritten as $81(9^x)=240+9^x$ , or $80(9^x)=240$ , or $9^x=3$ . Therefore, $x=\boxed{0.5}$


Answer: 20.0

If $A$ and $B$ are vertices of a polyhedron, define the distance $d(A,B)$ to be the minimum number of edges of the polyhedron one must traverse in order to connect $A$ and $B$ . For example, if $\overline{AB}$ is an edge of the polyhedron, then $d(A, B) = 1$ , but if $\overline{AC}$ and $\overline{CB}$ are edges and $\overline{AB}$ is not an edge, then $d(A, B) = 2$ . Let $Q$ $R$ , and $S$ be randomly chosen distinct vertices of a regular icosahedron (regular polyhedron made up of 20 equilateral triangles). What is the probability that $d(Q, R) > d(R, S)$
$\textbf{(A) } \frac{7}{22} \qquad \textbf{(B) } \frac{1}{3} \qquad \textbf{(C) } \frac{3}{8} \qquad \textbf{(D) } \frac{5}{12} \qquad \textbf{(E) } \frac{1}{2}$

Solution:
Since the icosahedron is symmetric polyhedron, we can rotate it so that R is on the topmost vertex. Since Q and
S basically the same, we can first count the probability that $d(Q,R) = d(R,S)$
$\mathfrak{Case} \ \mathfrak{1}: d(Q,R) = d(R,S) = 1$
There are 5 points $P$ such that $d(Q,P) = 1$ . There is $5 \times 4 = \boxed{20}$ ways to choose Q and S in this case.


Answer: 20.0

Solution:
Since the icosahedron is symmetric polyhedron, we can rotate it so that R is on the topmost vertex. Since Q and
S basically the same, we can first count the probability that $d(Q,R) = d(R,S)$
$\mathfrak{Case} \ \mathfrak{1}: d(Q,R) = d(R,S) = 1$
There are 5 points $P$ such that $d(Q,P) = 1$ . There is $5 \times 4 = \boxed{20}$ ways to choose Q and S in this case.


Answer: 52.0

Solution:
Notice that $\frac{2F(n)+1}{2}=F(n)+\frac{1}{2}.$
This means that for every single increment $n$ goes up from $1$ $F(n)$ will increase by $\frac{1}{2}.$ Since $101$ is $100$ increments from $1$ $F(n)$ will increase $\frac{1}{2}\times100=50.$
Since $F(1)=2,$ $F(101)$ will equal $2+50=\boxed{52}.$


Answer: 1.0

If $P = \frac{s}{(1 + k)^n}$ then $n$ equals:
$\textbf{(A)}\ \frac{\log{\left(\frac{s}{P}\right)}}{\log{(1 + k)}}\qquad  \textbf{(B)}\ \log{\left(\frac{s}{P(1 + k)}\right)}\qquad  \textbf{(C)}\ \log{\left(\frac{s - P}{1 + k}\right)}\qquad \\ \textbf{(D)}\ \log{\left(\frac{s}{P}\right)} + \log{(1 + k)}\qquad  \textbf{(E)}\ \frac{\log{(s)}}{\log{(P(1 + k))}}$

Solution:
\[P=\frac{s}{(1+k)^n}\]
\[(1+k)^n=\frac{s}{P}\]
Take the $\log$ of each side.
\[n \log(1+k) = \log\left(\frac{s}{P}\right)\]
\[n = \frac{\log\left(\frac{s}{P}\right)}{\log(1+k)} \to \boxed{1}\]


Answer: 3.0

If $S = i^n + i^{-n}$ , where $i = \sqrt{-1}$ and $n$ is an integer, then the total number of possible distinct values for $S$ is:
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 3\qquad \textbf{(D)}\ 4\qquad \textbf{(E)}\ \text{more than 4}$

Solution:
We first use the fact that $i^{-n}=\frac1{i^n}=\left(\frac1i\right)^n=(-i)^n$ . Note that $i^4=1$ and $(-i)^4=1$ , so $i^n$ and $(-i)^n$ have are periodic with periods at most 4. Therefore, it suffices to check for $n=0,1,2,3$
For $n=0$ , we have $i^0+(-i)^0=1+1=2$
For $n=1$ , we have $i^1+(-i)^1=i-i=0$
For $n=2$ , we have $i^2+(-i)^2=-1-1=-2$
For $n=3$ , we have $i^3+(-i)^3=-i+i=0$
Hence, the answer is $\boxed{3}$

Solution:
Notice that the powers of $i$ cycle in cycles of 4. So let's see if $S$ is periodic.
For $n=0$ : we have $2$
For $n=1$ : we have $0$
For $n=2$ : we have $-2$
For $n=3$ : we have $0$
For $n=4$ : we have $2$ again. Well, it can be seen that $S$ cycles in periods of 4. Select $\boxed{3}$


Answer: 80.0

Solution:
As a strategy, think of how $\angle B + \angle D$ would be determined, particularly without determining either of the angles individually, since it may not be possible to determine $\angle B$ or $\angle D$ alone.  If you see $\triangle BFD$ , then you can see that the problem is solved quickly after determining $\angle BFD$
But start with $\triangle AGF$ , since that's where most of our information is.  Looking at $\triangle AGF$ , since $\angle F = \angle G$ , and $\angle A = 20$ , we can write:
$\angle A + \angle G + \angle F = 180$
$20 + 2\angle F = 180$
$\angle AFG = 80$
By noting that $\angle AFG$ and $\angle GFD$ make a straight line, we know
$\angle AFG + \angle GFD = 180$
$80 + \angle GFD = 180$
$\angle GFD = 100$
Ignoring all other parts of the figure and looking only at $\triangle BFD$ , you see that $\angle B + \angle D + \angle F = 180$ .  But $\angle F$ is the same as $\angle GFD$ .  Therefore:
$\angle B + \angle D + \angle GFD = 180$ $\angle B + \angle D + 100 = 180$ $\angle B + \angle D = 80^\circ$ , and the answer is thus $\boxed{80}$


Answer: 50.0

If $\angle A = 60^\circ$ $\angle E = 40^\circ$ and $\angle C = 30^\circ$ , then $\angle BDC =$
[asy] pair A,B,C,D,EE; A = origin; B = (2,0); C = (5,0); EE = (1.5,3); D = (1.75,1.5); draw(A--C--D); draw(B--EE--A); dot(A); dot(B); dot(C); dot(D); dot(EE); label("$A$",A,SW); label("$B$",B,S); label("$C$",C,SE); label("$D$",D,NE); label("$E$",EE,N); [/asy]
$\text{(A)}\ 40^\circ \qquad \text{(B)}\ 50^\circ \qquad \text{(C)}\ 60^\circ \qquad \text{(D)}\ 70^\circ \qquad \text{(E)}\ 80^\circ$

Solution:
The sum of the angles in a triangle is $180^\circ$ . We can find $\angle ABE = 80^\circ$ , so $\angle CBD = 180-80=100^\circ$
\[\angle BDC = 180-100-30=\boxed{50}\]


Answer: 18.0

If $\angle A$ is four times $\angle B$ , and the complement of $\angle B$ is four times the complement of $\angle A$ , then $\angle B=$
$\textbf{(A)}\ 10^{\circ} \qquad\textbf{(B)}\ 12^{\circ} \qquad\textbf{(C)}\ 15^{\circ} \qquad\textbf{(D)}\ 18^{\circ} \qquad\textbf{(E)}\ 22.5^{\circ}$

Solution: Let $\angle A=x$ and $\angle B=y$ . From the first condition, we have $x=4y$ . From the second condition, we have \[90-y=4(90-x).\] Substituting $x=4y$ into the previous equation and solving yields \begin{align*}90-y=4(90-4y)&\implies 90-y=360-16y\\&\implies 15y=270\\&\implies y=\boxed{18}


Answer: 50.0

If $\angle \text{CBD}$ is a right angle , then this protractor indicates that the measure of $\angle \text{ABC}$ is approximately
[asy] unitsize(36); pair A,B,C,D; A=3*dir(160); B=origin; C=3*dir(110); D=3*dir(20); draw((1.5,0)..(0,1.5)..(-1.5,0)); draw((2.5,0)..(0,2.5)..(-2.5,0)--cycle); draw(A--B); draw(C--B); draw(D--B); label("O",(-2.5,0),W); label("A",A,W); label("B",B,S); label("C",C,W); label("D",D,E); label("0",(-1.8,0),W); label("20",(-1.7,.5),NW); label("160",(1.6,.5),NE); label("180",(1.7,0),E); [/asy]
$\text{(A)}\ 20^\circ \qquad \text{(B)}\ 40^\circ \qquad \text{(C)}\ 50^\circ \qquad \text{(D)}\ 70^\circ \qquad \text{(E)}\ 120^\circ$

Solution: We have that $20^{\circ}+\angle ABD +\angle CBD=160^{\circ}$ , or $\angle ABD +\angle CBD=140^{\circ}$ . Since $\angle CBD$ is a right angle, we have $\angle ABD=140^{\circ}-90^{\circ}=50^{\circ}\Rightarrow \boxed{50}$


Answer: 2.0

If $\begin{tabular}{r|l}a&b \\ \hline c&d\end{tabular} = \text{a}\cdot \text{d} - \text{b}\cdot \text{c}$ , what is the value of $\begin{tabular}{r|l}3&4 \\ \hline 1&2\end{tabular}$
$\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$

Solution:
Plugging in values for $a$ $b$ $c$ , and $d$ , we get
$a=3$ $b=4$ $c=1$ $d=2$
$a\times d=3\times2=6$
$b\times c=4\times1=4$
$6-4=2$
$\boxed{2}$


Answer: 8.0

If $\dfrac{\frac{x}{4}}{2}=\dfrac{4}{\frac{x}{2}}$ , then $x=$
$\text{(A)}\ \pm\frac{1}{2}\qquad\text{(B)}\ \pm 1\qquad\text{(C)}\ \pm 2\qquad\text{(D)}\ \pm 4\qquad\text{(E)}\ \pm 8$

Solution:
Cross-multiplying leaves
\begin{align*}\dfrac{x^2}{8} &= 8\\ x^2 &= 64\\ \sqrt{x^2} &= \sqrt{64}\\ x &= \pm 8\end{align*}
So the answer is $\boxed{8}$


Answer: 1991.0

Solution: Note that for all integers $n\neq 0$ \[\frac{(n-1)+n+(n+1)}{n}=3.\] Thus, we must have $N=1991\rightarrow \boxed{1991}$

Solution: As we know that $\frac{1990+1991+1992}{n}$ has to be some multiple of $\frac{2+3+4}{3}$ , then we know that the first equation is $995$ (1990/2) times bigger than the second one(in my solution), so the bottom must be $3\cdot995=\boxed{1991}$ -fn106068


Answer: 127.0

If $\frac{3}{5}=\frac{M}{45}=\frac{60}{N}$ , what is $M+N$
$\textbf{(A)}\ 27\qquad \textbf{(B)}\ 29 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 105\qquad \textbf{(E)}\ 127$

Solution: Separate into two equations $\frac35 = \frac{M}{45}$ and $\frac35 = \frac{60}{N}$ and solve for the unknowns. $M=27$ and $N=100$ , therefore $M+N=\boxed{127}$


Answer: 2.0

If $\frac{xy}{x+y}= a,\frac{xz}{x+z}= b,\frac{yz}{y+z}= c$ , where $a, b, c$ are other than zero, then $x$ equals:
$\textbf{(A)}\ \frac{abc}{ab+ac+bc}\qquad\textbf{(B)}\ \frac{2abc}{ab+bc+ac}\qquad\textbf{(C)}\ \frac{2abc}{ab+ac-bc}$ $\textbf{(D)}\ \frac{2abc}{ab+bc-ac}\qquad\textbf{(E)}\ \frac{2abc}{ac+bc-ab}$

Solution:
Note that $\frac{1}{a}=\frac{1}{x}+\frac{1}{y}$ $\frac{1}{b}=\frac{1}{x}+\frac{1}{z}$ , and $\frac{1}{c}=\frac{1}{y}+\frac{1}{z}$ . Therefore
\[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{2}\]
Therefore
\[\frac{1}{x}=\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}-\frac{1}{c}=\frac{1}{2a}+\frac{1}{2b}-\frac{1}{2c}\]
A little algebraic manipulation yields that
\[x=\boxed{2}\]


Answer: 0.0

If $\left(r+\frac1r\right)^2=3$ , then $r^3+\frac1{r^3}$ equals
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 6$

Solution:
We know $r+\frac1r=\sqrt3$ . Cubing this gives $r^3+3r+\frac3r+\frac1{r^3}=3\sqrt3$ . But $3r+\frac3r=3\left(r+\frac1r\right)=3\sqrt3$ , so subtracting this from the first equation gives $r^3+\frac1{r^3}=\boxed{0}$ .
(Actually, $r+\frac1r$ could have been equal to $-\sqrt3$ instead of $\sqrt3$ , but this would have led to the same answer. Also, this answer implies that $r^6=-1$ , which means that $r$ is a complex number.)


Answer: 35.0

If $\log (xy^3) = 1$ and $\log (x^2y) = 1$ , what is $\log (xy)$
$\mathrm{(A)}\ -\frac 12  \qquad\mathrm{(B)}\ 0  \qquad\mathrm{(C)}\ \frac 12 \qquad\mathrm{(D)}\ \frac 35  \qquad\mathrm{(E)}\ 1$

Solution: We rewrite the logarithms in the problem. \[\log(x) + 3\log(y) = 1\] \[2\log(x) + \log(y) = 1\] \[\log(x) + \log(y) = c\] where $c$ is the desired quantity. Set $u = \log(x)$ and $v = \log(y)$ . Then we have that \[u + 3y = 1 \textbf{(1)}\] \[2u + y = 1 \textbf{(2)}\] \[u + v = c\] . Notice that \[2 \cdot \textbf{(2)} + \textbf{(1)} = 5u + 5v = 3 \implies u + v = \frac{3}{5} \implies c = \boxed{35}\]


Answer: 1.47

If $\log 2 = .3010$ and $\log 3 = .4771$ , the value of $x$ when $3^{x+3} = 135$ is approximately
$\textbf{(A) \ } 5  \qquad \textbf{(B) \ } 1.47 \qquad \textbf{(C) \ } 1.67 \qquad \textbf{(D) \ } 1.78 \qquad \textbf{(E) \ } 1.63$

Solution: Taking the logarithm in base $3$ of both sides, we get $x+3 = \log_3 135$ . Using the property $\log ab = \log a + \log b$ , we get $x+3 = \log_3 5 + \log_3 3^3$ , or $x = \log_3 5$ . Converting into base $10$ gives $x = \frac{\log 5}{\log 3} = \frac{1 - \log 2}{\log 3}$ . Now, plugging in the values yeilds $\boxed{1.47}$


Answer: 2.43

If $\log x-5 \log 3=-2$ , then $x$ equals:
$\textbf{(A)}\ 1.25\qquad\textbf{(B)}\ 0.81\qquad\textbf{(C)}\ 2.43\qquad\textbf{(D)}\ 0.8\qquad\textbf{(E)}\ \text{either 0.8 or 1.25}$

Solution:
$\log x-5 \log 3=-2$
$\log x- \log 3^5=-2$
$\log x- \log 243 =-2$
$\log x / 243 = -2$
$x/243 = 10^{-2}$
$x=\frac{243}{100}$ $x=\boxed{2.43}$


Answer: 10.0

If $\log_{10}{m}= b-\log_{10}{n}$ , then $m=$
$\textbf{(A)}\ \frac{b}{n}\qquad\textbf{(B)}\ bn\qquad\textbf{(C)}\ 10^{b}n\qquad\textbf{(D)}\ b-10^{n}\qquad\textbf{(E)}\ \frac{10^{b}}{n}$

Solution: We have $b=\log_{10}{10^b}$ . Substituting, we find $\log_{10}{m}= \log_{10}{10^b}-\log_{10}{n}$ . Using $\log{a}-\log{b}=\log{\dfrac{a}{b}}$ , the left side becomes $\log_{10}{\dfrac{10^b}{n}}$ . Because $\log_{10}{m}=\log_{10}{\dfrac{10^b}{n}}$ $m=\boxed{10}$

Solution: adding $\log_{10} n$ to both sides: \[\log_{10} m + \log_{10} n=b\] using the logarithm property: $\log_a {b} + \log_a {c}=\log_a{bc}$ \[\log_{10} {mn}=b\] rewriting in exponential notation: \[10^b=mn\] \[m=\boxed{10}\] ~Vndom


Answer: 18.0

If $\sin{2x}\sin{3x}=\cos{2x}\cos{3x}$ , then one value for $x$ is
$\mathrm{(A) \ }18^\circ \qquad \mathrm{(B) \ }30^\circ \qquad \mathrm{(C) \ } 36^\circ \qquad \mathrm{(D) \ }45^\circ \qquad \mathrm{(E) \ } 60^\circ$

Solution:
We divide both sides of the equation by $\cos{2x}\times\cos{3x}$ to get $\frac{\sin{2x}\times\sin{3x}}{\cos{2x}\times\cos{3x}}=1$ , or $\tan{2x}\times\tan{3x}=1$
This looks a lot like the formula relating the slopes of two perpendicular lines , which is $m_1\times m_2=-1$ , where $m_1$ and $m_2$ are the slopes . It's made even more relatable by the fact that the tangent of an angle can be defined by the slope of the line that makes that angle with the x-axis
We can make this look even more like the slope formula by multiplying both sides by $-1$
$\tan{2x}\times-\tan{3x}=-1$ , and using the trigonometric identity $-\tan{x}=\tan{-x}$ , we have $\tan{2x}\times\tan{-3x}=-1$
Now it's time for a diagram:
 Since the product of the two slopes, $\tan{2x}$ and $\tan{-3x}$ , is $-1$ , the lines are perpendicular, and the angle between them is $\frac{\pi}{2}$ . The angle between them is also $2x+3x=5x$ , so $5x=\frac{\pi}{2}$ and $x=\frac{\pi}{10}$ , or $18^\circ, \boxed{18}$


Answer: 18.0

Solution: Start by subtracting $\sin{2x}\sin{3x}$ from both sides to get $\cos{2x}\cos{3x}-\sin{2x}\sin{3x}=0$ . We recognize that this is of the form $\cos{a}\cos{b}-\sin{a}\sin{b}=\cos{(a+b)}$ , so $\cos{2x}\cos{3x}-\sin{2x}\sin{3x}=\cos{(2x+3x)}=0$ $\cos{90^\circ}=0$ , so $x=\boxed{18}$


Answer: 16.0

If $\sqrt{x+2}=2$ , then $(x+2)^2$ equals:
$\textbf{(A)}\ \sqrt{2}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 16$

Solution: If we square both sides of the $\sqrt{x+2} = 2$ , we will get $x+2 = 4$ , if we square that again, we get $(x+2)^2 = \boxed{16}$

Solution: We can immediately get that $x = 2$ , after we square $(2+2)$ , we get $\boxed{16}$


Answer: 150.0

If $\tan x+\tan y=25$ and $\cot x + \cot y=30$ , what is $\tan(x+y)$

Solution:
Since $\cot$ is the reciprocal function of $\tan$
$\cot x + \cot y = \frac{1}{\tan x} + \frac{1}{\tan y} = \frac{\tan x + \tan y}{\tan x \cdot \tan y} = 30$
Thus, $\tan x \cdot \tan y = \frac{\tan x + \tan y}{30} = \frac{25}{30} = \frac{5}{6}$
Using the tangent addition formula:
$\tan(x+y) = \frac{\tan x + \tan y}{1-\tan x \cdot \tan y} = \frac{25}{1-\frac{5}{6}} = \boxed{150}$

Solution:
Using the formula for tangent of a sum, $\tan(x+y)=\frac{\tan x + \tan y}{1-\tan x \tan y} = \frac{25}{1-\tan x \tan y}$ . We only need to find $\tan x \tan y$
We know that $25 = \tan x + \tan y = \frac{\sin x}{\cos x} + \frac{\sin y}{\cos y}$ . Cross multiplying, we have $\frac{\sin x \cos y + \cos x \sin y}{\cos x \cos y} = \frac{\sin(x+y)}{\cos x \cos y} = 25$
Similarly, we have $30 = \cot x + \cot y = \frac{\cos x}{\sin x} + \frac{\cos y}{\sin y} = \frac{\cos x \sin y + \sin x \cos y}{\sin x \sin y} = \frac{\sin(x+y)}{\sin x \sin y}$
Dividing:
$\frac{25}{30} = \frac{\frac{\sin(x+y)}{\cos x \cos y}}{\frac{\sin(x+y)}{\sin x \sin y}} = \frac{\sin x \sin y}{\cos x \cos y} = \tan x \tan y = \frac{5}{6}$ . Plugging in to the earlier formula, we have $\tan(x+y) = \frac{25}{1-\frac{5}{6}} = \frac{25}{\frac{1}{6}} = \boxed{150}$

Solution:
Let $a=\tan x$ and $b=\tan y$ . This simplifies the equations to:
\[a + b = 25\]
\[\frac{1}{a} + \frac{1}{b} = 30\]
Taking the tangent of a sum formula from Solution 2, we get $\tan(x+y) = \frac{25}{1 - ab}$
We can use substitution to solve the system of equations from above: $b = -a + 25$ , so $\frac{1}{a} + \frac{1}{-a + 25} = 30$
Multiplying by $-a(a-25)$ ,  we get $a + (-a + 25) = -30a(a-25)$ , which is $-30a^2 + 750a = 25$ . Dividing everything by 5 and shifting everything to one side gives $6a^2 - 150a + 5 = 0$
Using the quadratic formula gives $a = \frac{150 \pm \sqrt {22380}}{12}$ . Since this looks too hard to simplify, we can solve for $b$ using $a + b = 25$ , which turns out to also be $b = \frac{150 \pm \sqrt {22380}}{12}$ , provided that the sign of the radical in $a$ is opposite the one in $b$
WLOG, assume $a = \frac{150 + \sqrt{22380}}{12}$ and $b = \frac{150 - \sqrt{22380}}{12}$ . Multiplying them gives $ab = \frac{22500 - 22380}{144}$ which simplifies to $\frac{5}{6}$
THe denominator of $\frac{25}{1 - ab}$ ends up being $\frac{1}{6}$ , so multiplying both numerator and denominator by 6 gives $\boxed{150}$


Answer: 2.0

If $\tan{\alpha}$ and $\tan{\beta}$ are the roots of $x^2 - px + q = 0$ , and $\cot{\alpha}$ and $\cot{\beta}$ are the roots of $x^2 - rx + s = 0$ , then $rs$ is necessarily
$\textbf{(A)} \ pq \qquad  \textbf{(B)} \ \frac{1}{pq} \qquad  \textbf{(C)} \ \frac{p}{q^2} \qquad  \textbf{(D)}\ \frac{q}{p^2}\qquad \textbf{(E)}\ \frac{p}{q}$

Solution:
By Vieta's Formulae, we have $\tan(\alpha)\tan(\beta)=q$ and $\cot(\alpha)\cot(\beta)=s$ . Recalling that $\cot\theta=\frac{1}{\tan\theta}$ , we have $\frac{1}{\tan(\alpha)\tan(\beta)}=\frac{1}{q}=s$
Also by Vieta's Formulae, we have $\tan(\alpha)+\tan(\beta)=p$ and $\cot(\alpha)+\cot(\beta)=r$ , and again using $\cot\theta=\frac{1}{\tan\theta}$ , we have $\tan(\alpha)+\tan(\beta)=r(\tan(\alpha)\tan(\beta))$ . Using $\tan(\alpha)\tan(\beta)=q$ and $\tan(\alpha)+\tan(\beta)=p$ , we therefore deduce that $r=\frac{p}{q}$ , which yields $rs = \frac{1}{q}\cdot\frac{p}{q}=\frac{p}{q^2}$
Thus, the answer is $\boxed{2}$


Answer: 17.0

If $\texttt{a,b,}$ and $\texttt{c}$ are digits for which
then $\texttt{a+b+c =}$
$\mathrm{(A) \  }14 \qquad \mathrm{(B) \  }15 \qquad \mathrm{(C) \  }16 \qquad \mathrm{(D) \  }17 \qquad \mathrm{(E) \  }18$

Solution:
Working from right to left, we see that $2 - b = 3$ .  Clearly if $b$ is a single digit integer, this cannot be possible.  Therefore, there must be some borrowing from $a$ .  Borrow $1$ from the digit $a$ , and you get $12 - b = 3$ , giving $b = 9$
Since $1$ was borrowed from $a$ , we have from the tens column $(a-1) - 8 = 7$ .  Again for single digit integers this will not work.  Again, borrow $1$ from $7$ , giving $10 + (a-1) - 8 = 7$ .  Solving for $a$
$10 + a - 1 - 8 = 7$
$1 + a = 7$
$a = 6$
Finally, since $1$ was borrowed from the hundreds column, we have $7 - 1 - 4 = c$ , giving $c = 2$
As a check, the problem is $762 - 489 = 273$ , which is a true sentence.
The desired quantity is $a + b + c = 6 + 9 + 2 = 17$ , and the answer is $\boxed{17}$


Answer: 7.0

If $\texttt{a}$ and $\texttt{b}$ are digits for which
$\begin{array}{ccc}& 2 & a\\ \times & b & 3\\ \hline & 6 & 9\\ 9 & 2\\ \hline 9 & 8 & 9\end{array}$
then $\texttt{a+b =}$
$\mathrm{(A)\ } 3 \qquad \mathrm{(B) \ }4 \qquad \mathrm{(C) \  } 7 \qquad \mathrm{(D) \  } 9 \qquad \mathrm{(E) \  }12$

Solution:
From the units digit calculation, we see that the units digit of $a\times 3$ is $9$ .  Since $0 \le a \le 9$ and $a$ is an integer, the only value of $a$ that works is is $a=3$ .  As a double-check, that does work, since $23 \times 3 = 69$ , which is the first line of the multiplication.
The second line of the multiplication can be found by doing the multiplication $23\times b = 92$ .  Dividing both sides by $23$ gives $b=4$
Thus, $a + b = 3 + 4 = 7$ , and the answer is $\boxed{7}$


Answer: 6.0

If $\text{A}*\text{B}$ means $\frac{\text{A}+\text{B}}{2}$ , then $(3*5)*8$ is
$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 16\qquad \text{(E)}\ 30$

Solution:
We just plug in and evaluate: \begin{align*} (3*5)*8 &= \left( \frac{3+5}{2}\right) *8 \\ &= 4*8 \\ &= \frac{4+8}{2} \\ &= 6 \\ \end{align*}
$\boxed{6}$