Answer: 9.0

"If a whole number $n$ is not prime, then the whole number $n-2$ is not prime." A value of $n$ which shows this statement to be false is
$\textbf{(A)}\ 9 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 23$

Solution:
To show this statement to be false, we need a non-prime value of $n$ such that $n-2$ is prime.  Since $13$ and $23$ are prime, they won't prove anything relating to the truth of the statement.
Now we just check the statement for $n=9,12,16$ .  If $n=12$ or $n=16$ , then $n-2$ is $10$ or $14$ , which aren't prime.  However, $n=9$ makes $n-2=7$ , which is prime, so $n=9$ proves the statement false.
Therefore, the answer is $\boxed{9}$ , 9.


Answer: 250.0

Solution: We make use of the associative and commutative properties of addition to rearrange the sum as \begin{align*} (1+49)+(11+39)+(21+29)+(31+19)+(41+9) &= 50+50+50+50+50 \\ &= 250 \Longrightarrow \boxed{250}

Solution:
Or, we can just combine the first, second, third, fourth, and last terms in each parentheses. This gives us:
\[(1+9)+(11+19)+(21+29)+(31+39)+(41+49),\] which gives us \[10+30+50+70+90 = 250 = \boxed{250}.\]


Answer: 167400.0

Solution:
We see that $1901=1800+101$ $1902=1800+102$ , etc. Each term in the first set of numbers is $1800$ more than the corresponding term in the second set; Because there are $93$ terms in the first set, the expression can be paired up as follows and simplified:
\[(1901-101) + (1902-102) + (1903-103) + \cdots + (1993-193)\\ =1800 + 1800 + \cdots + 1800\\ =(1800)(93)\\ =\boxed{167,400}\]


Answer: 26.0

Solution: We use the distributive property to get \[3\times 4+2\times 4+2\times 3 = 26 \rightarrow \boxed{26}\]

Solution: Since $\frac12+\frac13+\frac14 > \frac12+\frac14+\frac14 = 1$ , we have \[(2\times 3\times 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) > 2\times 3\times 4 \times 1 = 24\] The only answer choice greater than $24$ is $\boxed{26}$

Solution: We can just bash it out, getting $24(\frac12+\frac13+\frac14)= 12 + 8 + 6 = 26 \Longrightarrow \boxed{26}$ -fn106068


Answer: 3.0

Solution:
We use the order of operations here to get
\begin{align*} -15+9\times (6\div 3) &= -15+9\times 2 \\ &= -15+18 \\ &= 3 \rightarrow \boxed{3}


Answer: 0.426

Solution: $.4+.02+.006 = .400 + .020 + .006 = .426\rightarrow \boxed{.426}$


Answer: 222222222223.0

Solution: \begin{align*} 1,000,000,000,000-777,777,777,777 &= 999,999,999,999-777,777,777,777+1 \\ &= 222,222,222,222+1 \\ &= 222,222,222,223 \rightarrow \boxed{222,222,222,223}


Answer: 0.0

Solution:
Put the numbers in groups of $4$
$(1-2-3+4)+(5-6-7+8)+(9-10-11+ 12) + \cdots + (1993-1994-1995+1996)$
The first group has a sum of $0$
The second group increases the two positive numbers on the end by $1$ , and decreases the two negative numbers in the middle by $1$ .  Thus, the second group also has a sum of $0$
Continuing the pattern, every group has a sum of $0$ , and thus the entire sum is $0$ , giving an answer of $\boxed{0}$

Solution: Let any term of the series be $t_n$ . Realize that at every $n\equiv0 \pmod4$ , the sum of the series is 0. For $t_{1996}$ we know $1996\equiv0 \pmod4$ so the solution is $\boxed{0}$


Answer: 168.0

$10\times10\times10$ grid of points consists of all points in space of the form $(i,j,k)$ , where $i$ $j$ , and $k$ are integers between $1$ and $10$ , inclusive. Find the number of different lines that contain exactly $8$ of these points.

Solution:
$Case \textrm{ } 1:$ The lines are not parallel to the faces
A line through the point $(a,b,c)$ must contain $(a \pm 1, b \pm 1, c \pm 1)$ on it as well, as otherwise, the line would not pass through more than 5 points. This corresponds to the 4 diagonals of the cube.
We look at the one from $(1,1,1)$ to $(10,10,10)$ . The lower endpoint of the desired lines must contain both a 1 and a 3, so it can be $(1,1,3), (1,2,3), (1,3,3)$ . If $\textrm{min} > 0$ then the point $(a-1,b-1,c-1)$ will also be on the line for example, 3 applies to the other end.
Accounting for permutations, there are $12$ ways, so there are $12 \cdot 4 = 48$ different lines for this case.
$Case \textrm{ } 2:$ The lines where the $x$ $y$ , or $z$ is the same for all the points on the line.
WLOG, let the $x$ value stay the same throughout. Let the line be parallel to the diagonal from $(1,1,1)$ to $(1,10,10)$ . For the line to have 8 points, the $y$ and $z$ must be 1 and 3 in either order, and the $x$ value can be any value from 1 to 10. In addition, this line can be parallel to 6 face diagonals. So we get $2 \cdot 10 \cdot 6 = 120$ possible lines for this case.
The answer is, therefore, $120 + 48 = \boxed{168}$

Solution:
Look at one pair of opposite faces of the cube. There are $4$ lines say $l_1, l_2, l_3, l_4$ with exactly $8$ collinear points on the top face. For each of these lines, draw a rectangular plane that consists of one of the $l_i$ for $1 \leq i \leq 4$ and perpendicular to the top face.
There are $16$ lines in total on this plane. $10$ of which are parallel to one of the edges of the rectangular plane and $6$ of which are diagonals. There are $3$ pairs of opposite faces. So $3 \cdot 4 \cdot 16=192$ lines.
But we are overcounting the lines of the diagonals of those rectangular planes twice. There are $4$ rectangular planes perpendicular to one pair of opposite faces. Thus $4 \cdot 6=24$ lines are overcounted.
So the answer is $192-24=\boxed{168}$

Solution: Considered the cases $(1, 2, ..., 8), (2, 3, ...,9), (3, 4, ..., 10)$ and reverse. Also, consider the constant subsequences of length 8 $(1, 1, ..., 1), (2, 2, ..., 2), ..., (10, 10, ..., 10)$ . Of all the triplets that work they cannot be extended to form another point on the line in the $10 \times 10 \times 10$ grid but  we need to divide by 2 because reversing all the subsequences gives the same line. Thus the answer is \[\frac{16^3 - 14^3 - 14^3 + 12^3}{2} = \boxed{168}\]

Solution:
The lines can be defined as starting from $(a, b, c)$ with "slope" (vector) $(d, e, f)$ . We impose the condition that at least one of $a - d, b - e, \textrm{ or } c - f$ is outside the range of $[1, 10]$ in order to ensure that $(a, b, c)$ is the first valid point on this line. Then, the line ranges from $(a, b, c), (a + d, b + e, c + f), \ldots, (a + 7d, b + 7e, c + 7f)$ , where $1 \le a + 7d, b + 7e, c + 7f \le 10$ , in which case at least one of $a + 8d, b + 8e, \textrm{ or } c + 8f$ is outside of the range $[1, 10]$ to ensure the line does not contain more than 8 points. For $1 \le a + 7d, b + 7e, c + 7f \le 10$ to be satisfied, the pairs $(a, d), (b, e), \textrm{ and } (c, f)$ can only be $(1, 0), (2, 0), \ldots (10, 0), (1, 1), (2, 1), (3, 1), (10, -1), (9, -1), \textrm{ and } (8, -1)$ . Notice that there are only two pairs such that $a + 8d, b + 8e, \textrm{ or } c + 8f \not \in [1,  10]$ , namely $(3, 1)$ and $(8, -1)$ . Thus, our line must contain at least one of these two pairs. In addition, only the two pairs $(1, 1)$ and $(10, -1)$ satisfy $a - d, b - e, \textrm{ or } c - f \not \in [1, 10]$ . Thus, we must also include at least one of these two pairs as well. Let us call these 4 pairs "important" pairs. Finally, we can include any valid pair for our third pair.
Case 1: We repeat one of the important pairs
There are 2 ways to choose from $(3, 1)$ and $(8, -1)$ , and 2 ways to choose from $(1, 1)$ and $(10, -1)$ . Then, there are 2 ways to choose the repeated pair. Next, we can arrange these pairs $3!/2! = 3$ ways. So, we have $2 \cdot 2 \cdot 2 \cdot 3 = 24.$
Case 2: We use 3 distinct important pairs
There are ${4 \choose 3}$ ways to choose the pairs (note that by pigeonhole principle, this guarantees we get at least one of each required pair). Then, there are $3! = 6$ ways to arrange it. We obtain $4 \cdot 6 = 24$
Case 3: We use 3 unique pairs, one that is not important.
Again, there are $2 \cdot 2 = 4$ ways to choose the important pairs. Then, there are 12 ways to choose the non-important pair. Again, there are $3! = 6$ ways to arrange it. So, we get $4 \cdot 12 \cdot 6 = 288$
Summing all of it up and dividing by two (since we over counted each line and its reversal), we get $\frac{24 + 24 + 288}{2} = \boxed{168}.$


Answer: 768.0

$150\times 324\times 375$ rectangular solid is made by gluing together $1\times 1\times 1$ cubes. An internal diagonal of this solid passes through the interiors of how many of the $1\times 1\times 1$ cubes

Solution:
Place one corner of the solid at $(0,0,0)$ and let $(a,b,c)$ be the general coordinates of the diagonally opposite corner of the rectangle, where $a, b, c \in \mathbb{Z_{+}}$
We consider the vector equation of the diagonal segment represented by: $(x, y, z) =m (a, b, c)$ , where $m \in \mathbb{R}, 0 < m \le 1$
Consider a point on the diagonal with coordinates $(ma, mb, mc)$ . We have 3 key observations as this point moves from $(0,0,0)$ towards $(a,b,c)$
The number of cubes the diagonal passes is equal to the number of points on the diagonal that has one or more positive integers as coordinates.
If we slice the solid up by the $x$ -planes defined by $x=1,2,3,4, \ldots, a$ , the diagonal will cut these $x$ -planes exactly $a$ times (plane of $x=0$ is not considered since $m \ne 0$ ). Similar arguments for slices along $y$ -planes and $z$ -planes give diagonal cuts of $b$ , and $c$ times respectively. The total cuts by the diagonal is therefore $a+b+c$ , if we can ensure that no more than $1$ positive integer is present in the x, y, or z coordinate in all points $(ma,mb,mc)$ on the diagonal. Note that point $(a,b,c)$ is already one such exception.
But for each diagonal point $(ma,mb,mc)$ with 2 (or more) positive integers occurring at the same time, $a+b+c$ counts the number of cube passes as $2$ instead of $1$ for each such point.  There are $\gcd(a,b)+\gcd(b,c)+\gcd(c,a)$ points in such over-counting. We therefore subtract one time such over-counting from $a+b+c$
And for each diagonal point $(ma,mb,mc)$ with exactly $3$ integers occurring at the same time, $a+b+c$ counts the number of cube passes as $3$ instead of $1$ ; ie $a+b+c$ over-counts each of such points by $2$ . But since $\gcd(a,b)+\gcd(b,c)+\gcd(c,a)$ already subtracted three times for the case of $3$ integers occurring at the same time (since there are $3$ of these gcd terms that represent all combinations of 3 edges of a cube meeting at a vertex), we  have the final count for each such point as $1 = 3-3+k \Rightarrow k=1$ , where $k$ is our correction term. That is, we need to add $k=1$ time $\gcd(a,b,c)$ back to account for the case of 3 simultaneous integers.
Therefore, the total diagonal cube passes is: $D = a+b+c-\left[ \gcd(a,b)+\gcd(b,c)+\gcd(c,a) \right]+\gcd(a,b,c)$
For $(a,b,c) = (150, 324, 375)$ , we have: $\gcd(150,324)=6$ $\gcd(324,375)=3$ $\gcd(150,375)=75$ $\gcd(150,324,375)=3$
Therefore $D = 150+324+375-(6+3+75)+3 = \boxed{768}$

Solution:
Consider a point travelling across the internal diagonal, and let the internal diagonal have a length of $d$ . The point enters a new unit cube in the $x,y,z$ dimensions at multiples of $\frac{d}{150}, \frac{d}{324}, \frac{d}{375}$ respectively. We proceed by using PIE.
The point enters a new cube in the $x$ dimension $150$ times, in the $y$ dimension $324$ times and in the $z$ dimension, $375$ times.
The point enters a new cube in the $x$ and $y$ dimensions whenever a multiple of $\frac{d}{150}$ equals a multiple of $\frac{d}{324}$ . This occurs $\gcd(150, 324)$ times. Similarly, a point enters a new cube in the $y,z$ dimensions $\gcd(324, 375)$ times and a point enters a new cube in the $z,x$ dimensions $\gcd(375, 150)$ times.
The point enters a new cube in the $x,y$ and $z$ dimensions whenever some multiples of $\frac{d}{150}, \frac{d}{324}, \frac{d}{375}$ are equal. This occurs $\gcd(150, 324, 375)$ times.
The total number of unit cubes entered is then $150+324+375-[\gcd(150, 324)+\gcd(324, 375)+\gcd(375, 150)] + \gcd(150, 324, 375) = \boxed{768}$


Answer: 1000.0

Solution:
In the middle, we have $\cdots + 1010-1000+990 -\cdots$
If we match up the back with the front, and then do the same for the rest, we get pairs with $2000$ and $-2000$ , so these will cancel out.  In the middle, we have $2000-1000$ which doesn't cancel, but gives us $1000 \rightarrow \boxed{1000}$

Solution: We can see that there are $199$ terms in total. We can also see that the first $198$ numbers form groups of two that add to $10$ each. Dividing to see how many pairs we have, $198$ $2$ $99$ groups of ten, or $990$ . However, we have to remember to add the 199th term ( $10$ ), so we get $990$ $10$ $1000$ , which gives us $\boxed{1000}$


Answer: 25.0

$2 \times 3$ rectangle and a $3 \times 4$ rectangle are contained within a square without overlapping at any point, and the sides of the square are parallel to the sides of the two given rectangles. What is the smallest possible area of the square?
$\textbf{(A) } 16\qquad \textbf{(B) } 25\qquad \textbf{(C) } 36\qquad \textbf{(D) } 49\qquad \textbf{(E) } 64$

Solution:
By placing the $2 \times 3$ rectangle adjacent to the $3 \times 4$ rectangle with the 3 side of the $2 \times 3$ rectangle next to the 4 side of the $3 \times 4$ rectangle, we get a figure that can be completely enclosed in a square with a side length of 5. The area of this square is $5^2 = 25$
Since placing the two rectangles inside a $4 \times 4$ square must result in overlap, the smallest possible area of the square is $25$
So the answer is $\boxed{25}$


Answer: 6.0

$2$ by $2$ square is divided into four $1$ by $1$ squares.  Each of the small squares is to be painted either green or red.  In how many different ways can the painting be accomplished so that no green square shares its top or right side with any red square?  There may be as few as zero or as many as four small green squares.
$\text{(A)}\  4 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 16$

Solution:
If a green square cannot share its top or right side with a red square, then a red square can not share its bottom or left side with a green square. Let us split this up into several cases.
Case 1: There are no green squares. This can be done in $1$ way.
Case 2: There is one green square and three red squares. This can only be done when the green square's top and right edges are against the edge, so there is $1$ way.
Case 3: There are two green squares and two red squares. This happens when the two green squares are in the two top squares or two right squares, so there are $2$ ways.
Case 4: There are three green squares and one red square. Similar to case 2, this happens when the red square's left and bottom edges are against the edge, so there is $1$ way.
Case 5: There are four green squares and zero red squares. $1$ way.
\[1+1+2+1+1 = \boxed{6}\]


Answer: 1800.0

Solution: Find that \[(81+83+85+87+89+91+93+95+97+99) = 5 \cdot 180\] Which gives us \begin{align*} 2(5 \cdot 180) &= 10 \cdot 180\\ &= 1800 & \text{ Thus \boxed{1800}

Solution: $2(81+83+85+87+89+91+93+95+97+99)$ Pair the least with the greatest, second least with the second greatest, etc, until you have five pairs, each adding up to $81+99$ $83+97$ $85+95$ $87+93$ $89+91$ $180$ .  Since we have $5$ pairs, we multiply $180$ by $5$ to get $900$ .  But since we have to multiply by 2 (remember the 2 at the beginning of the parentheses!), we get $1800$ , which is $\boxed{1800}$


Answer: 8.0

$25$ foot ladder is placed against a vertical wall of a building. The foot of the ladder is $7$ feet from the base of the building. If the top of the ladder slips $4$ feet, then the foot of the ladder will slide:
$\textbf{(A)}\ 9\text{ ft} \qquad \textbf{(B)}\ 15\text{ ft} \qquad \textbf{(C)}\ 5\text{ ft} \qquad \textbf{(D)}\ 8\text{ ft} \qquad \textbf{(E)}\ 4\text{ ft}$

Solution: By the Pythagorean triple $(7,24,25)$ , the point where the ladder meets the wall is $24$ feet above the ground. When the ladder slides, it becomes $20$ feet above the ground. By the $(15,20,25)$ Pythagorean triple, The foot of the ladder is now $15$ feet from the building. Thus, it slides $15-7 = \boxed{8}$

Solution:
We can observe that the above setup forms a right angled triangles whose base is 7ft and whose hypotenuse is 25ft taking the height to be x ft.
\[x^2 + 7^2 = 25^2\] \[x^2 = 625 - 49\] \[x^2 = 576\] \[x = 24\]
Since the top of the ladder slipped by 4 ft the new height is $24 - 4 = 20 ft$ . The base of the ladder has moved so the new base is say $(7+y)$ . The hypotenuse remains the same at 25ft. So,
\[20^2 + (7+y)^2 = 25^2\] \[400 + 49 + y^2 + 14y = 625\] \[y^2 + 14y - 176 = 0\] \[y^2 + 22y - 8y - 176\] \[x(y+22) - 8(y+22)\] \[(y-8)(y+22)\]
Disregarding the negative solution to equation the solution to the problem is $\boxed{8}$


Answer: 45.0

Solution:
Taking the first product, we have
$\left(1-\frac{1}{2}\right)=\frac{1}{2}$
$\frac{1}{2}\times2=1$
Looking at the second, we get
$\left(1-\frac{1}{3}\right)=\frac{2}{3}$
$\frac{2}{3}\times3=2$
We seem to be going up by $1$
Just to check,
$1-\frac{1}{n}=\frac{n-1}{n}$
$\frac{n-1}{n}\times n=n-1$
Now that we have discovered the pattern, we have to find the last term.
$1-\frac{1}{10}=\frac{9}{10}$
$\frac{9}{10}\times10=9$
The sum of all numbers from $1$ to $9$ is
$\frac{9\cdot10}{2}=45=\boxed{45}$


Answer: 5.0

$3 \times 7$ rectangle is covered without overlap by 3 shapes of tiles: $2 \times 2$ $1\times4$ , and $1\times1$ , shown below. What is the minimum possible number of $1\times1$ tiles used?
$\textbf{(A) } 1\qquad\textbf{(B)} 2\qquad\textbf{(C) } 3\qquad\textbf{(D) } 4\qquad\textbf{(E) } 5$

Solution: We can eliminate B, C, and D, because they are not $21$ subtracted by any multiple of $4$ . Finally, we see that there is no way to have A, so the solution is $\boxed{5}$

Solution: Let $x$ be the number of $1x1$ tiles. There are $21$ squares and each $2x2$ or $1x4$ tile takes up 4 squares, so $x \equiv 1 \pmod{4}$ , so it is either $1$ or $5$ . Color the columns, starting with red, then blue, and alternating colors, ending with a red column. There are $12$ red squares and $9$ blue squares, but each $2x2$ and $1x4$ shape takes up an equal number of blue and red squares, so there must be $3$ more $1x1$ tiles on red squares than on blue squares, which is impossible if there is just one, so the answer is $\boxed{5}$ , which can easily be confirmed to work.

Solution:
Suppose there are $a$ different $2\times 2$ tiles, $b$ different $4\times 1$ tiles and $c$ different $1\times 1$ tiles. Since the areas of these tiles must total up to $21$ (area of the whole grid), we have \[4a + 4b + c = 21.\] Reducing modulo $4$ gives $c\equiv 1\pmod{4}$ , or $c = 1$ or $c = 5$
If $c = 1$ , then $a + b = 5$ . After some testing, there is no valid pair $(a, b)$ that works, so the answer must be $\boxed{5}$ , which can be constructed in many ways.


Answer: 2989.0

Solution:
We can make use of the distributive property as follows: \begin{align*} 4(299)+3(299)+2(299)+298 &= 4(299)+3(299)+2(299)+1(299)-1 \\ &= (4+3+2+1)(299)-1 \\ &= 10(299)-1 \\ &= 2989 \\ \end{align*}
$\boxed{2989}$


Answer: 4.0

$4\times 4$ block of calendar dates is shown. First, the order of the numbers in the second and the fourth rows are reversed. Then, the numbers on each diagonal are added. What will be the positive difference between the two diagonal sums?
$\begin{tabular}[t]{|c|c|c|c|} \multicolumn{4}{c}{}\\\hline 1&2&3&4\\\hline 8&9&10&11\\\hline 15&16&17&18\\\hline 22&23&24&25\\\hline \end{tabular}$
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10$

Solution:
After reversing the numbers on the second and fourth rows, the block will look like this:
$\begin{tabular}[t]{|c|c|c|c|} \multicolumn{4}{c}{}\\\hline 1&2&3&4\\\hline 11&10&9&8\\\hline 15&16&17&18\\\hline 25&24&23&22\\\hline \end{tabular}$
The positive difference between the two diagonal sums is then $(4+9+16+25)-(1+10+17+22)=3-1-1+3=\boxed{4}$

Solution: Notice that at baseline the diagonals sum to the same number ( $52$ ).  Therefore we need only compute the effect of the swap.  The positive difference between $9$ and $10$ is $1$ and the positive difference between $22$ and $25$ is $3$ . Adding gives $1+3=\boxed{4}$


Answer: 4.0

Solution:
After reversing the numbers on the second and fourth rows, the block will look like this:
$\begin{tabular}[t]{|c|c|c|c|} \multicolumn{4}{c}{}\\\hline 1&2&3&4\\\hline 11&10&9&8\\\hline 15&16&17&18\\\hline 25&24&23&22\\\hline \end{tabular}$
The positive difference between the two diagonal sums is then $(4+9+16+25)-(1+10+17+22)=3-1-1+3=\boxed{4}$

Solution: Notice that at baseline the diagonals sum to the same number ( $52$ ).  Therefore we need only compute the effect of the swap.  The positive difference between $9$ and $10$ is $1$ and the positive difference between $22$ and $25$ is $3$ . Adding gives $1+3=\boxed{4}$


Answer: 52.0

$4\times 4\times 4$ cubical box contains 64 identical small cubes that exactly fill the box.  How many of these small cubes touch a side or the bottom of the box?
$\text{(A)}\ 48 \qquad \text{(B)}\ 52 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 80$

Solution:
Each small cube would have dimensions $1\times 1\times 1$ making each cube a unit cube.
If there are $16$ cubes per face and there are $5$ faces we are counting, we have $16\times 5= 80$ cubes.
Some cubes are on account of overlap between different faces.
We could reduce this number by subtracting the overlap areas, which could mean subtracting 4 cubes from each side and 12 from the bottom.
$80-(4\times 4+12)=80-28=52=\boxed{52}$

Solution:
You can imagine removing the cubes that do not fit the description of the problem, forming a "square cup".
There are $4$ cubes in the center of the top face that do not fit the description.  Remove those.
Once you remove those cubes on top, you must go down one level and remove the cubes in the same position on the second layer.  Thus, $4$ more cubes are removed.
Finally, you repeat this on the third layer, for $4$ more cubes.
Once you do the top three layers, you will be on the bottom layer, and you don't remove any more cubes.
This means you removed $4 + 4 + 4 = 12$ cubes of the $4 \times 4 \times 4 = 64$ cubes.
Thus, $64 - 12 = 52$ cubes remain, and the answer is $\boxed{52}$

Solution:
We can use casework to solve this.
Case $1$ : The cubes on the sides
There are four sides of a cube which means $64$ cubes, but there will be an overlap of $16$ cubes ( $4$ sides $\times$ $4$ cubes on each side).
So there are $16 \times 4-16=64-16=48$ cubes on the sides.
Case $2$ : The cubes on the bottom
There are $16$ cubes on the bottom, but by counting the cubes on the sides, $12$ of the cubes are already accounted for. $16-12=4$ cubes on the bottom that aren't already counted.
$48+4=\boxed{52}$ cubes touching at least one side. Since there is no top, we don't need to count the top face cubes.


Answer: 106.0

$7\times 1$ board is completely covered by $m\times 1$ tiles without overlap; each tile may cover any number of consecutive squares, and each tile lies completely on the board. Each tile is either red, blue, or green. Let $N$ be the number of tilings of the $7\times 1$ board in which all three colors are used at least once. For example, a $1\times 1$ red tile followed by a $2\times 1$ green tile, a $1\times 1$ green tile, a $2\times 1$ blue tile, and a $1\times 1$ green tile is a valid tiling. Note that if the $2\times 1$ blue tile is replaced by two $1\times 1$ blue tiles, this results in a different tiling. Find the remainder when $N$ is divided by $1000$

Solution:
Firstly, we consider how many different ways possible to divide the $7\times 1$ board.
We ignore the cases of 1 or 2 pieces since we need at least one tile of each color.
Secondly, we use Principle of Inclusion-Exclusion to consider how many ways to color them:
Finally, we combine them together: $15\times 6+20\times 36+15\times 150+6\times 540+1\times 1806= 8106$
So the answer is $\boxed{106}$

Solution:
This solution is basically solution 1 with more things done at once. The game plan:
$\sum_{i=0}^{7} ($ the amount of ways to divide the board into $i$ pieces $) \cdot ($ the amount of ways to color the respective divisions)
The amount of ways to divide the board is determined using stars and bars. The colorings are found using PIE giving $3^i-3\cdot2^i+3$ . Plus, we don't have to worry about the cases where $i=1$ or $I=2$ since they both give no solutions. So our equation becomes:
$\sum_{i=3}^{7} \left(\dbinom{6}{I}\right)\cdot\left(3^i-3\cdot2^i+3\right)$
Writing it all out and keeping the numbers small with mod 1000, we will eventually arrive at the answer of $\boxed{106}$

Solution:
3 colors is a lot. How many ways can we tile an $n \times 1$ board with one color? It's going to be $2^{n-1}$ because if you draw out the $n$ spaces, you can decide whether each of the borders between the tiles are either there or not there. There are $n-1$ borders so there are $2^{n-1}$ tilings. Define a one-tiling of an mx1 as $f_1(m)$
Now let's look at two colors. Let's see if we can get a two-tiling of an $(n+1) \times 1$ based off a $n \times 1$ . There are 2 cases we should consider:
1) The $n \times 1$ was a one-coloring and the block we are going to add consists of the second color. The number of ways we can do this is $2f_1(n)$
2) The $n \times 1$ was a two-color tiling so now we've got 3 cases to form the $(n+1) \times 1$ : We can either add a block of the first color, the second color, or we can adjoin a block to the last block in the $n \times 1$
This gives us $f_2(n+1)=2f_1(n)+3f_2(n)$
Time to tackle the 3-color tiling. Again, we split into 2 cases:
1) The $n \times 1$ was a two-color tiling, and the block we are adding is of the 3rd color. This gives $f_2(n)$ ways but we have to multiply by $3C2 = 3$ because we have to pick 2 different colors for the two-color tiling.
2) The $n \times 1$ was a 3-color tiling, and we have to consider what we can do with the block that we are adding. It can be any of the 3 colors, or we can adjoin it to whatever was the last block in the $n \times 1$ . This gives $4f_3(n)$
So in total we have $f_3(n+1)=3f_2(n)+4f_3(n)$
Finally, we just sorta bash through the computation to get $f_3(7)=8\boxed{106}$

Solution:
Let $n$ be the length of the board and $x$ be the number of colors. We will find the number of ways to tile the $n \times 1$ board with no color restrictions (some colors may be unused) and then use PIE.
By stars and bars, the number of ways to divide the board into $k$ pieces is ${n-1 \choose k-1}$ . There are $x^k$ ways to color each of these divisions. Therefore, the total number of ways to divide and color the board is \begin{align*} \chi(n, x) &:= \sum_{k=1}^n {n-1 \choose k-1} x^k \\ &= x\sum_{k=0}^{n-1} {n-1 \choose k} x^k \\ &= x(x+1)^{n-1}. \end{align*}
In the given problem, we have $n=7$ . By PIE, we have \begin{align*} &\quad {3 \choose 3} \chi(7, 3) - {3 \choose 2} \, \chi(7, 2) + {3 \choose 1} \, \chi(7, 1) \\ &= 3 \cdot 4^6 - 3(2 \cdot 3^6) + 3(1 \cdot 2^6) \\ &= 8106 \rightarrow \boxed{106}


Answer: 945.0

Solution:
To simplify the problem, we can group 90’s together: [mathjax]90 + 91 + ... + 98 + 99 = 90 \cdot 10 + 1 + 2 + 3 + ... + 8 + 9[/mathjax].
[mathjax]90\cdot10=900[/mathjax], and finding [mathjax]1 + 2 + ... + 8 + 9[/mathjax] has a trick to it.
Rearranging the numbers so each pair sums up to 10, we have:
[mathjax display=true](1 + 9)+(2+8)+(3+7)+(4+6)+5[/mathjax]. [mathjax]4\cdot10+5 = 45[/mathjax], and [mathjax]900+45=\boxed{945}[/mathjax].

Solution: We can express each of the terms as a difference from $100$ and then add the negatives using $\frac{n(n+1)}{2}=1+2+3+\cdots+(n-1)+n$ to get the answer. \begin{align*} (100-10)+(100-9)+\cdots+(100-1) &= 100\cdot10 -(1+2+\cdots+9+10)\\ &= 1000 - 55\\ &= \boxed{945}

Solution: Instead of breaking the sum then rearranging, we can rearrange directly: \begin{align*} 90+91+92+\cdots +98+99 &= (90+99)+(91+98)+(92+97)+(93+96)+(94+95) \\ &= 189+189+189+189+189 \\ &= \boxed{945}

Solution:
The finite arithmetic sequence formula states that the sum in the sequence is equal to $\frac{n}{2}\cdot(a_1+a_n)$ where $n$ is the number of terms in the sequence, $a_1$ is the first term and $a_n$ is the last term.
Applying the formula, we have: \[\frac{10}{2}\cdot(90+99)=\boxed{945}\]

Solution: The expression is equal to the sum of integers from $1$ to $99$ minus the sum of integers from $1$ to $89$ , so it is equal to $\frac{99(100)}{2} - \frac{89(90)}{2} = 4950 - 4005 = \boxed{945}$


Answer: 59.0

$A$ $B$ $C$ are three piles of rocks. The mean weight of the rocks in $A$ is $40$ pounds, the mean weight of the rocks in $B$ is $50$ pounds, the mean weight of the rocks in the combined piles $A$ and $B$ is $43$ pounds, and the mean weight of the rocks in the combined piles $A$ and $C$ is $44$ pounds. What is the greatest possible integer value for the mean in pounds of the rocks in the combined piles $B$ and $C$
$\textbf{(A)} \ 55 \qquad \textbf{(B)} \ 56 \qquad \textbf{(C)} \ 57 \qquad \textbf{(D)} \ 58 \qquad \textbf{(E)} \ 59$

Solution:
Let the number of rocks in $A$ be $a$ $B$ be $b$ $C$ be $c$ . The total weight of $A$ be $40a$ $B$ be $50b$ $C$ be $kc$
We can write the information given as, $\frac{40a + 50b}{a+b} = 43$ $\frac{40a + kc}{a+c} = 44$ $\frac{50b + kc}{b+c} = ?$
$40a + 50b = 43 a + 43 b$ $3a = 7b$
$40a + kc = 44a + 44 c$ $kc = 4a + 44c = \frac{28}{3}b + 44c$
$\frac{50b + kc}{b+c} = \frac{50 \cdot \frac{3}{7}a + kc}{\frac{3}{7}a+c} = \frac{150a + 7kc}{3a + 7c} = \frac{150a + 28a + 308c}{3a+7c} = \frac{178a + 308c}{3a+7c} = \frac{44(3a+7c)+46a}{3a+7c}$
$= 44 + \frac{46a}{3a+7c} = 44 + \frac{46}{3 + \frac{7}{a}} < 44 + \frac{46}{3} \approx 59.3$
$\frac{50b + kc}{b+c} \le \boxed{59}$


Answer: 59.0

Solution:
Let the total number of rocks in pile $A$ be $A_n$ , and the total number of rocks in pile $B$ be $B_n$ . Then, by restriction 3 (the average of $A$ and $B$ ), we can establish the equation: \[\frac{40A_n+50B_n}{A_n+B_n}=43\] .
Cross-multiplying, we get: \[40A_n+50B_n=43A_n+43B_n \implies 3A_n=7B_n\] .
Let's say we have $7k$ rocks in $A$ and $3k$ rocks in $B$ . Hence, we have $280k$ and $150k$ as the total weight of piles $A$ and $B$ , respectively. Let the total weight of $C$ be $m$ , and the total number of rocks in $C$ be $n$ .
Using the last restriction regarding the average of piles $A$ and $C$ , we have: \[\frac{280k+m}{7k+n}=44 \implies 280k + m=280k + 28k + 44n \implies m=28k+44n\] .
To find the average of piles $B$ and $C$ , we can establish the expression: \[\frac{150k+28k+44n}{3k+n}=\frac{178k+44n}{3k+n}=\frac{132k+44n+46k}{3k+n}=\frac{44(3k+n)+46k}{3k+n}=44+\frac{46k}{3k+n}.\] When we let the final expression equal $59$ , we get: \[44+\frac{46k}{3k+n}=59\implies\frac{46k}{3k+n}=15\] Cross-multiplying, we get: \[46k=45k+14n\implies k=14n\] $k$ is still positive here, so 59 works. As this is the greatest option, we can circle $\textbf{(E)}$ immediately.
To show why $59$ is the greatest, consider the following:
When we let the final expression equal $60$ , we get: \[44+\frac{46k}{3k+n}=60\implies\frac{46k}{3k+n}=16\] Cross-multiplying, we get: \[46k=48k+14n\implies k=-7n\] Since $k$ is positive, the final expression could not equal 60. It further implies that the final expression could not equal any other integer greater than 60. Therefore, we have our final answer $\boxed{59}$


Answer: 3.0

$A$ and $B$ together can do a job in $2$ days; $B$ and $C$ can do it in four days; and $A$ and $C$ in $2\frac{2}{5}$ days.
The number of days required for A to do the job alone is:
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 2.8$

Solution: Let $A$ do $r_A$ of the job per day, $B$ do $r_B$ of the job per day, and $C$ do $r_C$ of the job per day. These three quantities have unit $\frac{\text{job}}{\text{day}}$ . Therefore our three conditions give us the three equations: \begin{align*} (2\text{ days})(r_A+r_B)&=1\text{ job},\nonumber\\ (4\text{ days})(r_B+r_C)&=1\text{ job},\nonumber\\ (2.4\text{ days})(r_C+r_A)&=1\text{ job}.\nonumber \end{align*} We divide the three equations by the required constant so that the coefficients of the variables become 1: \begin{align*} r_A+r_B&=\frac{1}{2}\cdot\frac{\text{job}}{\text{day}},\nonumber\\ r_B+r_C&=\frac{1}{4}\cdot\frac{\text{job}}{\text{day}},\nonumber\\ r_C+r_A&=\frac{5}{12}\cdot\frac{\text{job}}{\text{day}}.\nonumber \end{align*} If we add these three new equations together and divide the result by two, we obtain an equation with left-hand side $r_A+r_B+r_C$ , so if we subtract $r_B+r_C$ (the value of which we know) from both equations, we obtain the value of $r_A$ , which is what we wish to determine anyways. So we add these three equations and divide by two: \[r_A+r_B+r_C=\frac{1}{2}\cdot\left(\frac{1}{2}+\frac{1}{4}+\frac{5}{12}\right)\cdot\frac{\text{job}}{\text{day}}=\frac{7}{12}\cdot\frac{\text{job}}{\text{day}}.\] Hence: \begin{align*} r_A &= (r_A+r_B+r_C)-(r_B+r_C)\nonumber\\ &=\frac{7}{12}\cdot\frac{\text{job}}{\text{day}}-\frac{1}{4}\cdot\frac{\text{job}}{\text{day}}\nonumber\\ &=\frac{1}{3}\cdot\frac{\text{job}}{\text{day}}.\nonumber \end{align*} This shows that $A$ does one third of the job per day. Therefore, if $A$ were to do the entire job himself, he would require $\boxed{3}$ days.


Answer: 293.0

$ABCD$ is a rectangular sheet of paper that has been folded so that corner $B$ is matched with point $B'$ on edge $AD.$ The crease is $EF,$ where $E$ is on $AB$ and $F$ is on $CD.$ The dimensions $AE=8, BE=17,$ and $CF=3$ are given. The perimeter of rectangle $ABCD$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
[asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(25,0), C=(25,70/3), D=(0,70/3), E=(8,0), F=(22,70/3), Bp=reflect(E,F)*B, Cp=reflect(E,F)*C; draw(F--D--A--E); draw(E--B--C--F, linetype("4 4")); filldraw(E--F--Cp--Bp--cycle, white, black); pair point=( 12.5, 35/3 ); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); label("$B^\prime$", Bp, dir(point--Bp)); label("$C^\prime$", Cp, dir(point--Cp));[/asy]

Solution: Since $EF$ is the perpendicular bisector of $\overline{BB'}$ , it follows that $BE = B'E$ (by SAS). By the Pythagorean Theorem , we have $AB' = 15$ . Similarly, from $BF = B'F$ , we have \begin{align*} BC^2 + CF^2 = B'D^2 + DF^2 &\Longrightarrow BC^2 + 9 = (BC - 15)^2 + 484 \\ BC  &= \frac{70}{3} \end{align*} Thus the perimeter of $ABCD$ is $2\left(25 + \frac{70}{3}\right) = \frac{290}{3}$ , and our answer is $m+n=\boxed{293}$

Solution:
Let $A = (0,0), B=(0,25)$ , so $E = (0,8)$ and $F = (l,22)$ , and let $l = AD$ be the length of the rectangle. The slope of $EF$ is $\frac{14}{l}$ and so the equation of $EF$ is $y -8 = \frac{14}{l}x$ . We know that $EF$ is perpendicular to and bisects $BB'$ . The slope of $BB'$ is thus $\frac{-l}{14}$ , and so the equation of $BB'$ is $y -25 = \frac{-l}{14}x$ . Let the point of intersection of $EF, BB'$ be $G$ . Then the y-coordinate of $G$ is $\frac{25}{2}$ , so \begin{align*} \frac{14}{l}x &= y-8 = \frac{9}{2}\\ \frac{-l}{14}x &= y-25 = -\frac{25}{2}\\ \end{align*} Dividing the two equations yields
The answer is $\boxed{293}$ as above.

Solution:
Firstly, observe that if we are given that $AE=8$ and $BE=17$ , the length of the triangle is given and the height depends solely on the length of $CF$ . Let Point $A = (0,0)$ . Since $AE=8$ , point E is at (8,0). Next, point $B$ is at $(25,0)$ since $BE=17$ and point $B'$ is at $(0,-15)$ since $BE=AE$ by symmetry. Draw line segment $BB'$ . Notice that this is perpendicular to $EF$ by symmetry. Next, find the slope of EB, which is $\frac{15}{25}=\frac{3}{5}$ . Then, the slope of $EF$ is - $\frac{5}{3}$
Line EF can be written as y= $-\frac{5}{3}x+b$ . Plug in the point $(8,0)$ , and we get the equation of EF to be y= $_\frac{5}{3}x+\frac{40}{3}$ . Since the length of $AB$ =25, a point on line $EF$ lies on $DC$ when $x=25-3=22$ . Plug in $x=22$ into our equation to get $y=-\frac{70}{3}$ $|y|=BC=\frac{70}{3}$ . Therefore, our answer is $2(AB+BC)=2\left(25+\frac{70}{3}\right)=2\left(\frac{145}{3}\right)=\frac{290}{3}= \boxed{293}$

Solution:
Firstly, note that $B'E=BE=17$ , so $AB'=\sqrt{17^2-8^2}=15$ . Then let $\angle BEF=\angle B'EF=\theta$ , so $\angle B'EA = \pi-2\theta$ . Then $\tan(\pi-2\theta)=\frac{15}{8}$ , or
\[\frac{2\tan(\theta)}{\tan^2(\theta)-1}=\frac{15}{8}\] using supplementary and double angle identities. Multiplying though and factoring yields
\[(3\tan(\theta)-5)(5\tan(\theta)+3)=0\]
It is clear from the problem setup that $0<\theta<\frac\pi2$ , so the correct value is $\tan(\theta)=\frac53$ . Next, extend rays $\overrightarrow{BC}$ and $\overrightarrow{EF}$ to intersect at $C'$ . Then $\tan(\theta)=\frac{BC'}{17}=\frac53$ , so $BC'=\frac{85}{3}$ . By similar triangles, $CC'=\frac{3}{17}BC'=\frac{15}{3}$ , so $BC=\frac{70}{3}$ . The perimeter is $\frac{140}{3}+50=\frac{290}{3}\Longrightarrow \boxed{293}$

Solution:
Use the prepared diagram for this solution.
Call the intersection of DF and B'C' G. AB'E is an 8-15-17 right triangle, and so are B'DG and C'FG. Since C'F is 3, then using the properties of similar triangles GF is 51/8. DF is 22, so DG is 125/8. Finally, DB can to calculated to be 25/3.
Add all the sides together to get $\boxed{293}$

Solution:
Call the intersection of $B'C'$ $BC$ , and $EF$ $G$ . Since $FCBE$ and $FC'B'E$ are congruent, we know that the three lines intersect.
We already know $AB$ so we just need to find $CB$ , call it $x$ . Drop an altitude from $F$ to $AB$ and call it $H$ $EH=EB-FC=14$ . Using Pythagorean Theorem, we have $EF=\sqrt{x^2+14^2}$ . Triangles $EFH$ and $EGB$ are similar (AA), so we get \[\frac{HF}{BG}=\frac{EH}{EB}\] \[\frac{x}{x+GC}=\frac{14}{17}\] Simplify and we get $GC=\frac{3x}{14}$
We find the area of $FCBE$ by using the fact that it is a trapezoid. $[FCBE]=\frac{(3+17)x}{2}=10x$
A different way to find the area: $[FCBE]=\frac{1}{2} EG\cdot($ height of $EGB$ with $EG$ as base $)-[FGC]$
Since $GBE$ and $G'B'E$ are congruent(SAS), their height from $EG$ is the same. $B'B=\sqrt{AB'^2+AB^2}=5\sqrt{34}$ $EG=\sqrt{EB^2+BG^2}=\sqrt{(\frac{17x}{14})^2+17^2}=17\sqrt{\frac{x^2}{196}+1}$
\[[FCBE]=\frac{1}{2} \cdot 17 \cdot \sqrt{\frac{x^2}{196}+1} \cdot \frac{5\sqrt{34}}{2}-\frac{9x}{28}\] \[280x+9x=7\cdot 5 \cdot \sqrt{34} \cdot 17 \cdot \sqrt{\frac{x^2}{196}+1}\] \[17^4 x^2=49 \cdot 25 \cdot 34 \cdot 17^2 \cdot (\frac{x^2}{196}+1)\] \[17x^2=\frac{25}{2}x^2+2450\] \[x=\frac{70}{3}\]
The perimeter is $\frac{140}{3}+50=\frac{290}{3},$ so our answer is $\boxed{293}$

Solution:
Let the endpoint of the intersection of the fold near $F$ be $G$ . Since trapezoid $BCFE$ is folded, it is congruent to trapezoid $B'C'FE$ . Therefore, $BE=B'E=17$ . Since $\triangle AB'E$ is a right triangle, $AB'=15$ from the pythagorean theorem. From here, we can see that triangles $\triangle AEB \sim \triangle DGB' \sim \triangle C'GF$ by AA similarity. From here, we find $BC$ from a lot of similarities. Let $BC=x$
Since $\triangle ABE' \sim \triangle DGB'$
\[\frac {AE}{AB'} = \frac{DB}{DG}\]
\[\frac {8}{15} = \frac {x-15}{DG}\]
\[DG = \frac {15(x-15)}{8}\]
\[GF = DC-DG-FC\]
\[GF = \frac{-15x+401}{8}\]
Since $\triangle ABE' \sim \triangle C'GF'$
\[\frac {AE}{B'E} = \frac {C'F}{GF}\]
\[\frac {8}{17} = \frac{3}{\frac {-15x+401}{8}}\]
from which we get $x= \frac {70}{3}$
Finally, our answer is $2(\frac {70}{3}) + 2(25)=\frac {290}{3}$ , which is $290+3=\boxed{293}$


Answer: 19.0

$ABCD$ is a square of side length $\sqrt{3} + 1$ . Point $P$ is on $\overline{AC}$ such that $AP = \sqrt{2}$ . The square region bounded by $ABCD$ is rotated $90^{\circ}$ counterclockwise with center $P$ , sweeping out a region whose area is $\frac{1}{c} (a \pi + b)$ , where $a$ $b$ , and $c$ are positive integers and $\text{gcd}(a,b,c) = 1$ . What is $a + b + c$
$\textbf{(A)} \ 15 \qquad \textbf{(B)} \ 17 \qquad \textbf{(C)} \ 19 \qquad \textbf{(D)} \ 21 \qquad \textbf{(E)} \ 23$

Solution:
We first note that diagonal $\overline{AC}$ is of length $\sqrt{6} + \sqrt{2}$ . It must be that $\overline{AP}$ divides the diagonal into two segments in the ratio $\sqrt{3}$ to $1$ . It is not difficult to visualize that when the square is rotated, the initial and final squares overlap in a rectangular region of dimensions $2\sqrt{3}$ by $\sqrt{3} + 1$ . The area of the overall region (of the initial and final squares) is therefore twice the area of the original square minus the overlap, or $2 (\sqrt{3} + 1)^2 - 2 (\sqrt{3} + 1) = 2 (4 + 2 \sqrt{3}) - 2 \sqrt{3} - 2 = 6 + 2 \sqrt{3}$
The area also includes $4$ circular segments. Two are quarter-circles centered at $P$ of radii $\sqrt{2}$ (the segment bounded by $\overline{PA}$ and $\overline{PA'}$ ) and $\sqrt{6}$ (that bounded by $\overline{PC}$ and $\overline{PC'}$ ). Assuming $A$ is the bottom-left vertex and $B$ is the bottom-right one, it is clear that the third segment is formed as $B$ swings out to the right of the original square [recall that the square is rotated counterclockwise], while the fourth is formed when $D$ overshoots the final square's left edge. To find these areas, consider the perpendicular from $P$ to $\overline{BC}$ . Call the point of intersection $E$ . From the previous paragraph, it is clear that $PE = \sqrt{3}$ and $BE = 1$ . This means $PB = 2$ , and $B$ swings back inside edge $\overline{BC}$ at a point $1$ unit above $E$ (since it left the edge $1$ unit below). The triangle of the circular sector is therefore an equilateral triangle of side length $2$ , and so the angle of the segment is $60^{\circ}$ . Imagining the process in reverse, it is clear that the situation is the same with point $D$
The area of the segments can be found by subtracting the area of the triangle from that of the sector; it follows that the two quarter-segments have areas $\frac{1}{4} \pi (\sqrt{2})^2 - \frac{1}{2} \sqrt{2} \sqrt{2} = \frac{\pi}{2} - 1$ and $\frac{1}{4} \pi (\sqrt{6})^2 - \frac{1}{2} \sqrt{6} \sqrt{6} = \frac{3 \pi}{2} - 3$ . The other two segments both have area $\frac{1}{6} \pi (2)^2 - \frac{(2)^2 \sqrt{3}}{4} = \frac{2 \pi}{3} - \sqrt{3}$
The total area is therefore \[(6 + 2 \sqrt{3}) + (\frac{\pi}{2} - 1) + (\frac{3 \pi}{2} - 3) + 2 (\frac{2 \pi}{3} - \sqrt{3})\] \[= 2 + 2 \sqrt{3} + 2 \pi + \frac{4 \pi}{3} - 2 \sqrt{3}\] \[= \frac{10 \pi}{3} + 2\] \[= \frac{1}{3} (10 \pi + 6)\]
Since $a = 10$ $b = 6$ , and $c = 3$ , the answer is $a + b + c = 10 + 6 + 3 = \boxed{19}$


Answer: 4.0

$ABCDE$ is a regular pentagon. $AP, AQ$ and $AR$ are the perpendiculars dropped from $A$ onto $CD, CB$ extended and $DE$ extended,
respectively. Let $O$ be the center of the pentagon. If $OP = 1$ , then $AO + AQ + AR$ equals
[asy] defaultpen(fontsize(10pt)+linewidth(.8pt)); pair O=origin, A=2*dir(90), B=2*dir(18), C=2*dir(306), D=2*dir(234), E=2*dir(162), P=(C+D)/2, Q=C+3.10*dir(C--B), R=D+3.10*dir(D--E), S=C+4.0*dir(C--B), T=D+4.0*dir(D--E); draw(A--B--C--D--E--A^^E--T^^B--S^^R--A--Q^^A--P^^rightanglemark(A,Q,S)^^rightanglemark(A,R,T)); dot(O); label("$O$",O,dir(B)); label("$1$",(O+P)/2,W); label("$A$",A,dir(A)); label("$B$",B,dir(B)); label("$C$",C,dir(C)); label("$D$",D,dir(D)); label("$E$",E,dir(E)); label("$P$",P,dir(P)); label("$Q$",Q,dir(Q)); label("$R$",R,dir(R)); [/asy]
$\textbf{(A)}\ 3\qquad \textbf{(B)}\ 1 + \sqrt{5}\qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 2 + \sqrt{5}\qquad \textbf{(E)}\ 5$

Solution:
To solve the problem, we compute the area of regular pentagon $ABCDE$ in two different ways. First, we can divide regular pentagon $ABCDE$ into five congruent triangles.
If $s$ is the side length of the regular pentagon, then each of the triangles $AOB$ $BOC$ $COD$ $DOE$ , and $EOA$ has base $s$ and height 1, so the area of regular pentagon $ABCDE$ is $5s/2$
Next, we divide regular pentagon $ABCDE$ into triangles $ABC$ $ACD$ , and $ADE$
 Triangle $ACD$ has base $s$ and height $AP = AO + 1$ . Triangle $ABC$ has base $s$ and height $AQ$ . Triangle $ADE$ has base $s$ and height $AR$ . Therefore, the area of regular pentagon $ABCDE$ is also \[\frac{s}{2} (AO + AQ + AR + 1).\] Hence, \[\frac{s}{2} (AO + AQ + AR + 1) = \frac{5s}{2},\] which means $AO + AQ + AR + 1 = 5$ , or $AO + AQ + AR = \boxed{4}$


Answer: 4.0

Solution: Now, we know that angle $D$ has measure $\frac{180 \cdot 3}{5} = 108$ . Since \[\sin 54 = \frac{OP}{DO} = \frac{1}{DO}, DO = \frac{1}{\sin 54}\] \[\tan 54 = \frac{OP}{DP} = \frac{1}{DP}, DP = \frac{1}{\tan 54}\] Therefore, $AB = 2DP = \frac{2}{\tan 54}$ \[\sin 72 = \frac{AQ}{AB} = AQ \tan 54 \cdot \frac{1}{2}, AQ = \frac{2 \sin 72}{\tan 54}\] Therefore, $AO + AQ + AR = AO + 2AQ = \frac{1}{\sin 54}+\frac{4 \sin 72}{\tan 54} = \frac{1}{\sin 54} + 8 \sin 36 \cos 54 = \frac{1}{\cos 36} + 8-8\cos^2(36)$ . Recalling that $\cos 36 = \frac{1 + \sqrt{5}}{4}$ gives a final answer of $\boxed{4}$


Answer: 30.0

$R$ varies directly as $S$ and inversely as $T$ . When $R = \frac{4}{3}$ and $T = \frac {9}{14}$ $S = \frac37$ . Find $S$ when $R = \sqrt {48}$ and $T = \sqrt {75}$
$\textbf{(A)}\ 28\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 42\qquad\textbf{(E)}\ 60$

Solution:
\[R=c\cdot\frac{S}T\]
for some constant $c$
You know that
\[\frac43=c\cdot\frac{3/7}{9/14}=c\cdot\frac37\cdot\frac{14}9=c\cdot\frac23\,,\]
so
\[c=\frac{4/3}{2/3}=2\,.\]
When $R=\sqrt{48}$ and $T=\sqrt{75}$ we have
\[\sqrt{48}=\frac{2S}{\sqrt{75}}\,,\]
so
\[S=\frac12\sqrt{48\cdot75}=30\,.\] $\boxed{30}$


Answer: 2.0

$[x-(y-z)] - [(x-y) - z] =$
$\textbf{(A)}\ 2y \qquad \textbf{(B)}\ 2z \qquad \textbf{(C)}\ -2y \qquad \textbf{(D)}\ -2z \qquad \textbf{(E)}\ 0$

Solution: The expression becomes $(x-y+z)-(x-y-z) = x-y+z-x+y+z = 2z$ , which is $\boxed{2}$


Answer: 35.0

Solution:
Using the left triangle, we have:
$\angle 1 + 70 + 40 = 180$
$\angle 1 = 180 - 110$
$\angle 1 = 70$
Using the given fact that $\angle 1 + \angle 2 = 180$ , we have $\angle 2 = 180 - 70 = 110$
Finally, using the right triangle, and the fact that $\angle 3 = \angle 4$ , we have:
$\angle 2 + \angle 3 + \angle 4 = 180$
$110 + \angle 3 + \angle 4 = 180$
$110 + \angle 4+ \angle 4 = 180$
$2\angle 4 = 70$
$\angle 4 = 35$
Thus, the answer is $\boxed{35}$


Answer: 1.0

Solution: \begin{align*} \dfrac{10-9+8-7+6-5+4-3+2-1}{1-2+3-4+5-6+7-8+9} &= \dfrac{(10-9)+(8-7)+(6-5)+(4-3)+(2-1)}{1+(-2+3)+(-4+5)+(-6+7)+(-8+9)} \\ &= \dfrac{1+1+1+1+1}{1+1+1+1+1} \\ &= 1 \rightarrow \boxed{1}


Answer: 0.1997

Solution:
Convert each fraction to a decimal.
$\dfrac{1}{10} + \dfrac{9}{100} + \dfrac{9}{1000} + \dfrac{7}{10000}$
$0.1+0.09+0.009+0.0007=0.1997$
$\boxed{0.1997}$


Answer: 10.0

Solution:
$1+ 2+ 3 + 4 + 5 + 6 + 7 + 8 + 9 = \dfrac{(9)(10)}{2} = 45$
$\frac{45+55}{10} = \dfrac{100}{10} = \boxed{10}$


Answer: 37.0

$\diamondsuit$ and $\Delta$ are whole numbers and $\diamondsuit \times \Delta =36$ .  The largest possible value of $\diamondsuit + \Delta$ is
$\text{(A)}\ 12 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 20\ \qquad \text{(E)}\ 37$

Solution:
Since it doesn't take too long, we can just make a table with all the possible values of the sum \[\begin{array}{|c|c|c|} \multicolumn{3}{}{} \\ \hline \diamondsuit & \Delta & \diamondsuit + \Delta \\ \hline 36 & 1 & 37 \\ \hline 18 & 2 & 20 \\ \hline 12 & 3 & 15 \\ \hline 9 & 4 & 13 \\ \hline 6 & 6 & 12 \\ \hline \end{array}\]
Clearly $37\rightarrow \boxed{37}$ is the largest.


Answer: 18.0

$\emph{triangular number}$ is a positive integer that can be expressed in the form $t_n = 1+2+3+\cdots+n$ , for some positive integer $n$ . The three smallest triangular numbers that are also perfect squares are $t_1 = 1 = 1^2$ $t_8 = 36 = 6^2$ , and $t_{49} = 1225 = 35^2$ . What is the sum of the digits of the fourth smallest triangular number that is also a perfect square?
$\textbf{(A) } 6 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 27$

Solution:
We have $t_n = \frac{n (n+1)}{2}$ .
If $t_n$ is a perfect square, then it can be written as $\frac{n (n+1)}{2} = k^2$ ,
where $k$ is a positive integer.
Thus, $n (n+1) = 2 k^2$ . Rearranging, we get $(2n+1)^2-2(2k)^2=1$ , a Pell equation. So $\frac{2n+1}{2k}$ must be a truncation of the continued fraction for $\sqrt{2}$
\begin{eqnarray*} 1+\frac12&=&\frac{2\cdot1+1}{2\cdot1}\\ 1+\frac1{2+\frac1{2+\frac12}}&=&\frac{2\cdot8+1}{2\cdot6}\\ 1+\frac1{2+\frac1{2+\frac1{2+\frac1{2+\frac12}}}}&=&\frac{2\cdot49+1}{2\cdot35}\\ 1+\frac1{2+\frac1{2+\frac1{2+\frac1{2+\frac1{2+\frac1{2+\frac12}}}}}}&=&\frac{2\cdot288+1}{2\cdot204} \end{eqnarray*}
Therefore, $t_{288} = \frac{288\cdot289}{2} = 204^2 = 41616$ , so the answer is $4+1+6+1+6=\boxed{18}$

Solution: As mentioned above, $t_n = \frac{n (n+1)}{2}$ . If $t_n$ is a perfect square, one of two things must occur when the fraction is split into a product. Either $\frac{n}{2}$ and $n+1$ must both be squares, or $n$ and $\frac{n+1}{2}$ must both be squares, and thus the search for the next perfect square triangular number can be narrowed down by testing values of $n$ that are close to or are perfect squares. After some work, we reach $n = 288$ $1$ less than $289$ , and $t_{288} = \frac{288\cdot289}{2} = 144 \cdot 289 = 41616$ . This product is a perfect square, and thus the sum of the digits of the fourth smallest perfect square triangular number is therefore $4+1+6+1+6=\boxed{18}$

Solution:
We want to find integer $n_i$ and $m_i$ such that $t_{n_i} =\frac{n_i (n_i + 1)}{2}=m_i^2, n_0 = 0.$
We use the formula $\sqrt{n_{i+1}} = \sqrt{2n_i} + \sqrt{n_i + 1}$ and get
\[n_1 = ( \sqrt{2n_0} + \sqrt{n_0 + 1})^2 = (0+1)^2 = 1,\] \[n_2= ( \sqrt{2n_1} + \sqrt{n_1 + 1})^2 = ( \sqrt{2} + \sqrt{1 + 1})^2 = 8,\] \[n_3= ( \sqrt{2n_2} + \sqrt{n_2 + 1})^2 = ( \sqrt{16} + \sqrt{8 + 1})^2 = 49,\] \[n_4 = ( \sqrt{2n_3} + \sqrt{n_3 + 1})^2 = ( \sqrt{98} + \sqrt{49 + 1})^2 =  ((7 + 5)\sqrt{2})^2 = 288,\] \[n_5 = ( \sqrt{2n_4} + \sqrt{n_4 + 1})^2 = ( \sqrt{576} + \sqrt{289})^2 =  (24 + 17)^2 = 1681,\] \[n_6 = ( \sqrt{2n_5} + \sqrt{n_5 + 1})^2 = ( 41\sqrt{2} + \sqrt{1682})^2 =  ((41 + 29)\sqrt{2})^2 = 9800,...\] Therefore, $t_{n_4} = t_{288} = \frac{288\cdot289}{2} = 41616 \implies 4+1+6+1+6=\boxed{18}$


Answer: 18.0

Solution: According to the problem, we want to find integer $p$ such $\frac{n(n+1)}{2}=p^2$ , after expanding, we have $n^2+n=2p^2, 4n^2+4n=8p^2, (2n+1)^2-8p^2=1$ , we call $2n+1=q$ , the equation becomes $q^2-8p^2=1$ , obviously $(q,p)=(3,1)$ is the elementary solution for this pell equation, thus the forth smallest solution set $q_4+2\sqrt{2}p_4=(3+2\sqrt{2})^4=577+408\sqrt{2}$ , which indicates $p=204, p^2=41616$ leads to $\boxed{18}$

Solution:
If $n \choose 2$ is a square, then ${(2n-1)^2 \choose 2}$ is also a square. We can prove this quite simply:
\[{(2n-1)^2 \choose 2}\] \[= \frac{(2n-1)^2 \cdot ((2n-1)^2 - 1)}{2}\] \[= \frac{(2n-1)^2 \cdot (2n \cdot (2n - 2))}{2}\] \[= (2n-1)^2 \cdot 4{n \choose 2}.\]
Therefore, ${(2 \cdot 9 - 1)^2 \choose 2}$ is a square. Note that $T_n = {n+1 \choose 2}$ . We can easily check all smaller possibilities using a bit of casework, and they don't work. Our solution is thus ${289 \choose 2} = 204^2 = 41616$ , and so the answer is $\boxed{18}$


Answer: 120.0

Solution:
The numerator is $(3!)! = 6!$
The denominator is $3! = 6$
Using the property that $6! = 6 \cdot 5!$ in the numerator, the sixes cancel, leaving $5! = 120$ , which is answer $\boxed{120}$


Answer: 200.0

Solution:
We immediately see some canceling. We see powers of ten on the top and on the bottom of the fraction, and we make quick work of this: \[\frac{10^7}{5 \times 10^4} = \frac{10^3}{5}\]
We know that $10^3 = 10 \times 10 \times 10$ , so
\begin{align*} \frac{10^3}{5} &= \frac{10\times 10\times 10}{5} \\ &= 2\times 10\times 10 \\ &= 200 \\ \end{align*}
So the answer is $\boxed{200}$


Answer: 12.0

Solution: \begin{align*} \frac{16+8}{4-2} &= \frac{24}{2} \\ &= 12\rightarrow \boxed{12}


Answer: 0.3

Solution: Each of the fractions simplify to $\frac{1}{10}$ , so this sum is \begin{align*} \frac{1}{10}+\frac{1}{10}+\frac{1}{10} &= \frac{3}{10} \\ &= .3 \rightarrow \boxed{.3}


Answer: 0.246

Solution: \begin{align*} \frac{2}{10}+\frac{4}{100}+\frac{6}{1000} &= \frac{200}{1000}+\frac{40}{1000}+\frac{6}{1000} \\ &= \frac{246}{1000} \\ &= .246 \rightarrow \boxed{.246}


Answer: 0.08

Solution: $\frac{2}{25}=\frac{2\cdot 4}{25\cdot 4} = \frac{8}{100} = 0.08\rightarrow \boxed{.08}$


Answer: 1.0

Solution:
First, note that we can pull the exponents out of every factor, since they are all squared. This results in $\left(\frac{(x+1)(x^{2}-x+1)}{x^{3}+1}\right)^{4}\cdot\left(\frac{(x-1)(x^{2}+x+1)}{x^{3}-1}\right)^{4}$ Now, multiplying the numerators together gives $\left(\frac{x^3+1}{x^3+1}\right)^{4}\cdot\left(\frac{x^3-1}{x^3-1}\right)^{4}$ ,
which simplifies to $\boxed{1}$


Answer: 1.0

$\log p+\log q=\log(p+q)$ only if:
$\textbf{(A) \ }p=q=\text{zero}  \qquad \textbf{(B) \ }p=\frac{q^2}{1-q} \qquad \textbf{(C) \ }p=q=1 \qquad$
$\textbf{(D) \ }p=\frac{q}{q-1} \qquad \textbf{(E) \ }p=\frac{q}{q+1}$

Solution: $\log p+\log q=\log(p+q)\implies \log pq=\log(p+q)\implies pq=p+q\implies \boxed{1}$


Answer: 24.0

$\text{palindrome}$ , such as $83438$ , is a number that remains the same when its digits are reversed. The numbers $x$ and $x+32$ are three-digit and four-digit palindromes, respectively. What is the sum of the digits of $x$
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 23 \qquad \textbf{(E)}\ 24$

Solution:
$x$ is at most $999$ , so $x+32$ is at most $1031$ . The minimum value of $x+32$ is $1000$ . However, the only palindrome between $1000$ and $1032$ is $1001$ , which means that $x+32$ must be $1001$
It follows that $x$ is $969$ , so the sum of the digits is $\boxed{24}$

Solution: For $x+32$ to be a four-digit number, $x$ is in between $968$ and $999$ . The palindromes in this range are $969$ $979$ $989$ , and $999$ , so the sum of the digits of $x$ can be $24$ $25$ $26$ , or $27$ . Only $\boxed{24}$ is an option, and upon checking, $x+32=1001$ is indeed a palindrome.

Solution:
Since we know $x+32$ to be $1 a a 1$ and the only palindrome that works is $0 = a$ , that means $x+32 = 1001$ , and so $x = 1001 - 32 = 969$ . So, $9$ $6$ $9$ $\boxed{24}$ .
~songmath20


Answer: 24.0

Solution:
$x$ is at most $999$ , so $x+32$ is at most $1031$ . The minimum value of $x+32$ is $1000$ . However, the only palindrome between $1000$ and $1032$ is $1001$ , which means that $x+32$ must be $1001$
It follows that $x$ is $969$ , so the sum of the digits is $\boxed{24}$

Solution: For $x+32$ to be a four-digit number, $x$ is in between $968$ and $999$ . The palindromes in this range are $969$ $979$ $989$ , and $999$ , so the sum of the digits of $x$ can be $24$ $25$ $26$ , or $27$ . Only $\boxed{24}$ is an option, and upon checking, $x+32=1001$ is indeed a palindrome.

Solution:
Since we know $x+32$ to be $1 a a 1$ and the only palindrome that works is $0 = a$ , that means $x+32 = 1001$ , and so $x = 1001 - 32 = 969$ . So, $9$ $6$ $9$ $\boxed{24}$ .
~songmath20


Answer: 55.0

$\triangle ABC$ has a right angle at $C$ and $\angle A = 20^\circ$ . If $BD$ $D$ in $\overline{AC}$ ) is the bisector of $\angle ABC$ , then $\angle BDC =$
$\textbf{(A)}\ 40^\circ \qquad \textbf{(B)}\ 45^\circ \qquad \textbf{(C)}\ 50^\circ \qquad \textbf{(D)}\ 55^\circ\qquad \textbf{(E)}\ 60^\circ$

Solution: Since $\angle C = 90^{\circ}$ and $\angle A = 20^{\circ}$ , we have $\angle ABC = 70^{\circ}$ . Thus $\angle DBC = 35^{\circ}$ . It follows that $\angle BDC = 90^{\circ} - 35^{\circ} = 55^{\circ}$ , which is $\boxed{55}$


Answer: 84.0

$\triangle$ or $\bigcirc$ is placed in each of the nine squares in a $3$ -by- $3$ grid. Shown below is a sample configuration with three $\triangle$ s in a line. [asy] //diagram size(5cm); defaultpen(linewidth(1.5)); real r = 0.37; path equi = r * dir(-30) -- (r+0.03) * dir(90) -- r * dir(210) -- cycle; draw((0,0)--(0,3)--(3,3)--(3,0)--cycle); draw((0,1)--(3,1)--(3,2)--(0,2)--cycle); draw((1,0)--(1,3)--(2,3)--(2,0)--cycle); draw(circle((3/2,5/2),1/3)); draw(circle((5/2,1/2),1/3)); draw(circle((3/2,3/2),1/3)); draw(shift(0.5,0.38) * equi); draw(shift(1.5,0.38) * equi); draw(shift(0.5,1.38) * equi); draw(shift(2.5,1.38) * equi); draw(shift(0.5,2.38) * equi); draw(shift(2.5,2.38) * equi); [/asy] How many configurations will have three $\triangle$ s in a line and three $\bigcirc$ s in a line?
$\textbf{(A) } 39 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 78 \qquad \textbf{(D) } 84 \qquad \textbf{(E) } 96$

Solution:
Notice that diagonals and a vertical-horizontal pair can never work, so the only possibilities are if all lines are vertical or if all lines are horizontal. These are essentially the same, so we'll count up how many work with all lines of shapes vertical, and then multiply by 2 at the end.
We take casework:
Case 1: 3 lines :
In this case, the lines would need to be $2$ of one shape and $1$ of another, so there are $\frac{3!}{2} = 3$ ways to arrange the lines and $2$ ways to pick which shape has only one line. In total, this is $3\cdot 2 = 6.$
Case 2: 2 lines :
In this case, the lines would be one line of triangles, one line of circles, and the last one can be anything that includes both shapes. There are $3! = 6$ ways to arrange the lines and $2^3-2 = 6$ ways to choose the last line. (We subtract $2$ from the last line because one arrangement of the last line is all triangles and the other arrangement of the last line is all circles, which causes Case 2 to overlap with Case 1 and further complicating the solution.) In total, this is $6\cdot 6 = 36.$
Finally, we add and multiply: $2(36+6)=2(42)=\boxed{84}$

Solution: We will only consider cases where the three identical symbols are the same column, but at the end we shall double our answer as the same holds true for rows. There are $3$ ways to choose a column with all $\bigcirc$ 's and $2$ ways to choose a column with all $\triangle$ 's. The third column can be filled in $2^3=8$ ways. Therefore, we have a total of $3\cdot2\cdot8=48$ cases. However, we overcounted the cases with $2$ complete columns of with one symbol and $1$ complete column with another symbol. This happens in $2\cdot3=6$ cases. $48-6=42$ . However, we have to remember to double our answer, giving us $\boxed{84}$ ways to complete the grid.


Answer: 36.0

Solution:
The best way to solve this problem is to find patterns and to utilize them to our advantage. For example, we can't really do anything without knowing how many squares there are in the 37th row. But who wants to continue the diagram for 37 rows? And what if the problem said 100,000th row? It'll still be possible - but not if your method is to continue the diagram...
So hopefully there's a pattern. We find a pattern by noticing what is changing from row 1 to row 2. Basically, for the next row, we are just adding 2 squares (1 on each side) to the number of squares we had in the previous row.  So each time we're adding 2. So how can we find $N$ , if $N$ is the number of squares in the $a^\text{th}$ row of this diagram? We can't just say that $N = 1 + 2a$ , because it doesn't work for the first row. But since 1 is the first term, we have to EXCLUDE the first term, meaning that we must subtract 1 from a. Thus, $N = 1 + 2\times(a - 1) = 2a - 1$ . So in the 37th row we will have $2 \times 37 - 1 = 74 - 1 = 73$
You may now be thinking - aha, we're finished. But we're only half finished. We still need to find how many black squares there are in these 73 squares. Well let's see - they alternate white-black-white-black... but we can't divide by two - there aren't exactly as many white squares as black squares... there's always 1 more white square... aha! If we subtract 1 from the number of squares (1 white square), we will have exactly 2 times the number of black squares.
Thus, the number of black squares is $\frac{73 - 1}{2} = \frac{72}{2} = 36$
36 is $\boxed{36}$

Solution:
Note that each row adds one black and one white square to the end of the previous one, and then shifts the new row over a little.  Since the first row had no black squares, the number of black squares in any row is one less than the row number.
Now that this has been established, we just have $37-1=36\rightarrow \boxed{36}$

Solution: We can easily spot the pattern. In the first row, there are $0$ black squares. In the second row, there is $1$ black square. So the pattern is that in order to find the number of black squares in a row, you need to subtract $1$ from the row number. Therefore, the number of black squares in the $37\text{th}$ row we subtract 1. $37-1=36\rightarrow \boxed{36}$


Answer: 52.0

A 100 foot long moving walkway moves at a constant rate of 6 feet per second.  Al steps onto the start of the walkway and stands.  Bob steps onto the start of the walkway two seconds later and strolls forward along the walkway at a constant rate of 4 feet per second.  Two seconds after that, Cy reaches the start of the walkway and walks briskly forward beside the walkway at a constant rate of 8 feet per second.  At a certain time, one of these three persons is exactly halfway between the other two.  At that time, find the distance in feet between the start of the walkway and the middle person.

Solution:
Clearly we have people moving at speeds of $6,8$ and $10$ feet/second. Notice that out of the three people, Cy is at the largest disadvantage to begin with and since all speeds are close, it is hardest for him to catch up. Furthermore, Bob is clearly the farthest along. Thus it is reasonable to assume that there is some point when Al is halfway between Cy and Bob. At this time $s$ , we have that
$\frac{8(s-4)+10(s-2)}{2}=6s$ After solving, $s=\frac{26}{3}$ . At this time, Al has traveled $6\cdot\frac{26}{3}=52$ feet.
We could easily check that Al is in the middle by trying all three possible cases. $\frac{6s + 8(s-4)}{2} = 10(s-2)$ yields that $s = \frac 43$ , which can be disregarded since both Bob and Cy hadn't started yet. $\frac{6s + 10(s-2)}{2} = 8(s-4)$ yields that $-10=-32$ , a contradiction. Thus, the answer is $\boxed{52}$


Answer: 1698.0

A 16-step path is to go from $(-4,-4)$ to $(4,4)$ with each step increasing either the $x$ -coordinate or the $y$ -coordinate by 1. How many such paths stay outside or on the boundary of the square $-2 \le x \le 2$ $-2 \le y \le 2$ at each step?
$\textbf{(A)}\ 92 \qquad \textbf{(B)}\ 144 \qquad \textbf{(C)}\ 1568 \qquad \textbf{(D)}\ 1698 \qquad \textbf{(E)}\ 12,800$

Solution:
As stated in the solution, there are $6$ points along the line $y=-x$ that constitute a sort of "boundary". Once the ant reaches one of these $6$ points, it is exactly halfway to $(4, 4)$ . Also notice that the ant will only cross one of the $6$ points during any one of its paths. Therefore we can divide the problem into $3$ cases, focusing on $1$ quadrant; then multiplying the sum by $2$ to get the total (because there is symmetry).
For the sake of this explanation, we will focus on the fourth quadrant (it really doesn't matter which quadrant because, again the layout is symmetrical) The three cases are when the ant crosses $(4, -4), (3, -3),$ and $(2, -2)$
For each of the cases, notice that the path the ant takes can be expressed as a sequence of steps, such as:
right, right, up, right,..., etc.
Also notice that there are always $8$ steps per sequence (if there were more or less steps, the ant would be breaking the conditions given in the problem). This means we can figure out the number of ways to get to a point based on the particular sequence of steps that denote each path. For example, there is only one way for the ant to pass through $(4, -4);$ it MUST keep traveling right for all $8$ steps. This seems fairly obvious; however, notice that this is equivalent to
$\binom{8}{0}$
Now we consider the number of ways to get from $(4, -4)$ to $(4, 4)$ . by symmetry, there is only $1$ such way. So the number of paths containing $(4, -4)$ is $1^2,$ or $1$
Moving on to the next case, we see that the ant MUST travel right exactly $7$ times and up exactly once. So each sequence of this type will have $7$ "right"s and $1$ "up". So, the total number of paths that go through $(3, -3)$ is equivalent to the number of ways to arrange $1$ "up" into $8$ spots. This is
$\binom{8}{1} = 8$
Similarly to the first case, we square this value to account for the second half of the journey: $8^2 = 64$
Finally, for the third case (ant passes through $(2, -2)$ ) the ant must travel right exactly $6$ times and up exactly $2$ times. This is equivalent to the number of ways to arrange $2$ "ups" in a sequence of $8$ movements, or
$\binom{8}{2} = 28$
Again, we square $28$ $28^2 = 784$ . Adding up all of these cases we get
$1+64+784 = 849$
paths through the fourth quadrant. Doubling this number to account for the paths through the second quadrant, we have
$849*2=1698 \Rightarrow \boxed{1698}$


Answer: 1698.0

Solution:
Each path must go through either the second or the fourth quadrant.
Each path that goes through the second quadrant must pass through exactly one of the points $(-4,4)$ $(-3,3)$ , and $(-2,2)$
There is $1$ path of the first kind, ${8\choose 1}^2=64$ paths of the second kind, and ${8\choose 2}^2=28^2=784$ paths of the third type.
Each path that goes through the fourth quadrant must pass through exactly one of the points $(4,-4)$ $(3,-3)$ , and $(2,-2)$ .
Again, there is $1$ path of the first kind, ${8\choose 1}^2=64$ paths of the second kind, and ${8\choose 2}^2=28^2=784$ paths of the third type.
Hence the total number of paths is $2(1+64+784) = \boxed{1698}$


Answer: 90.0

A 3x3x3 cube is made of $27$ normal dice. Each die's opposite sides sum to $7$ . What is the smallest possible sum of all of the values visible on the $6$ faces of the large cube?
$\text{(A)}\ 60 \qquad \text{(B)}\ 72 \qquad \text{(C)}\ 84 \qquad \text{(D)}\ 90 \qquad \text{(E)} 96$

Solution: In a 3x3x3 cube, there are $8$ cubes with three faces showing, $12$ with two faces showing and $6$ with one face showing. The smallest sum with three faces showing is $1+2+3=6$ , with two faces showing is $1+2=3$ , and with one face showing is $1$ . Hence, the smallest possible sum is $8(6)+12(3)+6(1)=48+36+6=90$ . Our answer is thus $\boxed{90}$


Answer: 18.0

A bag contains only blue balls and green balls.  There are $6$ blue balls.  If the probability of drawing a blue ball at random from this bag is $\frac{1}{4}$ , then the number of green balls in the bag is
$\text{(A)}\ 12 \qquad \text{(B)}\ 18 \qquad \text{(C)}\ 24 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 36$

Solution: The total number of balls in the bag must be $4\times 6=24$ , so there are $24-6=18$ green balls $\rightarrow \boxed{18}$


Answer: 5.0

A ball is dropped from a height of $3$ meters. On its first bounce it rises to a height of $2$ meters. It keeps falling and bouncing to $\frac{2}{3}$ of the height it reached in the previous bounce. On which bounce will it not rise to a height of $0.5$ meters?
$\textbf{(A)}\  3 \qquad \textbf{(B)}\  4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$

Solution: Each bounce is $2/3$ times the height of the previous bounce. The first bounce reaches $2$ meters, the second $4/3$ , the third $8/9$ , the fourth $16/27$ , and the fifth $32/81$ . Half of $81$ is $40.5$ , so the ball does not reach the required height on bounce $\boxed{5}$


Answer: 5.0

A bank charges $\textdollar{6}$ for a loan of $\textdollar{120}$ . The borrower receives $\textdollar{114}$ and repays the loan in $12$ easy installments of $\textdollar{10}$ a month. The interest rate is approximately:
$\textbf{(A)}\ 5 \% \qquad \textbf{(B)}\ 6 \% \qquad \textbf{(C)}\ 7 \% \qquad \textbf{(D)}\ 9\% \qquad \textbf{(E)}\ 15 \%$

Solution: The borrower pays $\textdollar{120}$ in a single year for a loan of $\textdollar{114}$ . This means that the bank charges an interest of $\textdollar{6}$ for a loan of $\textdollar{114}$ over a single year, so that the annual interest rate is $100*\frac{\textdollar{6}}{\textdollar{114}} = \frac{100}{19} \approx 5$ percentage points. Therefore, our answer is $\boxed{5}$ and we are done.


Answer: 2.5

A bar graph shows the number of hamburgers sold by a fast food chain each season.  However, the bar indicating the number sold during the winter is covered by a smudge.  If exactly $25\%$ of the chain's hamburgers are sold in the fall, how many million hamburgers are sold in the winter?
[asy] size(250);  void bargraph(real X, real Y, real ymin, real ymax, real ystep, real tickwidth,  string yformat, Label LX, Label LY, Label[] LLX, real[] height,pen p=nullpen) { draw((0,0)--(0,Y),EndArrow); draw((0,0)--(X,0),EndArrow); label(LX,(X,0),plain.SE,fontsize(9)); label(LY,(0,Y),plain.NW,fontsize(9)); real yscale=Y/(ymax+ystep);  for(real y=ymin; y<ymax; y+=ystep) { draw((-tickwidth,yscale*y)--(0,yscale*y)); label(format(yformat,y),(-tickwidth,yscale*y),plain.W,fontsize(9)); }  int n=LLX.length; real xscale=X/(2*n+2); for(int i=0;i<n;++i) { real x=xscale*(2*i+1); path P=(x,0)--(x,height[i]*yscale)--(x+xscale,height[i]*yscale)--(x+xscale,0)--cycle; fill(P,p); draw(P); label(LLX[i],(x+xscale/2),plain.S,fontsize(10)); } for(int i=0;i<n;++i) draw((0,height[i]*yscale)--(X,height[i]*yscale),dashed); }  string yf="%#.1f"; Label[] LX={"Spring","Summer","Fall","Winter"}; for(int i=0;i<LX.length;++i) LX[i]=rotate(90)*LX[i]; real[] H={4.5,5,4,4};  bargraph(60,50,1,5.1,0.5,2,yf,"season","hamburgers (millions)",LX,H,yellow); fill(ellipse((45,30),7,10),brown); [/asy]
$\text{(A)}\ 2.5 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 3.5 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 4.5$

Solution:
What we want to find is the number of hamburgers sold in the winter. Since we don't know what it is, let's call it $x$ . From the graph, we know that in Spring, 4.5 million hamburgers were sold, in the Summer was 5 million and in the Fall was 4 million. We know that the number of hamburgers sold in Fall is exactly $\frac{1}{4}$ of the total number of hamburgers sold, so we can say that...
$4 \times \text{Fall} = \text{Spring} + \text{Winter} + \text{Fall} + \text{Summer}$
$4 \times 4 = 4.5 + 4 + x + 5$
$16 = x + 13.5$
$2.5 = x$
The answer is 2.5, or $\boxed{2.5}$


Answer: 590.0

A barn with a roof is rectangular in shape, $10$ yd. wide, $13$ yd. long and $5$ yd. high.  It is to be painted inside and outside, and on the ceiling, but not on the roof or floor. The total number of sq. yd. to be painted is:
$\mathrm{(A) \ } 360 \qquad \mathrm{(B) \ } 460 \qquad \mathrm{(C) \ } 490 \qquad \mathrm{(D) \ } 590 \qquad \mathrm{(E) \ } 720$

Solution:
The walls are $13*5=65$ and $10*5=50$ in area, and the ceiling has an area of $10*13=130$
$((65+50)2)2+130=590 \Rightarrow \boxed{590}$


Answer: 0.7

A base-10 three digit number $n$ is selected at random. Which of the following is closest to the probability that the base-9 representation and the base-11 representation of $n$ are both three-digit numerals?
$\mathrm{(A) \ } 0.3\qquad \mathrm{(B) \ } 0.4\qquad \mathrm{(C) \ } 0.5\qquad \mathrm{(D) \ } 0.6\qquad \mathrm{(E) \ } 0.7$

Solution:
To be a three digit number in base-10:
$10^{2} \leq n \leq 10^{3}-1$
$100 \leq n \leq 999$
Thus there are $900$ three-digit numbers in base-10
To be a three-digit number in base-9:
$9^{2} \leq n \leq 9^{3}-1$
$81 \leq n \leq 728$
To be a three-digit number in base-11:
$11^{2} \leq n \leq 11^{3}-1$
$121 \leq n \leq 1330$
So, $121 \leq n \leq 728$
Thus, there are $608$ base-10 three-digit numbers that are three digit numbers in base-9 and base-11.
Therefore the desired probability is $\frac{608}{900}\approx 0.7 \Rightarrow\boxed{0.7}$


Answer: 48.0

A baseball league consists of two four-team divisions. Each team plays every other team in its division $N$ games. Each team plays every team in the other division $M$ games with $N>2M$ and $M>4$ . Each team plays a $76$ game schedule. How many games does a team play within its own division?
$\textbf{(A) } 36 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 54 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 72$

Solution:
On one team they play $3N$ games in their division and $4M$ games in the other.  This gives $3N+4M=76$
Since $M>4$ we start by trying M=5. This doesn't work because $56$ is not divisible by $3$
Next, $M=6$ does not work because $52$ is not divisible by $3$
We try $M=7$ does work by giving $N=16$ $~M=7$ and thus $3\times 16=\boxed{48}$ games in their division.

Solution:
$76=3N+4M > 10M$ , giving $M \le 7$ .
Since $M>4$ , we have $M=5,6,7$ .
Since $4M$ is $1$ $\pmod{3}$ , we must have $M$ equal to $1$ $\pmod{3}$ , so $M=7$
This gives $3N=48$ , as desired. The answer is $\boxed{48}$

Solution:
Notice that each team plays $N$ games against each of the three other teams in its division. So that's $3N$
Since each team plays $M$ games against each of the four other teams in the other division, that's $4M$
So $3N+4M=76$ , with $M>4, N>2M$
Let's start by solving this Diophantine equation. In other words, $N=\frac{76-4M}{3}$
So $76-4M\equiv0 \pmod{3}$ (remember: $M$ must be divisible by 3 for $N$ to be an integer!). Therefore, after reducing $76$ to $1$ and $-4M$ to $2M$ (we are doing things in $\pmod{3}$ ), we find that $M\equiv1 \pmod{3}$
Since $M>4$ , so the minimum possible value of $M$ is $7$ . However, remember that $N>2M$ ! To find the greatest possible value of M, we assume that $N=2M$ and that is the upper limit of $M$ (excluding that value because $N>2M$ ). Plugging $N=2M$ in, $10M=76$ . So $M<7.6$ . Since you can't have $7.6$ games, we know that we can only check $M=7$ since we know that since $M>4, M<7.6, M\equiv1 \pmod{3}$ . After checking $M=7$ , we find that it works.
So $M=7, N=16$ . So each team plays 16 games against each team in its division. Since they are asking for games in it division, which equals $3n = 48$ . Select $\boxed{48}$


Answer: 660.0

A basketball player has a constant probability of $.4$ of making any given shot, independent of previous shots. Let $a_n$ be the ratio of shots made to shots attempted after $n$ shots. The probability that $a_{10} = .4$ and $a_n\le.4$ for all $n$ such that $1\le n\le9$ is given to be $p^aq^br/\left(s^c\right)$ where $p$ $q$ $r$ , and $s$ are primes, and $a$ $b$ , and $c$ are positive integers. Find $\left(p+q+r+s\right)\left(a+b+c\right)$

Solution: We graph the $10$ shots on a grid. Suppose that a made shot is represented by a step of $(0,1)$ , and a missed shot is represented by $(1,0)$ . Then the basketball player's shots can be represented by the number of paths from $(0,0)$ to $(6,4)$ that always stay below the line $y=\frac{2x}{3}$ . We can find the number of such paths using a Pascal's Triangle type method below, computing the number of paths to each point that only move right and up.  Therefore, there are $23$ ways to shoot $4$ makes and $6$ misses under the given conditions. The probability of each possible sequence occurring is $(.4)^4(.6)^6$ . Hence the desired probability is \[\frac{23\cdot 2^4\cdot 3^6}{5^{10}},\] and the answer is $(23+2+3+5)(4+6+10)=\boxed{660}$

Solution:
The first restriction is that $a_{10} = .4$ , meaning that the player gets exactly 4 out of 10 baskets.  The second restriction is $a_n\le.4$ .  This means that the player may never have a shooting average over 40%.  Thus, the first and second shots must fail, since $\frac{1}{1}$ and $\frac{1}{2}$ are both over $.4$ , but the player may make the third basket, since $\frac{1}{3} \le .4$ In other words, the earliest the first basket may be made is attempt 3.  Using similar reasoning, the earliest the second basket may be made is attempt 5, the earliest the third basket may be made is attempt 8, and the earliest the fourth basket may be made is attempt 10.
Using X to represent a basket and O to represent a failure, this 'earliest' solution may be represented as:
OOXOXOOXOX
To simplify counting, note that the first, second, and tenth shots are predetermined.  The first two shots must fail, and the last shot must succeed.  Thus, only slots 3-9 need to be counted, and can be abbreviated as follows:
XOXOOXO
The problem may be separated into five cases, since the first shot may be made on attempt 3, 4, 5, 6, or 7.  The easiest way to count the problem is to remember that each X may slide to the right, but NOT to the left.
First shot made on attempt 3:
XOXOOXO
XOXOOOX
XOOXOXO
XOOXOOX
XOOOXXO
XOOOXOX
XOOOOXX
Total - 7
First shot made on attempt 4:
Note that all that needs to be done is change each line in the prior case from starting with "XO....." to "OX.....".
Total - 7
First shot made on attempt 5:
OOXXOXO
OOXXOOX
OOXOXXO
OOXOXOX
OOXOOXX
Total - 5
First shot made on attempt 6:
OOOXXXO
OOOXXOX
OOOXOXX
Total - 3
First shot made on attempt 7:
OOOOXXX
Total - 1
The total number of ways the player may satisfy the requirements is $7+7+5+3+1=23$
The chance of hitting any individual combination (say, for example, OOOOOOXXXX) is $\left(\frac{3}{5}\right)^6\left(\frac{2}{5}\right)^4$
Thus, the chance of hitting any of these 23 combinations is $23\left(\frac{3}{5}\right)^6\left(\frac{2}{5}\right)^4 = \frac{23\cdot3^6\cdot2^4}{5^{10}}$
Thus, the final answer is $(23+3+2+5)(6+4+10)=\boxed{660}$

Solution:
Note $a_{10}=.4$ . Therefore the player made 4 shots out of 10. He must make the 10th shot, because if he doesn't, then $a_9=\frac{4}{9}>.4$ . Since $a_n\leq .4$ for all $n$ less than 11, we know that $a_1=a_2=0$ . Now we must look at the 3rd through 9th shot.
Now let's take a look at those un-determined shots. Let's put them into groups: the 3rd, 4th, and 5th shots in group A, and the 6th, 7th, 8th, and 9th shots in group B. The total number of shots made in groups A and B must be 3, since the player makes the 10th shot. We cannot have all three shots made in group A, since $a_5\leq .4$ . Therefore we can have two shots made, one shot made, or no shots made in group A.
Case 1: Group A contains no made shots.
The number of ways this can happen in group A is 1. Now we must arrange the shots in group B accordingly. There are four ways to arrange them total, and all of them work. There are $\textbf{4}$ possibilities here.
Case 2: Group A contains one made shot.
The number of ways this could happen in group A is 3.  Now we must arrange the shots in group B accordingly. There are six ways to arrange them total, but the arrangement "hit hit miss miss" fails, because that would mean $a_7=\frac{3}{7}>.4$ . All the rest work. Therefore there are $3\cdot5=\textbf{15}$ possibilities here.
Case 3: Group A contains two made shots.
The number of ways this could happen in group A is 2 (hit hit miss doesn't work but the rest do). Now we must arrange the shots in group B accordingly. Note hit miss miss miss and miss hit miss miss fail. Therefore there are only 2 ways to do this, and there are $2\cdot 2=\textbf{4}$ total possibilities for this case.
Taking all these cases into account, we find that there are $4+15+4=23$ ways to have $a_{10} = .4$ and $a_n\leq .4$ . Each of these has a probability of $.4^4\cdot.6^6=\frac{2^4\cdot 3^6}{5^{10}}$ . Therefore the probability that we have $a_{10} = .4$ and $a_n\leq .4$ is $\frac{23\cdot2^4\cdot3^6}{5^{10}}$ . Now we are asked to find the product of the sum of the primes and the sum of the exponents, which is $(23+2+3+5)(4+6+10)=33\cdot20=\boxed{660}$


Answer: 6.0

A basketball player made 5 baskets during a game. Each basket was worth either 2 or 3 points. How many different numbers could represent the total points scored by the player?
$\mathrm{(A)}\ 2\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 5\qquad\mathrm{(E)}\ 6$

Solution: The number of points could have been 10, 11, 12, 13, 14, or 15. This is because the minimum is 2*5=10 and the maximum is 3*5=15. The numbers between 10 and 15 are possible as well. Thus, the answer is $\boxed{6}$


Answer: 20.0

A basketball team's players were successful on 50% of their two-point shots and 40% of their three-point shots, which resulted in 54 points. They attempted 50% more two-point shots than three-point shots. How many three-point shots did they attempt?
$\textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }20\qquad\textbf{(D) }25\qquad\textbf{(E) }30$

Solution: Let $x$ be the number of two point shots attempted and $y$ the number of three point shots attempted. Because each two point shot is worth two points and the team made 50% and each three point shot is worth 3 points and the team made 40%, $0.5(2x)+0.4(3y)=54$ or $x+1.2y=54$ . Because the team attempted 50% more two point shots then threes, $x=1.5y$ . Substituting $1.5y$ for $x$ in the first equation gives $1.5y+1.2y=54$ , which equals $2.7y=54$ so $y=$ $\boxed{20}$


Answer: 71.0

A beam of light strikes $\overline{BC}\,$ at point $C\,$ with angle of incidence $\alpha=19.94^\circ\,$ and reflects with an equal angle of reflection as shown.  The light beam continues its path, reflecting off line segments $\overline{AB}\,$ and $\overline{BC}\,$ according to the rule: angle of incidence equals angle of reflection.  Given that $\beta=\alpha/10=1.994^\circ\,$ and $AB=BC,\,$ determine the number of times the light beam will bounce off the two line segments.  Include the first reflection at $C\,$ in your count.
AIME 1994 Problem 14.png

Solution: At each point of reflection, we pretend instead that the light continues to travel straight.  Note that after $k$ reflections (excluding the first one at $C$ ) the extended line will form an angle $k \beta$ at point $B$ . For the $k$ th reflection to be just inside or at point $B$ , we must have $k\beta \le 180 - 2\alpha \Longrightarrow k \le \frac{180 - 2\alpha}{\beta} = 70.27$ . Thus, our answer is, including the first intersection, $\left\lfloor \frac{180 - 2\alpha}{\beta} \right\rfloor + 1 = \boxed{071}$


Answer: 2024.0

A bee starts flying from point $P_0$ . She flies $1$ inch due east to point $P_1$ . For $j \ge 1$ , once the bee reaches point $P_j$ , she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$ . When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away from $P_0$ , where $a$ $b$ $c$ and $d$ are positive integers and $b$ and $d$ are not divisible by the square of any prime. What is $a+b+c+d$
$\textbf{(A)}\; 2016 \qquad\textbf{(B)}\; 2024 \qquad\textbf{(C)}\; 2032 \qquad\textbf{(D)}\; 2040 \qquad\textbf{(E)}\; 2048$

Solution:
Let $x=e^{i\pi/6}$ , a $30^\circ$ counterclockwise rotation centered at the origin. Notice that $P_k$ on the complex plane is:
\[1+2x+3x^2+\cdots+(k+1)x^k\]
We need to find the magnitude of $P_{2015}$ on the complex plane. This is an arithmetico-geometric series
\begin{align*} S &=1+2x+3x^2+\cdots+2015x^{2014} \\ xS &=x+2x^2+3x^3+\cdots+2015x^{2015} \\ (1-x)S &=1+x+x^2+\cdots+x^{2014}-2015x^{2015} \\ S &= \frac{1-x^{2015}}{(1-x)^2}-\frac{2015x^{2015}}{1-x} \end{align*}
We want to find $|S|$ . First, note that $x^{2015}=x^{11}=x^{-1}$ because $x^{12}=1$ . Therefore
\[S =\frac{1-\frac{1}{x}}{(1-x)^2}-\frac{2015}{x(1-x)}=-\frac{1}{x(1-x)}-\frac{2015}{x(1-x)}=-\frac{2016}{x(1-x)}.\]
Hence, since $|x|=1$ , we have $|S| = \frac{2016}{|1-x|}.$
Now we just have to find $|1-x|$ . This can just be computed directly:
\[1-x=1-\frac{\sqrt{3}}{2}-\frac{1}{2}i\]
\[|1-x|^2=\left(1-\sqrt{3}+\frac{3}{4}\right)+\frac{1}{4}=2-\sqrt{3}={\left(\frac{\sqrt{6}-\sqrt{2}}{2}\right)}^2\]
\[|1-x|=\frac{\sqrt{6}-\sqrt{2}}{2}\]
Therefore $|S|=2016\cdot\frac{2}{\sqrt{6}-\sqrt{2}}=2016\left(\frac{\sqrt{6}+\sqrt{2}}{2}\right)=1008\sqrt{2}+1008 \sqrt{6}$
Thus the answer is $1008+2+1008+6=\boxed{2024}$

Solution:
Here is an alternate solution that does not use complex numbers:
The distance from $P_{2015}$ to $P_0$ can be evaluated using the Pythagorean theorem . Assuming $P_0$ lies at the origin, we can calculate the distance the bee traveled to $P_{2015}$ by evaluating the distance the bee traveled in the x-direction and the y-direction. Let's start by summing each movement:
$x=1\cos{0}+2\cos{30}+3\cos{60}+\cdots+2014\cos{270}+2015\cos{300}$
A movement of $p$ units at $q$ degrees is the same thing as a movement of $-p$ units at $q-180$ degrees, so we can adjust all the cosines with arguments greater than $180$ as follows:
$x=1\cos{0}+2\cos{30}+3\cos{60}+4\cos{90}+5\cos{120}+6\cos{150}-7\cos{0}-8\cos{30}-\cdots-2015\cos{120}$
Grouping terms with like-cosines and factoring out the cosines:
$x=(1-7+13-\cdots+2005-2011)\cos{0}+\cdots+(6-12+18-\cdots-2004+2010)\cos{150}$
Each sum in the parentheses has $336$ terms (except the very last one, which has $335$ ). By pairing each term, we see there are $\frac{336}{2}$ pairs of $-6$ . Therefore, each sum equals $168\cdot-6=-1008$ except the very last sum, which has $167$ pairs of $-6$ plus an extra 2010 and equals $167\cdot-6+2010=1008$ . Plugging in these values:
$x=-1008\cos{0}-1008\cos{30}-1008\cos{60}-1008\cos{90}-1008\cos{120}+1008\cos{150}$ $x=1008(-1-\frac{\sqrt{3}}{2}-\frac{1}{2}-0+\frac{1}{2}-\frac{\sqrt{3}}{2})=-1008(1+\sqrt{3})$
We can find how far the bee traveled in the y-direction using the same logic as above, we arrive at the sum:
$y=-1008\sin{0}-1008\sin{30}-1008\sin{60}-1008\sin{90}-1008\sin{120}+1008\sin{150}$
$y=1008(0-\frac{1}{2}-\frac{\sqrt{3}}{2}-1-\frac{\sqrt{3}}{2}+\frac{1}{2})=-1008(1+\sqrt{3})$
Finally, we use the Pythagorean to find the distance from $P_0$ . This distance is given by:
$\sqrt{x^2+y^2}=\sqrt{(-1008(1+\sqrt{3}))^2+(-1008(1+\sqrt{3}))^2}=\sqrt{2\cdot1008^2\cdot(1+\sqrt{3})^2}=1008(1+\sqrt{3})\sqrt{2}=1008\sqrt{2}+1008\sqrt{6}$ , so the answer is $1008+2+1008+6=\boxed{2024}$

Solution:
Suppose that the bee makes a move of distance $i$ . After $6$ turns it will be facing the opposite direction and move $i+6$ units. Combining these opposite movements gives a total movement of $-6$ units in the original direction. This means that every $12$ moves, the bee will move $-6$ units in each direction of $0^\circ, 30^\circ, 60^\circ, 90^\circ, 120^\circ, 150^\circ$
We want to find the displacement vector for every $12$ moves. Factoring out the $-6$ for now (which flips the direction), we draw a quick diagram of one unit in each direction.  Using the 30-60-90 triangles, it is clear that the displacement vector (factoring back in the $-6$ ) is $-6\left\langle 1, 2+\sqrt{3}\right\rangle$
To compute the distance to $P_{2015}$ , we can compute the position of $P_{2016}$ (a multiple of $12$ moves) and then subtract the vector from $P_{2015}$ to $P_{2016}$
The bee reaches $P_{2016}$ after $\frac{2016}{12} = 168$ sets of $12$ moves, so the total displacement vector to $P_{2016}$ is $168(-6)\left\langle 1, 2+\sqrt{3}\right\rangle = \left\langle -1008, -2006-1008\sqrt{3}\right\rangle$
The bee moves at an angle of $-30^\circ$ from $P_{2015}$ to $P_{2016}$ , so subtracting it means moving an angle of $150^\circ$ . Since the vector is $2016$ units long, by a 30-60-90 triangle, it is $\left\langle -1008\sqrt{3}, 1008\right\rangle$
Therefore the total displacement vector to $P_{2015}$ is $\left\langle -1008, -2006-1008\sqrt{3}\right\rangle + \left\langle -1008\sqrt{3}, 1008\right\rangle = \left\langle -1008-1008\sqrt{3}, -1008-1008\sqrt{3}\right\rangle$ . The displacement is thus $\sqrt{2}\left|-1008-1008\sqrt{3}\right| = 1008\sqrt{2}+1008\sqrt{6} \implies 1008+2+1008+6 = \boxed{2024}$

Solution:
Let $P_0$ be the origin. East would be the real axis in the positive direction. Then we can assign each $P_n$ a complex value. The displacement would then be the magnitude of the complex number.
Notice that after the $n$ th move the value of $P_n$ is $P_{n-1}+ne^{\frac{i(n-1)\pi}{6}}$ . Also notice that after six moves the bee is facing in the opposite direction. And because we have found a recursion, we can add these up.
Then we have \[P_n=e^{0}+2e^{\frac{i\pi}{6}}+\cdots+2015e^{\frac{2014i\pi}{6}},\] and this becomes \[(1-7+13-\cdots-2011)e^0+(2-8+\cdots-2012)e^{\frac{i\pi}{6}}+\cdots+(6-12+\cdots+2010)e^{\frac{5i\pi}{6}}.\]
Simplifying, we have \[-6\cdot168-6\cdot168e^{\frac{i\pi}{6}}-6\cdot168e^{\frac{2i\pi}{6}}-6\cdot168e^{\frac{3i\pi}{6}}-6\cdot168e^{\frac{4i\pi}{6}}-(2010-6\cdot167)e^{\frac{5i\pi}{6}},\] which eventually simplifies to \[-1008-(2+\sqrt3)1008i,\] and this is a $15-75-90$ triangle which has ratios of $1:(2+\sqrt3):(\sqrt2+\sqrt6)$ so the magnitude is $1008\sqrt2+1008\sqrt6$ and the answer is $1008+2+1008+6=\boxed{2024}$

Solution: After each 12 moves, the bee will be facing the same direction as it started. Let $P_0$ be the origin and let $P_n$ (with $n$ divisible by $12$ ) be $(x,y)$ . We now notice that each of the move pairs with lengths $n+1$ $n+7$ $n+2$ $n+8$ $n+3$ $n+9$ $n+4$ $n+10$ $n+5$ $n+11$ $n+6$ $n+12$ will move the bee 6 units in the directions corresponding to the moves with lengths $n+7$ $n+8$ $n+9$ $n+10$ $n+11$ , and $n+12$ . This equates to moving the bee from $(x,y)$ to $(x-6,y-12-6\sqrt{3})$ , a move that repeats every 12 moves. Since $\frac{2016}{12} = 168$ , we have that $P_(2016) = (-1008, -2016-1008\sqrt{3})$ . It follows that $P_(2015) = (-1008-1008\sqrt{3}, -1008-1008\sqrt{3})$ so the distance to $P_0$ is $1008(\sqrt{2} + \sqrt{6})$ so the answer is $1008 + 2 + 1008 + 6 = \boxed{2024}$


Answer: 2024.0

Solution:
We first notice that if the bee is turning 30 degrees each turn, it will take 12 turns to be looking in the same direction when the bee initially left. This means we simply need to answer the question; how far will the bee be when the bee is facing in the same direction?
First we use the fact that after 3 turns, the bee will be facing in a direction perpendicular to the the initial direction. From here we can draw a perpendicular from $P_2$ to the line $\overline{P_0P_1}$ intersecting a point $C_0$ . We will also place the point $C_1$ at the intersection of $\overline{P_0P_1}$ and $\overline{P_3P_4}$ . In addition, the point $C_2$ is placed at the perpendicular dropped from $P_2$ to the line $\overline{P_3C_1}$ . We will also set the distance $\overline{P_0P_1} = n$ and thus $\overline{P_1P_2} = n+1$ . With this perpendicular we see that the triangle $\triangle{P_1P_2C_0}$ is a 30-60-90 triangle. This means that the length $\overline{P_1C_0} = \frac{(n+1)\sqrt{3}}{2}$ and the length $\overline{C_1C_2} = \frac{n+1}{2}$ . We can also see that the triangle $\triangle{P_2C_1P_3}$ is a 30-60-90 triangle and thus $\overline{C_0C_1} = \frac{n+2}{2}$ and $\overline{C_2P_3} = \frac{(n+2)\sqrt{3}}{2}$ . Now if we continue this across all $P_i$ and set the point $P_0$ to the coordinates $(0, 0)$ . As you can see, we are inherently putting a “box” around the figure. Doing similar calculations for all four “sides” of this spiral we get that the length
\[\overline{P_0C_1} = n + \frac{(n+1)\sqrt{3}}{2} + \frac{n+2}{2}\] \[\overline{C_1C_4} = \frac{(n+1)}{2} + \frac{(n+2)\sqrt{3}}{2} + (n+3) + \frac{(n+4)\sqrt{3}}{2} + \frac{n+5}{2}\] \[\overline{C_4C_7} = \frac{(n+4)}{2} + \frac{(n+5)\sqrt{3}}{2} + (n+6) + \frac{(n+7)\sqrt{3}}{2} + \frac{n+8}{2}\] \[\overline{C_7C_{10}} = \frac{(n+7)}{2} + \frac{(n+8)\sqrt{3}}{2} + (n+9) + \frac{(n+10)\sqrt{3}}{2} + \frac{n+11}{2}\] , and finally \[\overline{C_{10}P_{12}} = \frac{(n+10)}{2} + \frac{(n+11)\sqrt{3}}{2}\]
Here the point $C_4$ is defined as the intersection of lines $\overline{P_3P_4}$ and $\overline{P_6P_7}$ . The point $C_7$ is defined as the intersection of lines $\overline{P_6P_7}$ and $\overline{P_9P_{10}}$ . Finally, the point $C_10$ is defined as the intersection of lines $\overline{P_{9}P_{10}}$ and $\overline{P_{12}P_{13}}$ . Note that our spiral stops at $P_{12}$ before the next spiral starts. Calculating the offset from the x and the y direction, we see that the offset, or the new point $P_{12}$ , is $({-6}, {-6}-12 \sqrt{3})$ . This is an interesting property that the points’ coordinate changes by a constant offset no matter what $n$ is. Since the new point’s subscript changes by 12 each time and we see that 2016 is divisible by 12, the point $P_{2016} = ({-168} \cdot {6}, {168} \cdot ({-6} \sqrt{3} {-12}))$ . Using similar 30-60-90 triangle properties, we see that $P_{2015} = ({-6} \cdot 168-1008 \sqrt{3}, 168({-6} \sqrt{3} - 12) + 1008)$ . Using the distance formula, the numbers cancel out nicely (1008 is divisible by 168, so take 168 when using the distance formula) and we see that the final answer is $(1008)(1+\sqrt{3})(\sqrt{2})$ which gives us a final answer of $\boxed{2024}$


Answer: 109.0

A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c$ . (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. What is $p+q?$
$\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601$

Solution: We see that $a \, \diamondsuit \, a = 1$ , and think of division. Testing, we see that the first condition $a \, \diamondsuit \, (b \, \diamondsuit \, c) = (a \, \diamondsuit \, b) \cdot c$ is satisfied, because $\frac{a}{\frac{b}{c}} = \frac{a}{b} \cdot c$ . Therefore, division can be the operation $\diamondsuit$ . Solving the equation, \[\frac{2016}{\frac{6}{x}} = \frac{2016}{6} \cdot x = 336x = 100\implies x=\frac{100}{336} = \frac{25}{84},\] so the answer is $25 + 84 = \boxed{109}$

Solution:
If the given conditions hold for all nonzero numbers $a, b,$ and $c$
Let $a=b=c.$ From the first two givens, this implies that
\[a\diamondsuit\, (a\diamondsuit\, {a})=(a\diamondsuit\, a)\cdot{a}.\]
From $a\diamondsuit\,{a}=1,$ this equation simply becomes $a\diamondsuit\,{1}=a.$
Let $c=b.$ Substituting this into the first two conditions, we see that
\[a\diamondsuit\, (b\diamondsuit\, {c})=(a\diamondsuit\, {b})\cdot{c} \implies a\diamondsuit\, (b\diamondsuit\, {b})=(a\diamondsuit\, {b})\cdot{b}.\]
Substituting $b\diamondsuit\, {b} =1$ , the second equation becomes
\[a\diamondsuit\, {1}=(a\diamondsuit\, {b})\cdot{b} \implies a=(a\diamondsuit\,{b})\cdot{b}.\]
Since $a, b$ and $c$ are nonzero, we can divide by $b$ which yields,
\[\frac{a}{b}=(a\diamondsuit\, {b}).\]
Now we can find the value of $x$ straightforwardly:
\[\frac{2016}{(\frac{6}{x})}=100 \implies 2016=\frac{600}{x} \implies x=\frac{600}{2016} = \frac{25}{84}.\]
Therefore, $a+b=25+84=\boxed{109}$

Solution:
One way to eliminate the $\diamondsuit$ in this equation is to make $a = b$ so that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = c$ . In this case, we can make $b = 2016$
\[2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100\implies  (2016\, \diamondsuit\, 6) \cdot x = 100\]
By multiplying both sides by $\frac{6}{x}$ , we get:
\[(2016\, \diamondsuit\, 6) \cdot 6 = \frac{600}{x}\implies  2016 \, \diamondsuit\, (6\, \diamondsuit\, 6) = \frac{600}{x}\]
Because $6\, \diamondsuit\, 6 = 2016\, \diamondsuit\, 2016 = 1:$
\[2016 \, \diamondsuit\, (2016\, \diamondsuit\, 2016) = \frac{600}{x}\implies  (2016\, \diamondsuit\, 2016) \cdot 2016 = \frac{600}{x}\implies  2016 = \frac{600}{x}\]
Therefore, $x = \frac{600}{2016} = \frac{25}{84}$ , so the answer is $25 + 84 = \boxed{109.}$

Solution:
We can manipulate the given identities to arrive at a conclusion about the binary operator $\diamondsuit$ .  Substituting $b = c$ into the first identity yields \[( a\ \diamondsuit\ b) \cdot b = a\ \diamondsuit\ (b\ \diamondsuit\  b) = a\ \diamondsuit\  1 = a\ \diamondsuit\ ( a\ \diamondsuit\ a) = ( a\ \diamondsuit\ a) \cdot a = a.\] Hence, $( a\ \diamondsuit\ b) \cdot b = a,$ or, dividing both sides of the equation by $b,$ $( a\ \diamondsuit\ b) = \frac{a}{b}.$
Hence, the given equation becomes $\frac{2016}{\frac{6}{x}} = 100$ .  Solving yields $x=\frac{100}{336} = \frac{25}{84},$ so the answer is $25 + 84 = \boxed{109.}$

Solution:
$2016 \diamondsuit (6 \diamondsuit x) = (2016 \diamondsuit 6) \cdot x = 100$
$2016 \diamondsuit (2016 \diamondsuit 1) = (2016 \diamondsuit 2016) \cdot 1 = 1 \cdot 1 = 1$
$2016 \diamondsuit 2016 = 1$ $2016 \diamondsuit (2016 \diamondsuit 1) = 1$ , so $2016 \diamondsuit 1 = 2016$
$2016  \diamondsuit 1 = (2016  \diamondsuit 6) \cdot 6$ $2016 \diamondsuit 6 = \frac{2016  \diamondsuit 1}{6} = 336$
$x = \frac{100}{2016 \diamondsuit 6} = \frac{100}{336} = \frac{25}{84}$ $24 + 85 = \boxed{109}$

Solution: Notice that $2016 \diamondsuit (6 \diamondsuit 6)=(2016 \diamondsuit 6) \cdot 6 = 2016$ . Hence, $2016 \diamondsuit 6 = 336$ . Thus, $2016 \diamondsuit (6 \diamondsuit x)=100 \implies (2016 \diamondsuit 6) \cdot x = 100 \implies 336x=100 \implies x=\frac{25}{84}$ . Therefore, the answer is $\boxed{109}$