Answer: 2.37

A circle with center $O$ is tangent to the coordinate axes and to the hypotenuse of the $30^\circ$ $60^\circ$ $90^\circ$ triangle $ABC$ as shown, where $AB=1$ . To the nearest hundredth, what is the radius of the circle?
[asy] defaultpen(linewidth(.8pt)); dotfactor=3; pair A = origin; pair B = (1,0); pair C = (0,sqrt(3)); pair O = (2.33,2.33); dot(A);dot(B);dot(C);dot(O); label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,W);label("$O$",O,NW); label("$1$",midpoint(A--B),S);label("$60^\circ$",B,2W + N); draw((3,0)--A--(0,3)); draw(B--C); draw(Arc(O,2.33,163,288.5));[/asy]
$\textbf{(A)}\ 2.18\qquad\textbf{(B)}\ 2.24\qquad\textbf{(C)}\ 2.31\qquad\textbf{(D)}\ 2.37\qquad\textbf{(E)}\ 2.41$

Solution:
Draw radii $OE$ and $OD$ to the axes, and label the point of tangency to triangle $ABC$ point $F$ .  Let the radius of the circle $O$ be $r$ .  Square $OEAD$ has side length $r$
Because $BD$ and $BF$ are tangents from a common point $B$ $BD = BF$
$AD = AB + BD$
$r = 1 + BD$
$r = 1 + BF$
Similarly, $CF = CE$ , and we can write:
$AE = AC + CE$
$r = \sqrt{3} + CF$
Equating the radii lengths, we have $1 + BF = \sqrt{3} + CF$
This means $BF - CF = \sqrt{3} - 1$
$BF + CF = 2$ by the 30-60-90 triangle.
Therefore, $2BF = 2 + \sqrt{3} - 1$ , and we get $BF = \frac{1}{2} + \frac{\sqrt{3}}{2}$
The radius of the circle is $AD$ , which is $BF + 1 = \frac{3}{2} + \frac{\sqrt{3}}{2}$
Using decimal approximations, $r \approx 1.5 + \frac{1.73^+}{2} \approx 2.37$ , and the answer is $\boxed{2.37}$


Answer: 312.0

A circle with diameter $\overline{PQ}$ of length 10 is internally tangent at $P$ to a circle of radius 20. Square $ABCD$ is constructed with $A$ and $B$ on the larger circle, $\overline{CD}$ tangent at $Q$ to the smaller circle, and the smaller circle outside $ABCD$ . The length of $\overline{AB}$ can be written in the form $m + \sqrt{n}$ , where $m$ and $n$ are integers. Find $m + n$
1994 AIME Problem 2.png
Note: The diagram was not given during the actual contest.

Solution:
1994 AIME Problem 2 - Solution.png
Call the center of the larger circle $O$ . Extend the diameter $\overline{PQ}$ to the other side of the square (at point $E$ ), and draw $\overline{AO}$ . We now have a right triangle , with hypotenuse of length $20$ . Since $OQ = OP - PQ = 20 - 10 = 10$ , we know that $OE = AB - OQ = AB - 10$ . The other leg, $AE$ , is just $\frac 12 AB$
Apply the Pythagorean Theorem
The quadratic formula shows that the answer is $\frac{16 \pm \sqrt{16^2 + 4 \cdot 240}}{2} = 8 \pm \sqrt{304}$ . Discard the negative root, so our answer is $8 + 304 = \boxed{312}$


Answer: 17.0

A circle with integer radius $r$ is centered at $(r, r)$ . Distinct line segments of length $c_i$ connect points $(0, a_i)$ to $(b_i, 0)$ for $1 \le i \le 14$ and are tangent to the circle, where $a_i$ $b_i$ , and $c_i$ are all positive integers and $c_1 \le c_2 \le \cdots \le c_{14}$ . What is the ratio $\frac{c_{14}}{c_1}$ for the least possible value of $r$
$\textbf{(A)} ~\frac{21}{5} \qquad\textbf{(B)} ~\frac{85}{13} \qquad\textbf{(C)} ~7 \qquad\textbf{(D)} ~\frac{39}{5} \qquad\textbf{(E)} ~17$

Solution:
Case 1: The tangent and the origin are on the opposite sides of the circle.
In this case, $a, b > 2r$
We can easily prove that \[a + b - 2 r = c . \hspace{1cm} (1)\]
Recall that $c = \sqrt{a^2 + b^2}$
Taking square of (1) and reorganizing all terms, (1) is converted as \[\left( a - 2 r \right) \left( b - 2 r \right) = 2 r^2 .\]
Case 2: The tangent and the origin are on the same sides of the circle.
In this case, $0 < a, b < r$
We can easily prove that \[2 r - a - b  = c . \hspace{1cm} (2)\]
Recall that $c = \sqrt{a^2 + b^2}$
Taking square of (2) and reorganizing all terms, (2) is converted as \[\left( a - 2 r \right) \left( b - 2 r \right) = 2 r^2 .\]
Putting both cases together, for given $r$ , we look for solutions of $a$ and $b$ satisfying \[\left( a - 2 r \right) \left( b - 2 r \right) = 2 r^2 ,\] with either $a, b > 2r$ or $0 < a, b < r$
Now, we need to find the smallest $r$ , such that the number of feasible solutions of $(a, b)$ is at least 14.
For equation \[uv = 2 r^2 ,\] we observe that the R.H.S. is a not a perfect square. Thus, the number of positive $(u, v)$ is equal to the number of positive divisors of $2 r^2$
Second, for each feasible positive solution $(u, v)$ , its opposite $(-u, -v)$ is also a solution. However, $(u,v)$ corresponds to a feasible solution if $(a, b)$ with $a = u + 2r$ and $b = v + 2r$ , but $(-u, -v)$ may not lead to a feasible solution if $(a, b)$ with $a = 2 r - u$ and $b = 2 r - v$
Recall that we are looking for $r$ that leads to at least 14 solutions.
Therefore, the above observations imply that we must have $r$ , such that $2 r^2$ has least 7 positive divisors.
Following this guidance, we find the smallest $r$ is 6. This leads to the following solutions:
$\left( a_1, b_1, c_1 \right) = \left( 3, 4, 5 \right)$ $\left( a_2, b_2, c_2 \right) = \left( 4, 3, 5 \right)$
$\left( a_3, b_3, c_3 \right) = \left( 20, 21, 29 \right)$ $\left( a_4, b_4, c_4 \right) = \left( 21, 20, 29 \right)$
$\left( a_5, b_5, c_5 \right) = \left( 18, 24, 30 \right)$ $\left( a_6, b_6, c_6 \right) = \left( 24, 18, 30 \right)$
$\left( a_7, b_7, c_7 \right) = \left( 16, 30, 34 \right)$ $\left( a_8, b_8, c_8 \right) = \left( 30, 16, 34 \right)$
$\left( a_9, b_9, c_9 \right) = \left( 15, 36, 39 \right)$ $\left( a_{10}, b_{10}, c_{10} \right) = \left( 36, 15, 39 \right)$
$\left( a_{11}, b_{11}, c_{11} \right) = \left( 14, 48, 50 \right)$ $\left( a_{12}, b_{12}, c_{12} \right) = \left( 48, 14, 50 \right)$
$\left( a_{13}, b_{13}, c_{13} \right) = \left( 13, 84, 85 \right)$ $\left( a_{14}, b_{14}, c_{14} \right) = \left( 84, 13, 85 \right)$
Therefore, $\frac{c_{14}}{c_1} = \boxed{17}$

Solution:
As $c_i$ is the length of the segment $(0,a_i)$ and $(b_i,0)$ $a_i^2+b_i^2=c_i^2$ . The equation for the line that passes through $(0,a_i)$ and $(b_i,0)$ is $a_ix+b_iy-a_ib_i=0$
By the point-line distance formula from point $(r,r)$ to line $a_ix+b_iy-a_ib_i=0$
\[r = \frac{ |a_ir+b_ir-a_ib_i| }{ \sqrt{a_i^2+b_i^2} }, \quad c_i = \frac{ |a_ir+b_ir-a_ib_i| }{r} = |a_i+b_i- \frac{a_ib_i}{r}|\]
\[\because c_i \quad \text{is an integer}, \quad \therefore r|a_ib_i\]
To find $c_1$ and $c_{14}$ for the smallest $r$ , we will list Pythagorean triples $(a_i,b_i,c_i)$ with relatively prime elements from the smallest. Notice that we only need to find $7$ triples with $a_i<b_i$ , the $7$ other triples will simply be $(b_i,a_i,c_i)$ $a_i$ will not equal $b_i$ because then $c_i = a_i \cdot \sqrt{2}$ , meaning that $a_i$ $b_i$ , and $c_i$ cannot all be integers.
The $7$ smallest triples are: $(3,4,5)$ $(5,12,13)$ $(7,24,25)$ $(8,15,17)$ $(11,60,61)$ $(12,35,37)$ $(13,84,85)$
Therefore, $\frac{c_{14}}{c_1} = \frac{85}{5} = \boxed{17}$


Answer: 17.0

Solution:
Suppose that with a pair $(a_i,b_i)$ the circle is an excircle. Then notice that the hypotenuse must be $(r-x)+(r-y)$ , so it must be the case that \[a_i^2+b_i^2=(2r-a_i-b_i)^2.\] Similarly, if with a pair $(a_i,b_i)$ the circle is an incircle, the hypotenuse must be $(x-r)+(y-r)$ , leading to the same equation.
Notice that this equation can be simplified through SFFT to \[(a_i-2r)(b_i-2r)=2r^2.\] Thus, we want the smallest $r$ such that this equation has at least $14$ distinct pairs $(a_i,b_i)$ for which this holds. The obvious choice to check is $r=6$ . In this case, since $2r^2=2^3\cdot 3^2$ has $12$ positive factors, we get $12$ pairs, and we get another two if the factors are $-8,-9$ or vice versa. One can check that for smaller values of $r$ , we do not even get close to $14$ possible pairs.
When $r=6$ , the smallest possible $c$ -value is clearly when the factors are negative. When this occurs, $a_i=4, b_i=3$ (or vice versa), so the mimimal $c$ is $5$ . The largest possible $c$ -value occurs when the largest of $a_i$ and $b_i$ are maximized. This occurs when the factors are $72$ and $1$ , leading to $a_i=84, b_i=13$ (or vice-versa), leading to a maximal $c$ of $85$
Hence the answer is $\frac{85}5=\boxed{17}$


Answer: 12.0

A circle with radius $1$ is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle's shaded area to the area between the two squares?
[asy] filldraw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle,gray,black); filldraw(Circle((0,0),1), mediumgray,black); filldraw((-1,0)--(0,1)--(1,0)--(0,-1)--cycle,white,black);[/asy]
$\textbf{(A)}\ \frac{1}2\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{3}2\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{5}2$

Solution:
The area of the smaller square is one half of the product of its diagonals. Note that the distance from a corner of the smaller square to the center is equivalent to the circle's radius so the diagonal is equal to the diameter: $2 \cdot  2 \cdot \frac{1}{2}=2.$
The circle's shaded area is the area of the smaller square subtracted from the area of the circle: $\pi - 2.$
If you draw the diagonals of the smaller square, you will see that the larger square is split $4$ congruent half-shaded squares. The area between the squares is equal to the area of the smaller square: $2.$
Approximating $\pi$ to $3.14,$ the ratio of the circle's shaded area to the area between the two squares is about
\[\frac{\pi-2}{2} \approx \frac{3.14-2}{2} = \frac{1.14}{2} \approx \boxed{12}\]

Solution: For the ratio of the circle's shaded area to the area between the squares to be $1,$ they would have to be approximately the same size. For any ratio larger than that, the circle's shaded area must be greater. However, we can clearly see that the circle's shaded area is part of the area between the squares, and is approximately $\boxed{12}$

Solution:
Set the side length of the bigger square to be $8$ .
Then the area of the big square is $8^2 =64$ and
the area of the small square $(4\sqrt{2})^2  = 32$ .
The difference is $32$ .
The area of the circle is $4^2$ times $\pi$ which is $16 \pi$ or about $48$ .
Knowing the area of the small square is $32$ $48-32$ is $16$ .
The area of the big square is $64$ .
So $32/64$ is $1/2$ , or $\boxed{12}$


Answer: 192.0

A circle with radius $6$ is externally tangent to a circle with radius $24$ . Find the area of the triangular region bounded by the three common tangent lines of these two circles.

Solution:
$r_1 = O_1A = 24$ $r_2 = O_2B = 6$ $AG = BO_2 = r_2 = 6$ $O_1G = r_1 - r_2 = 24 - 6 = 18$ $O_1O_2 = r_1 + r_2 = 30$
$\triangle O_2BD \sim \triangle O_1GO_2$ $\frac{O_2D}{O_1O_2} = \frac{BO_2}{GO_1}$ $\frac{O_2D}{30} = \frac{6}{18}$ $O_2D = 10$
$CD = O_2D + r_2 = 10 + 6 = 16$
$EF = 2EC = EA + EB = AB = GO_2 = \sqrt{(O_1O_2)^2-(O_1G)^2} = \sqrt{30^2-18^2} = 24$
$DEF = \frac12 \cdot EF \cdot CD = \frac12 \cdot 24 \cdot 16 = \boxed{192}$

Solution: Let the center of the circle with radius $6$ be labeled $A$ and the center of the circle with radius $24$ be labeled $B$ . Drop perpendiculars on the same side of line $AB$ from $A$ and $B$ to each of the tangents at points $C$ and $D$ , respectively. Then, let line $AB$ intersect the two diagonal tangents at point $P$ . Since $\triangle{APC} \sim \triangle{BPD}$ , we have \[\frac{AP}{AP+30}=\frac14 \implies AP=10.\] Next, throw everything on a coordinate plane with $A=(0, 0)$ and $B = (30, 0)$ . Then, $P = (-10, 0)$ , and if $C = (x, y)$ , we have \[(x+10)^2+y^2=64,\] \[x^2+y^2=36.\] Combining these and solving, we get $(x, y)=\left(-\frac{18}5, \frac{24}5\right)$ . Notice now that $P$ $C$ , and the intersections of the lines $x=6$ (the vertical tangent) with the tangent containing these points are collinear, and thus every slope between a pair of points will have the same slope, which in this case is $\frac{-\frac{18}5+10}{\frac{24}5}=\frac34$ . Thus, the other two vertices of the desired triangle are $(6, 12)$ and $(6, -12)$ . By the Shoelace Formula, the area of a triangle with coordinates $(-10, 0)$ $(6, 12)$ , and $(6, -12)$ is \[\frac12|-120-0-72-72+0-120|=\boxed{192}.\]

Solution: First, we want to find $O_2D$ . We know that $\angle O_1AD = \angle O_2BD = 90^{\circ}$ , so by AA similarity, $\triangle O_1AD \sim \triangle O_2BD$ . We want to find the length of $x$ , and using the similar triangles, we write an equation: $\frac{30 + x}{4} = x$ . Solving, we get $x=10$ . Therefore, $CD = 10 + 6 = 16$ . Next, we find that using AA similarity, $\triangle O_2BD \sim \triangle HO_2D \sim \triangle ECD$ and they are 3-4-5 triangles. We can quickly compute $EF = 2EC = 2 \cdot \left( \frac{3}{4} \cdot 16 \right) = 2 \cdot 12 = 24$ . Therefore, the area is $\frac{1}{2} \cdot 16 \cdot 24 = \boxed{192}$


Answer: 32.0

A circular disc with diameter $D$ is placed on an $8\times 8$ checkerboard with width $D$ so that the centers coincide. The number of checkerboard squares which are completely covered by the disc is
$\textbf{(A) }48\qquad \textbf{(B) }44\qquad \textbf{(C) }40\qquad \textbf{(D) }36\qquad  \textbf{(E) }32$

Solution: Consider the upper right half of the grid, which consists of a $4\times4$ section of the checkerboard and a quarter-circle of radius $4$ . We can draw this as a coordinate grid and shade in the complete squares. There are $8$ squares in the upper right corner, so there are $8 \cdot 4 = \boxed{32}$ whole squares in total.


Answer: 9.0

A class collects $50$ dollars to buy flowers for a classmate who is in the hospital. Roses cost $3$ dollars each, and carnations cost $2$ dollars each. No other flowers are to be used. How many different bouquets could be purchased for exactly $50$ dollars?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 17$

Solution: The class could send $25$ carnations and no roses, $22$ carnations and $2$ roses, $19$ carnations and $4$ roses, and so on, down to $1$ carnation and $16$ roses. There are 9 total possibilities (from 0 to 16 roses, incrementing by 2 at each step), $\Rightarrow \boxed{9}$


Answer: 9.0

A class collects 50 dollars to buy flowers for a classmate who is in the hospital. Roses cost 3 dollars each, and carnations cost 2 dollars each. No other flowers are to be used. How many different bouquets could be purchased for exactly 50 dollars?
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 7 \qquad \mathrm{(C)}\ 9 \qquad \mathrm{(D)}\ 16 \qquad \mathrm{(E)}\ 17$

Solution: The cost of a rose is odd, hence we need an even number of roses. Let there be $2r$ roses for some $r\geq 0$ . Then we have $50-3\cdot 2r = 50-6r$ dollars left. We can always reach the sum exactly $50$ by buying $(50-6r)/2 = 25-3r$ carnations. Of course, the number of roses must be such that the number of carnations is non-negative. We get the inequality $25-3r \geq 0$ which solves to $r\leq 8 \frac13$ $r$ must be an integer, so there are $\boxed{9}$ possible values of $r$ , and each gives us one solution.

Solution: Let $x$ and $y$ be the number of roses and carnations bought. The equation should be $3x+2y = 50$ . Since $50$ is an even number, the product of $3x$ must be even and smaller than $50$ . You can try nonnegative even integers for $x$ and you will end up with the numbers $0$ $2$ $4$ $6$ $8$ $10$ $12$ $14$ , and $16$ . There are $9$ numbers in total, so the answer is $\boxed{9}$

Solution:
Let $r$ represent the number of roses, and let $c$ represent the number of carnations. Then, we get the linear Diophantine equation, $3r+2c=50$ .
Using the Euclidean algorithm, we get the initial solutions to be $r_0=50$ and $c_0=-50$ , meaning the complete solution will be, $r=50+\frac{2}{\gcd(2,3)}$ $k=50+2k$ $c=-50-\frac{3}{\gcd(2,3)}k=-50-3k$
The solution range for which both $r$ and $c$ are positive is $17$ $\leq k$ $\leq$ $25$ . There are $\boxed{9}$ possible values for $k$


Answer: 9.0

A clock chimes once at $30$ minutes past each hour and chimes on the hour according to the hour. For example, at $1 \text{PM}$ there is one chime and at noon and midnight there are twelve chimes. Starting at $11:15 \text{AM}$ on $\text{February 26, 2003},$ on what date will the $2003^{\text{rd}}$ chime occur?
$\textbf{(A) } \text{March 8} \qquad\textbf{(B) } \text{March 9} \qquad\textbf{(C) } \text{March 10} \qquad\textbf{(D) } \text{March 20} \qquad\textbf{(E) } \text{March 21}$

Solution:
First, find how many chimes will have already happened before midnight (the beginning of the day) of $\text{February 27, 2003}.$ $13$ half-hours have passed, and the number of chimes according to the hour is $1+2+3+\cdots+12.$ The total number of chimes is $13+78=91.$
Every day, there will be $24$ half-hours and $2(1+2+3+\cdots+12)$ chimes according to the arrow, resulting in $24+156=180$ total chimes.
On $\text{February 26},$ the number of chimes that still need to occur is $2003-91=1912.$ $1912 \div 180=10 \text{R}112.$ Rounding up, it is $11$ days past $\text{February 26},$ which is $\boxed{9}$


Answer: 81.0

A club consisting of $11$ men and $12$ women needs to choose a committee from among its members so that the number of women on the committee is one more than the number of men on the committee. The committee could have as few as $1$ member or as many as $23$ members. Let $N$ be the number of such committees that can be formed. Find the sum of the prime numbers that divide $N.$

Solution:
Let $k$ be the number of women selected. Then, the number of men not selected is $11-(k-1)=12-k$ .
Note that the sum of the number of women selected and the number of men not selected is constant at $12$ . Each combination of women selected and men not selected corresponds to a committee selection. Since choosing 12 individuals from the total of 23 would give $k$ women and $12-k$ men, the number of committee selections is $\binom{23}{12}$ .
The answer is $\boxed{081}$ .
~awang11's sol

Solution:
We casework on the amount of men on the committee.
If there are no men in the committee, there are $\dbinom{12}{1}$ ways to pick the women on the committee, for a total of $\dbinom{11}{0} \cdot \dbinom{12}{1}$ . Notice that $\dbinom{11}{0}$ is equal to $\dbinom{11}{11}$ , so the case where no men are picked can be grouped with the case where all men are picked. When all men are picked, all women must also be picked, for a total of $\dbinom{12}{12}$ . Therefore, these cases can be combined to \[\dbinom{11}{0} \cdot \left(\dbinom{12}{1} + \dbinom{12}{12}\right)\] Since $\dbinom{12}{12} = \dbinom{12}{0}$ , and $\dbinom{12}{0} + \dbinom{12}{1} = \dbinom{13}{1}$ , we can further simplify this to \[\dbinom{11}{0} \cdot \dbinom{13}{1}\]
All other cases proceed similarly. For example, the case with one men or ten men is equal to $\dbinom{11}{1} \cdot \dbinom{13}{2}$ . Now, if we factor out a $13$ , then all cases except the first two have a factor of $121$ , so we can factor this out too to make our computation slightly easier. The first two cases (with $13$ factored out) give $1+66=67$ , and the rest gives $121(10+75+270+504) = 103,939$ . Adding the $67$ gives $104,006$ . Now, we can test for prime factors. We know there is a factor of $2$ , and the rest is $52,003$ . We can also factor out a $7$ , for $7,429$ , and the rest is $17 \cdot 19 \cdot 23$ . Adding up all the prime factors gives $2+7+13+17+19+23 = \boxed{081}$


Answer: 11.0

A coin that comes up heads with probability $p > 0$ and tails with probability $1 - p > 0$ independently on each flip is flipped $8$ times. Suppose that the probability of three heads and five tails is equal to $\frac {1}{25}$ of the probability of five heads and three tails. Let $p = \frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution:
The probability of three heads and five tails is $\binom {8}{3}p^3(1-p)^5$ and the probability of five heads and three tails is $\binom {8}{3}p^5(1-p)^3$
\begin{align*} 25\binom {8}{3}p^3(1-p)^5&=\binom {8}{3}p^5(1-p)^3 \\ 25(1-p)^2&=p^2 \\ 25p^2-50p+25&=p^2 \\ 24p^2-50p+25&=0 \\ p&=\frac {5}{6}\end{align*}
Therefore, the answer is $5+6=\boxed{011}$

Solution:
We start as shown above. However, when we get to $25(1-p)^2=p^2$ , we square root both sides to get $5(1-p)=p$ . We can do this because we know that both $p$ and $1-p$ are between $0$ and $1$ , so they are both positive. Now, we have:
\begin{align*} 5(1-p)&=p \\ 5-5p&=p \\ 5&=6p \\ p&=\frac {5}{6}\end{align*}
Now, we get $5+6=\boxed{011}$

Solution:
Rewrite it as : $(P)^3$ $(1-P)^5=\frac {1}{25}$ $(P)^5$ $(1-P)^3$
This can be simplified as $24P^2 -50P + 25 = 0$
This can be factored into $(4P-5)(6P-5)$
This yields two solutions: $\frac54$ (ignored because it would result in $1-p<0$ ) or $\frac56$
Therefore, the answer is $5+6$ $\boxed{011}$


Answer: 458.0

A collection of 8 cubes consists of one cube with edge length $k$ for each integer $k, 1 \le k \le 8.$ A tower is to be built using all 8 cubes according to the rules:
Let $T$ be the number of different towers than can be constructed. What is the remainder when $T$ is divided by 1000?

Solution: We proceed recursively .  Suppose we can build $T_m$ towers using blocks of size $1, 2, \ldots, m$ .  How many towers can we build using blocks of size $1, 2, \ldots, m, m + 1$ ?  If we remove the block of size $m + 1$ from such a tower (keeping all other blocks in order), we get a valid tower using blocks $1, 2, \ldots, m$ .  Given a tower using blocks $1, 2, \ldots, m$ (with $m \geq 2$ ), we can insert the block of size $m + 1$ in exactly 3 places: at the beginning, immediately following the block of size $m - 1$ or immediately following the block of size $m$ .  Thus, there are 3 times as many towers using blocks of size $1, 2, \ldots, m, m + 1$ as there are towers using only $1, 2, \ldots, m$ .  There are 2 towers which use blocks $1, 2$ , so there are $2\cdot 3^6 = 1458$ towers using blocks $1, 2, \ldots, 8$ , so the answer is $\boxed{458}$


Answer: 20.0

A collector offers to buy state quarters for 2000% of their face value. At that rate how much will Bryden get for his four state quarters?
$\text{(A)}\ 20\text{ dollars} \qquad \text{(B)}\ 50\text{ dollars} \qquad \text{(C)}\ 200\text{ dollars} \qquad \text{(D)}\ 500\text{ dollars} \qquad \text{(E)}\ 2000\text{ dollars}$

Solution: $2000\%$ is equivalent to $20\times100\%$ . Therefore, $2000\%$ of a number is the same as $20$ times that number. $4$ quarters is $1$ dollar, so Bryden will get $20\times1={20}$ dollars, $\boxed{20}$

Solution: Since $2000\%$ is just $\frac{2000}{100}$ , we can multiply that by $100$ , because four quarters is a $100$ cents. After the multiplication, we get $2000$ . Since our answer is in cents right now, we need to convert it to dollars, which would be $\boxed{20}$ dollars.


Answer: 24.0

A college student drove his compact car $120$ miles home for the weekend and averaged $30$ miles per gallon. On the return trip the student drove his parents' SUV and averaged only $20$ miles per gallon. What was the average gas mileage, in miles per gallon, for the round trip?
$\textbf{(A) } 22 \qquad\textbf{(B) } 24 \qquad\textbf{(C) } 25 \qquad\textbf{(D) } 26 \qquad\textbf{(E) } 28$

Solution: The trip was $240$ miles long and took $\dfrac{120}{30}+\dfrac{120}{20}=4+6=10$ gallons. Therefore, the average mileage was $\dfrac{240}{10}= \boxed{24}$

Solution: Alternatively, we can use the harmonic mean to get $\frac{2}{\frac{1}{20} + \frac{1}{30}} = \frac{2}{\frac{1}{12}} = \boxed{24}$


Answer: 24.0

Solution: The trip was $240$ miles long and took $\dfrac{120}{30}+\dfrac{120}{20}=4+6=10$ gallons. Therefore, the average mileage was $\dfrac{240}{10}= \boxed{24}$

Solution: Alternatively, we can use the harmonic mean to get $\frac{2}{\frac{1}{20} + \frac{1}{30}} = \frac{2}{\frac{1}{12}} = \boxed{24}$


Answer: 36.0

A company sells peanut butter in cylindrical jars.  Marketing research suggests that using wider jars will increase sales.  If the diameter of the jars is increased by $25\%$ without altering the volume , by what percent must the height be decreased?
$\mathrm{(A) \ } 10 \qquad \mathrm{(B) \ } 25 \qquad \mathrm{(C) \ } 36 \qquad \mathrm{(D) \ } 50 \qquad \mathrm{(E) \ } 60$

Solution:
When the diameter is increased by $25\%$ , it is increased by $\dfrac{5}{4}$ , so the area of the base is increased by $\left(\dfrac54\right)^2=\dfrac{25}{16}$
To keep the volume the same, the height must be $\dfrac{1}{\frac{25}{16}}=\dfrac{16}{25}$ of the original height, which is a $36\%$ reduction. $\boxed{36}$


Answer: 36.0

Solution:
When the diameter is increased by $25\%$ , it is increased by $\dfrac{5}{4}$ , so the area of the base is increased by $\left(\dfrac54\right)^2=\dfrac{25}{16}$
To keep the volume the same, the height must be $\dfrac{1}{\frac{25}{16}}=\dfrac{16}{25}$ of the original height, which is a $36\%$ reduction. $\boxed{36}$


Answer: 36.0

A computer can do $10,000$ additions per second.  How many additions can it do in one hour?
$\text{(A)}\ 6\text{ million} \qquad \text{(B)}\ 36\text{ million} \qquad \text{(C)}\ 60\text{ million} \qquad \text{(D)}\ 216\text{ million} \qquad \text{(E)}\ 360\text{ million}$

Solution:
There are $3600$ seconds per hour, so we have \begin{align*} \frac{3600\text{ seconds}}{\text{hour}}\cdot \frac{10,000\text{ additions}}{\text{second}} &= \frac{36,000,000\text{ additions}}{\text{hour}} \\ &= 36\text{ million additions per hour}  \end{align*}
$\boxed{36}$


Answer: 52.0

A container in the shape of a right circular cone is $12$ inches tall and its base has a $5$ -inch radius . The liquid that is sealed inside is $9$ inches deep when the cone is held with its point down and its base horizontal. When the liquid is held with its point up and its base horizontal, the height of the liquid is $m - n\sqrt [3]{p},$ from the base where $m,$ $n,$ and $p$ are positive integers and $p$ is not divisible by the cube of any prime number. Find $m + n + p$

Solution:
The scale factor is uniform in all dimensions, so the volume of the liquid is $\left(\frac{3}{4}\right)^{3}$ of the container. The remaining section of the volume is $\frac{1-\left(\frac{3}{4}\right)^{3}}{1}$ of the volume, and therefore $\frac{\left(1-\left(\frac{3}{4}\right)^{3}\right)^{1/3}}{1}$ of the height when the vertex is at the top.
So, the liquid occupies $\frac{1-\left(1-\left(\frac{3}{4}\right)^{3}\right)^{1/3}}{1}$ of the height, or $12-12\left(1-\left(\frac{3}{4}\right)^{3}\right)^{1/3}=12-3\left(37^{1/3}\right)$ . Thus $m+n+p=\boxed{052}$

Solution:
(Computational) The volume of a cone can be found by $V = \frac{\pi}{3}r^2h$ . In the second container, if we let $h',r'$ represent the height, radius (respectively) of the air (so $12 -h'$ is the height of the liquid), then the volume of the liquid can be found by $\frac{\pi}{3}r^2h - \frac{\pi}{3}(r')^2h'$
By similar triangles , we find that the dimensions of the liquid in the first cone to the entire cone is $\frac{3}{4}$ , and that $r' = \frac{rh'}{h}$ ; equating,
\begin{align*}\frac{\pi}{3}\left(\frac{3}{4}r\right)^2 \left(\frac{3}{4}h\right) &= \frac{\pi}{3}\left(r^2h - \left(\frac{rh'}{h}\right)^2h'\right)\\ \frac{37}{64}r^2h &= \frac{r^2}{h^2}(h')^3 \\ h' &= \sqrt[3]{\frac{37}{64} \cdot 12^3} = 3\sqrt[3]{37}\end{align*}
Thus the answer is $12 - h' = 12-3\sqrt[3]{37}$ , and $m+n+p=\boxed{052}$

Solution: From the formula $V=\frac{\pi r^2h}{3}$ , we can find that the volume of the container is $100\pi$ . The cone formed by the liquid is similar to the original, but scaled down by $\frac{3}{4}$ in all directions, so its volume is $100\pi*\frac{27}{64}=\frac{675\pi}{16}$ . The volume of the air in the container is the volume of the container minus the volume of the liquid, which is $\frac{925\pi}{16}$ , which is $\frac{37}{64}$ of the volume of the container. When the point faces upwards, the air forms a cone at the top of the container. This cone must have $\sqrt[3]{\frac{37}{64}}=\frac{\sqrt[3]{37}}{4}$ of the height of the container. This means that the height of the liquid is $12\left(1-\frac{\sqrt[3]{37}}{4}\right)=12-3\sqrt[3]{37}$ inches, so our answer is $\boxed{052}$ . Solution by Zeroman

Solution:
We find that the volume of the cone is $100\pi$
The volume of the cone with height 9 is $\frac{675}{16}\pi$
The difference between the two volumes is $\frac{925}{16}\pi$ . Note that this is the volume of the cone essentially 'on top of' the frustum described in the problem when the liquid is held with the base horizontal.
We can express the volume as $x^2\pi\cdot\frac{12}{5}x\cdot\frac{1}{3}$ , where x is the radius of the cone.
Solving this equation, we get $x=\frac{5\sqrt[3]{37}}{4}$ . The height of this cone is $\frac{12}{5}$ of the radius.
Then we subtract the value from the height, 12, to get our answer: $12-3\sqrt[3]{37}$
Therefore, our answer is $12+3+37=\boxed{52}$ . ~MC413551


Answer: 300.0

A convex polyhedron $Q$ has vertices $V_1,V_2,\ldots,V_n$ , and $100$ edges. The polyhedron is cut by planes $P_1,P_2,\ldots,P_n$ in such a way that plane $P_k$ cuts only those edges that meet at vertex $V_k$ . In addition, no two planes intersect inside or on $Q$ . The cuts produce $n$ pyramids and a new polyhedron $R$ . How many edges does $R$ have?
$\mathrm{(A)}\ 200\qquad \mathrm{(B)}\ 2n\qquad \mathrm{(C)}\ 300\qquad \mathrm{(D)}\ 400\qquad \mathrm{(E)}\ 4n$

Solution: Euler's Polyhedron Formula applied to $Q$ gives $n - 100 + F = 2$ , where F is the number of faces of $Q$ .  Each edge of $Q$ is cut by two planes, so $R$ has $200$ vertices.  Each cut by a plane $P_k$ creates an additional face on $R$ , so Euler's Polyhedron Formula applied to $R$ gives $200 - E + (F+n) = 2$ , where $E$ is the number of edges of $R$ .  Subtracting the first equation from the second gives $300 - E = 0$ , whence $E = \boxed{300}$

Solution:
Each edge connects two points. The plane cuts that edge so it splits into $2$ at each end (like two legs) for a total of $4$ new edges.

But because each new edge is shared by an adjacent original edge cut similarly, the additional edges are overcounted $\times 2$

Since there are $100$ edges to start with, $400/2=200$ new edges result. So there are $100+200=\boxed{300}$ edges in the figure.

Solution:
The question specifies the slices create as many pyramids as there are vertices, implying each vertex owns 4 edge ends. There are twice as many edge-ends as there are edges, and $2 \cdot 100 = 200$
$\frac{200}{4} = 50$ , so there are $50$ vertices.
The base of a pyramid has 4 edges, so each sliced vertex would add four edges to $R$
$100 + 4 \cdot 50$ $\boxed{300}$

Solution:
I doubt anyone has the mental capacity to imagine $Q$ in its full complexity in their mind. So, we begin by examining a simpler structure: a cube.
Let $E_C$ denote the number of edges of a polyhedron $C$ . Let $V_C$ denote the number of vertices of $C$ . Let $B_C$ denote the number of edges that meet at each vertex.
Suppose $C$ is a cube. Then
$E_C = 12$ $V_C = 8$ , and $B_C = 3$
(A cube has $12$ edges, $8$ vertices, and $3$ edges meet per vertex)
Notice that
$V_C*B_C = 2E_C$ $8*3 = 2*12$
Why? Well, imagine an ant walking along an edge of a cube. When it reaches the end of the edge, it will be standing on a vertex (by definition). But, what if it had walked the other way? Well, it would still be on the same edge, but it would reach the other endpoint of the edge, another vertex. Therefore, we can say that a polyhedron with $E$ edges has $2E$ total endpoints. In the case of $C$ , a cube, this means a cube has $2*12 = 24$ total endpoints. But a cube only has $8$ vertices, not $24$ ... This doesn't make any sense... Wait a minute, we're dealing with a polyhedron, which means multiple edges will be connected to the same vertex. In the case of a cube, $3$ edges meet at every vertex. So we have $24 \div 3$ , or $8$ vertices. The math works out. Let's apply this to $Q$ , our mystery polyhedron.
We are told that $E_Q = 100$ $V_Q = n$ and $B_Q$ isn't given explicitly. So we'll let $B_Q = b$ . Now our equation reads
$bn = 200$
What does this mean, exactly? Well $Q$ has $100$ edges, so it must has $200$ endpoints (endpoints, not vertices). So if we can figure out either $n$ , the number of vertices, or $b$ , the number of edges per vertex, we can figure out the other variable. Hmmmm... The problem doesn't give us any clues about either. How frustrating. Let's leave this alone and consider the other information given to us.
We are also told that "the polyhedron is cut by planes $P_1, P_2,...,P_n$ " (at this point I imagine $Q$ is a giant block of cheese, potato, etc.), and also that "plane $P_k$ cuts only those edges that meet at $V_k$ ". What does this mean?
Well, let's go back to our cube. Notice that there are exactly as many planes as vertices. Think of the planes as knives (or knife strokes) cutting into a big cube of cheese. We're cutting the cheese $8$ times, since there are $8$ vertices. Also, whenever we cut into the cheese, we have to direct the cuts near a vertex, cutting into only the edges connected to that vertex. How do we do that? Well, since we can't cut into any other edges, we're gonna have to settle for cutting corners (see what I did there? ...Ok, I admit it was cheesy... see what I did there, too?😂). We have to shave off just the corners of the cheese block, because this way it won't interfere with any of the other edges.
The problem also tells us that none of the planes overlap. In our cheese analogy, this means that every time we cut, we have to make sure the cuts are always into new cheese, and not into previously cut areas (all of the cuts are distinct, and don't overlap). Sure enough, this is in agreement with the information that exactly $n$ pyramids are cut off. Let's go back to $Q$
$Q$ has $n$ vertices, so we're gonna make $n$ cuts. The question we want to answer is: "After all of the cuts are made, how many total edges will $R$ (the figure that remains), have?". Well, we're gonna have to figure out either $n$ or $b$ ... or do we? We're told that $n$ pyramids were made, but it doesn't specify exactly how many sides the pyramid has. So... that implies that it doesn't matter, or else the problem wouldn't be solvable. Let's suppose $b = 4$ $4$ edges meet at every vertex). Then we would have $200 \div 4 = 50$ vertices, and we would make $50$ cuts. But how many more edges does it yield? (We are told that the planes don't overlap, so the original edges are all present.) Well, if we cut off the corner of a cube, it will create $3$ more edges... which is coincident with the fact that $3$ edges meet at every vertex. So for every vertex, we gain exactly $b$ new edges. Rewriting the equation to find the number of vertices, we have
$v = \frac{200}{b}$
But we just said that for every vertex/cut, we gain $b$ more edges. So the total number of new edges is $v*b \Rightarrow \frac{200}{b}*b \Rightarrow 200$ . It doesn't matter how many vertices $Q$ has, or how many edges meet at every vertex. Given the conditions in the problem, we will always gain $200$ more edges, regardless of the values of $n$ , or $b$ . We have $200$ new edges $+$ $100$ old edges to make $300$ edges $\Rightarrow \boxed{300}$


Answer: 300.0

Solution: Each edge of $Q$ is cut by two planes, so $R$ has $200$ vertices. Three edges of $R$ meet at each vertex, so $R$ has $\frac 12 \cdot 3 \cdot 200 = \boxed{300}$ edges.

Solution: At each vertex, as many new edges are created by this process as there are original edges meeting at that vertex.  Thus the total number of new edges is the total number of endpoints of the original edges, which is $200$ .  A middle portion of each original edge is also present in $R$ , so $R$ has $100 + 200 = \boxed{300}$ edges.


Answer: 47.0

A convex quadrilateral has area $30$ and side lengths $5, 6, 9,$ and $7,$ in that order. Denote by $\theta$ the measure of the acute angle formed by the diagonals of the quadrilateral. Then $\tan \theta$ can be written in the form $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution:
Since we are asked to find $\tan \theta$ , we can find $\sin \theta$ and $\cos \theta$ separately and use their values to get $\tan \theta$ . We can start by drawing a diagram. Let the vertices of the quadrilateral be $A$ $B$ $C$ , and $D$ . Let $AB = 5$ $BC = 6$ $CD = 9$ , and $DA = 7$ . Let $AX = a$ $BX = b$ $CX = c$ , and $DX = d$ . We know that $\theta$ is the acute angle formed between the intersection of the diagonals $AC$ and $BD$  We are given that the area of quadrilateral $ABCD$ is $30$ . We can express this area using the areas of triangles $AXB$ $BXC$ $CXD$ , and $DXA$ . Since we want to find $\sin \theta$ and $\cos \theta$ , we can represent these areas using $\sin \theta$ as follows: \begin{align*} 30 &=[ABCD] \\ &=[AXB] + [BXC] + [CXD] + [DXA] \\ &=\frac{1}{2} ab \sin (\angle AXB) + \frac{1}{2} bc \sin (\angle BXC) + \frac{1}{2} cd \sin (\angle CXD) + \frac{1}{2} da \sin (\angle AXD) \\ &=\frac{1}{2} ab \sin (180^\circ - \theta) + \frac{1}{2} bc \sin (\theta) + \frac{1}{2} cd \sin (180^\circ - \theta) + \frac{1}{2} da \sin (\theta). \end{align*} We know that $\sin (180^\circ - \theta) = \sin \theta$ . Therefore it follows that: \begin{align*} 30 &=\frac{1}{2} ab \sin (180^\circ - \theta) + \frac{1}{2} bc \sin (\theta) + \frac{1}{2} cd \sin (180^\circ - \theta) + \frac{1}{2} da \sin (\theta) \\ &=\frac{1}{2} ab \sin (\theta) + \frac{1}{2} bc \sin (\theta) + \frac{1}{2} cd \sin (\theta) + \frac{1}{2} da \sin (\theta) \\ &=\frac{1}{2}\sin\theta (ab + bc + cd + da). \end{align*} From here we see that $\sin \theta = \frac{60}{ab + bc + cd + da}$ . Now we need to find $\cos \theta$ . Using the Law of Cosines on each of the four smaller triangles, we get following equations: \begin{align*}   5^2 &= a^2 + b^2 - 2ab\cos(180^\circ-\theta), \\   6^2 &= b^2 + c^2 - 2bc\cos \theta, \\   9^2 &= c^2 + d^2 - 2cd\cos(180^\circ-\theta), \\   7^2 &= d^2 + a^2 - 2da\cos \theta.   \end{align*} We know that $\cos (180^\circ - \theta) = -\cos \theta$ for all $\theta$ . We can substitute this value into our equations to get: \begin{align*}   5^2 &= a^2 + b^2 + 2ab\cos \theta, &&(1) \\   6^2 &= b^2 + c^2 - 2bc\cos \theta, &&(2) \\   9^2 &= c^2 + d^2 + 2cd\cos \theta, &&(3) \\   7^2 &= d^2 + a^2 - 2da\cos \theta. &&(4) \end{align*} If we subtract $(2)+(4)$ from $(1)+(3)$ , the squared terms cancel, leaving us with: \begin{align*} 5^2 + 9^2 - 6^2 - 7^2 &= 2ab \cos \theta + 2bc \cos \theta + 2cd \cos \theta + 2da \cos \theta \\  21 &= 2\cos \theta (ab  + bc + cd + da). \end{align*} From here we see that $\cos \theta = \frac{21/2}{ab + bc + cd + da}$
Since we have figured out $\sin \theta$ and $\cos \theta$ , we can calculate $\tan \theta$ \[\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{60}{ab + bc + cd + da}}{\frac{21/2}{ab + bc + cd + da}} = \frac{60}{21/2} = \frac{120}{21} = \frac{40}{7}.\] Therefore our answer is $40 + 7 = \boxed{047}$

Solution: In convex quadrilateral $ABCD,$ let $AB=5,BC=6,CD=9,$ and $DA=7.$ Let $A'$ and $C'$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to $\overline{BD}.$ We obtain the following diagram:  Let $BC'=p,C'E=q,EA'=r,A'D=s,AA'=h_1,$ and $CC'=h_2.$ We apply the Pythagorean Theorem to right triangles $\triangle ABA',\triangle BCC',\triangle CDC',$ and $\triangle DAA',$ respectively: \[\begin{array}{ccccccccccccccccc} (p+q+r)^2&+&h_1^2&=&5^2, &&&&&&&&&&&&\hspace{36mm}(1) \\ [1ex] p^2&+&h_2^2&=&6^2, &&&&&&&&&&&&\hspace{36mm}(2) \\ [1ex] (q+r+s)^2&+&h_2^2&=&9^2, &&&&&&&&&&&&\hspace{36mm}(3) \\ [1ex] s^2&+&h_1^2&=&7^2. &&&&&&&&&&&&\hspace{36mm}(4) \end{array}\] Let the brackets denote areas. We get \begin{align*} [ABD]+[CBD]&=[ABCD] \\ \frac12(p+q+r+s)h_1+\frac12(p+q+r+s)h_2&=30 \\ \frac12(p+q+r+s)(h_1+h_2)&=30 \\ (p+q+r+s)(h_1+h_2)&=60. \hspace{49.25mm}(5) \end{align*} We subtract $(2)+(4)$ from $(1)+(3):$ \begin{align*} (p+q+r)^2+(q+r+s)^2-p^2-s^2&=21 \\ \left[(p+q+r)^2-s^2\right]+\left[(q+r+s)^2-p^2\right]&=21 \\ (p+q+r+s)(p+q+r-s)+(p+q+r+s)(-p+q+r+s)&=21 \\ (p+q+r+s)(2q+2r)&=21 \\ 2(p+q+r+s)(q+r)&=21 \\ (p+q+r+s)(q+r)&=\frac{21}{2}. \hspace{9.5mm}(6) \end{align*} From right triangles $\triangle AEA'$ and $\triangle CEC',$ we have $\tan\theta=\frac{h_1}{r}=\frac{h_2}{q}.$ It follows that \begin{alignat*}{8} \tan\theta&=\frac{h_1}{r}\qquad&\implies\qquad h_1&=r\tan\theta, \hspace{64mm}&(1\star)\\ \tan\theta&=\frac{h_2}{q}\qquad&\implies\qquad h_2&=q\tan\theta. &(2\star) \end{alignat*} Finally, we divide $(5)$ by $(6):$ \begin{align*} \frac{h_1+h_2}{q+r}&=\frac{40}{7} \\ \frac{r\tan\theta+q\tan\theta}{q+r}&=\frac{40}{7} \hspace{15mm} &&\text{by }(1\star)\text{ and }(2\star)\\ \frac{(r+q)\tan\theta}{q+r}&=\frac{40}{7} \\ \tan\theta&=\frac{40}{7}, \end{align*} from which the answer is $40+7=\boxed{047}.$

Solution: By Bretschneider's Formula, \[30=\tfrac{1}{4}\sqrt{4u^2v^2-(b^2+d^2-a^2-c^2)^2}=\tfrac{1}{4}\sqrt{4u^2v^2-441}.\] Thus, $uv=3\sqrt{1649}$ . Also, \[[ABCD]=\tfrac 12 \cdot uv\sin{\theta};\] solving for $\sin{\theta}$ yields $\sin{\theta}=\tfrac{40}{\sqrt{1649}}$ . Since $\theta$ is acute, $\cos{\theta}$ is positive, from which $\cos{\theta}=\tfrac{7}{\sqrt{1649}}$ . Solving for $\tan{\theta}$ yields \[\tan{\theta}=\frac{\sin{\theta}}{\cos{\theta}}=\frac{40}{7},\] for a final answer of $\boxed{047}$


Answer: 49.0

Solution: The same pattern is repeated for every $6 \times 6$ tile. Looking closer, there is also symmetry of the top $3 \times 3$ square, so the fraction of the entire floor in dark tiles is the same as the fraction in the square. Counting the tiles, there are $4$ dark tiles, and $9$ total tiles, giving a fraction of $\boxed{49}$


Answer: 108.0

A cryptographer devises the following method for encoding positive integers. First, the integer is expressed in base $5$ .
Second, a 1-to-1 correspondence is established between the digits that appear in the expressions in base $5$ and the elements of the set $\{V, W, X, Y, Z\}$ . Using this correspondence, the cryptographer finds that three consecutive integers in increasing
order are coded as $VYZ, VYX, VVW$ , respectively. What is the base- $10$ expression for the integer coded as $XYZ$
$\textbf{(A)}\ 48 \qquad \textbf{(B)}\ 71 \qquad \textbf{(C)}\ 82 \qquad \textbf{(D)}\ 108 \qquad \textbf{(E)}\ 113$

Solution: Since $VYX + 1 = VVW$ , i.e. adding $1$ causes the "fives" digit to change, we must have $X = 4$ and $W = 0$ . Now since $VYZ + 1 = VYX$ , we have $X = Z + 1 \implies Z = 4 - 1 = 3$ . Finally, note that in $VYX + 1 = VVW$ , adding $1$ will cause the "fives" digit to change by $1$ if it changes at all, so $V = Y + 1$ , and thus since $1$ and $2$ are the only digits left (we already know which letters are assigned to $0$ $3$ , and $4$ ), we must have $V = 2$ and $Y = 1$ . Thus $XYZ = 413_{5} = 4 \cdot 5^{2} + 1 \cdot 5 + 3 = 100 + 5 + 3 = 108$ , which is answer $\boxed{108}$


Answer: 17.0

A cube has edge length $2$ . Suppose that we glue a cube of edge length $1$ on top of the big cube so that one of its faces rests entirely on the top face of the larger cube. The percent increase in the surface area (sides, top, and bottom) from the original cube to the new solid formed is closest to
[asy] draw((0,0)--(2,0)--(3,1)--(3,3)--(2,2)--(0,2)--cycle); draw((2,0)--(2,2)); draw((0,2)--(1,3)); draw((1,7/3)--(1,10/3)--(2,10/3)--(2,7/3)--cycle); draw((2,7/3)--(5/2,17/6)--(5/2,23/6)--(3/2,23/6)--(1,10/3)); draw((2,10/3)--(5/2,23/6)); draw((3,3)--(5/2,3));[/asy]
$\text{(A)}\ 10\qquad\text{(B)}\ 15\qquad\text{(C)}\ 17\qquad\text{(D)}\ 21\qquad\text{(E)}\ 25$

Solution:
The original cube has $6$ faces, each with an area of $2\cdot 2 = 4$ square units.  Thus the original figure had a total surface area of $24$ square units.
The new figure has the original surface, with $6$ new faces that each have an area of $1$ square unit, for a total surface area of of $6$ additional square units added to it.  But $1$ square unit of the top of the bigger cube, and $1$ square unit on the bottom of smaller cube, is not on the surface, and does not count towards the surface area.
The total surface area is therefore $24 + 6 - 1 - 1 = 28$ square units.
The percent increase in surface area is $\frac{SA_{new} - SA_{old}}{SA_{old}}\cdot 100\% = \frac{28-24}{24}\cdot 100\% \approx 16.67\%$ , giving the closest answer as $\boxed{17}$


Answer: 16.0

A cube has eight vertices (corners) and twelve edges.  A segment, such as $x$ , which joins two vertices not joined by an edge is called a diagonal.  Segment $y$ is also a diagonal.  How many diagonals does a cube have?
[asy] draw((0,3)--(0,0)--(3,0)--(5.5,1)--(5.5,4)--(3,3)--(0,3)--(2.5,4)--(5.5,4)); draw((3,0)--(3,3)); draw((0,0)--(2.5,1)--(5.5,1)--(0,3)--(5.5,4),dashed); draw((2.5,4)--(2.5,1),dashed); label("$x$",(2.75,3.5),NNE); label("$y$",(4.125,1.5),NNE); [/asy]
$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16$

Solution:
On each face, there are $2$ diagonals like $x$ .  There are $6$ faces on a cube.  Thus, there are $2\times 6 = 12$ diagonals that are "x-like".
Every "y-like" diagonal must connect the bottom of the cube to the top of the cube.  Thus, for each of the $4$ bottom vertices of the cube, there is a different "y-like" diagonal.  So there are $4$ "y-like" diagonals.
This gives a total of $12 + 4 = 16$ diagonals on the cube, which is answer $\boxed{16}$

Solution:
There are $8$ vertices on a cube(A, B, C, D, E, F, G, H).  If you pick one vertice, it will have 7 line segments associated with it, so you will have $\frac{8\cdot7}{2} = 28$ segments within the cube.  The division by $2$ is necessary because you counted both the segment from $A$ to $B$ and the segment from $B$ to $A$
But not all $28$ of these segments are diagonals.  Some are edges.  There are $4$ edges on the top, $4$ edges on the bottom, and $4$ edges that connect the top to the bottom.  So there are $12$ edges total, meaning that there are $28 - 12 = 16$ segments that are not edges.  All of these segments are diagonals, and thus the answer is $\boxed{16}$

Solution:
Consider picking one point on the corner of the cube.  That point has $3$ "x-like" diagonals , and $1$ "y-like" diagonal that ends on the opposite vertex.  Thus, each vertex has $3 + 1 = 4$ diagonals associated with it.  There are $8$ vertices on the cube, giving a total of $8 \cdot 4 = 32$ diagonals.
However, each diagonal was counted as both a "starting point" and an "ending point".  So there are really $\frac{32}{2} = 16$ diagonals, giving an answer of $\boxed{16}$


Answer: 7.0

A cube is constructed from $4$ white unit cubes and $4$ blue unit cubes. How many different ways are there to construct the $2 \times 2 \times 2$ cube using these smaller cubes? (Two constructions are considered the same if one can be rotated to match the other.)
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\  8 \qquad\textbf{(C)}\  9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 11$

Solution:
This problem is about the relationships between the white unit cubes and the blue unit cubes, which can be solved by Graph Theory . We use a Planar Graph to represent the larger cube. Each vertex of the planar graph represents a unit cube. Each edge of the planar graph represents a shared face between $2$ neighboring unit cubes. Each face of the planar graph represents a face of the larger cube.
Now the problem becomes a Graph Coloring problem of how many ways to assign $4$ vertices blue and $4$ vertices white with Topological Equivalence . For example, in Figure $(1)$ , as long as the $4$ blue vertices belong to the same planar graph face, the different planar graphs are considered to be topological equivalent by rotating the larger cube.
Topology.jpg
Here is how the $4$ blue unit cubes are arranged:
In Figure $(1)$ $4$ blue unit cubes are on the same layer (horizontal or vertical).
In Figure $(2)$ $4$ blue unit cubes are in $T$ shape.
In Figure $(3)$ and $(4)$ $4$ blue unit cubes are in $S$ shape.
In Figure $(5)$ $3$ blue unit cubes are in $L$ shape, and the other is isolated without a shared face.
In Figure $(6)$ $2$ pairs of neighboring blue unit cubes are isolated from each other without a shared face.
In Figure $(7)$ $4$ blue unit cubes are isolated from each other without a shared face.
So the answer is $\boxed{7}$

Solution:
Let’s split the cube into two layers; a bottom and top. Note that there must be four of each color, so however many number of one color are in the bottom, there will be four minus that number of the color on the top. We do casework on the color distribution of the bottom layer.
Case 1: 4, 0
In this case, there is only one possibility for the top layer - all of the other color - $\binom{4}{4}$ . Therefore there is 1 construction from this case.
Case 2: 3, 1
In this case, the top layer has four possibilities, because there are four different ways to arrange it so that it also has a 3, 1 color distribution - $\binom{4}{3}$ . Therefore there are 4 constructions from this case.
Case 3: 2, 2
In this case, the top layer has six possibilities of arrangement - $\binom{4}{2}$ . However, having adjacent colors one way can be rotated to having adjacent colors any other way, so there is only one construction for the adjacent colors subcase and similarly, only one for the diagonal color subcase. Therefore the total number of constructions for this case is 2.
The total number of constructions for the cube is thus $1+4+2=7=\boxed{7}$

Solution:
Divide the $2 \times 2 \times 2$ cube into two layers, say, front and back. Any possible construction can be rotated such that the front layer has the same or greater number of white cubes than blue cubes, so we only need to count the number of cases given that is true.
Therefore, our answer is $6+1+0=\boxed{7}$

Solution:
Since rotations of a single pattern are considered indistinguishable, we can assume that the forward upper right corner of the 2-by-2-by-2 cube is a blue cube (since we can always rotate the big cube to place a blue cube in that spot).
Once we've assigned this cube to be blue, we note that 3 1-by-1-by-1 cubes share a side with it, 3 1-by-1-by-1 cubes share a corner with it, and 1 1-by-1-by-1 cube does not touch the assigned cube at all, from the perspective of someone who can only see the cube's faces. We'll call the first 3 "adjacent", the second 3 "cornering", and the last one "opposite."
We can use a little bit of intuition to confirm that due to the rotation condition, we should treat all adjacents as indistinguishable, all cornerings as indistinguishable, and of course the opposite one is unique from all the others. Thus, we can list out like so (keeping in mind that there are 3 adjacents, 3 cornerings, and 1 opposite, and that we're choosing the positions of the remaining 3 blue cubes):
OAA, OAC, OCC, CCC, CAA, CCA, AAA.
This gives the answer to be $\boxed{7}$


Answer: 7.0

Solution:
Burnside lemma is used to counting number of orbit where the element on the same orbit can be achieved by the defined operator, naming rotation, reflection and etc.
The fact for Burnside lemma are
1. the sum of stablizer on the same orbit equals to the # of operators;
2. the sum of stablizer can be counted as $fix(g)$
3. the sum of the $fix(g)/|G|$ equals the # of orbit.
Let's start with defining the operator for a cube,
1. $\textbf{e (identity)}$
For identity, there are $\frac{8!}{4!4!} = 70$
2. ${\bf r^{1}, r^{2}, r^{3}}$ to be the rotation axis along three pair of opposite face,
each contains $r^{i}_{90}, r^{i}_{180}, r^{i}_{270}$ where $i= 1, 2, 3$
$fix(r^{i}_{90}) = fix(r^{i}_{270}) = 2\cdot1 = 2$
$fix(r^{i}_{180}) = \frac{4!}{2!\cdot2!} = 6$
therefore $fix(\bf r^{i}) = 2+2+6 = 10$ , and $fix(\bf r^{1})+fix(\bf r^{2})+fix(\bf r^{3}) = 30$
3. ${\bf r^{4}, r^{5}, r^{6}, r^{7}}$ to the rotation axis along four cube diagnals.
each contains $r^{i}_{120}, r^{i}_{240}$ where $i= 4, 5, 6, 7$
$fix(r^{i}_{120}) = fix(r^{i}_{240}) = 2\cdot1\cdot2\cdot1 = 4$
therefore $fix(\bf r^{i}) = 4+4 = 8$ , and $fix(\bf r^{4})+fix(\bf r^{5})+fix(\bf r^{6})+fix(\bf r^{7}) = 32$
4. ${\bf r^{8}, r^{9}, r^{10}, r^{11}, r^{12}, r^{13}}$ to be the rotation axis along 6 pairs of diagnally opposite sides
each contains $r^{i}_{180}$ where $i= 8, 9, 10, 11, 12, 13$
$fix(r^{i}_{180}) = \frac{4!}{2!\cdot2!} = 6$
therefore $fix(\bf r^{8})+fix(\bf r^{9})+fix(\bf r^{10})+fix(\bf r^{11})+fix(\bf r^{12})+fix(\bf r^{13}) = 36$
5. The total number of operators are
$|G| = 1 + 3\cdot3 + 4\cdot2 + 6\cdot1 = 24$
Based on 1, 2, 3, 4 the total number of stablizer is $70 + 30 + 32 + 36 = 168$
therefore the number of orbit $= \frac{168}{G=24} = \boxed{7}$


Answer: 7.0

Solution:
This problem is about the relationships between the white unit cubes and the blue unit cubes, which can be solved by Graph Theory . We use a Planar Graph to represent the larger cube. Each vertex of the planar graph represents a unit cube. Each edge of the planar graph represents a shared face between $2$ neighboring unit cubes. Each face of the planar graph represents a face of the larger cube.
Now the problem becomes a Graph Coloring problem of how many ways to assign $4$ vertices blue and $4$ vertices white with Topological Equivalence . For example, in Figure $(1)$ , as long as the $4$ blue vertices belong to the same planar graph face, the different planar graphs are considered to be topological equivalent by rotating the larger cube.
Topology.jpg
Here is how the $4$ blue unit cubes are arranged:
In Figure $(1)$ $4$ blue unit cubes are on the same layer (horizontal or vertical).
In Figure $(2)$ $4$ blue unit cubes are in $T$ shape.
In Figure $(3)$ and $(4)$ $4$ blue unit cubes are in $S$ shape.
In Figure $(5)$ $3$ blue unit cubes are in $L$ shape, and the other is isolated without a shared face.
In Figure $(6)$ $2$ pairs of neighboring blue unit cubes are isolated from each other without a shared face.
In Figure $(7)$ $4$ blue unit cubes are isolated from each other without a shared face.
So the answer is $\boxed{7}$

Solution:
Let’s split the cube into two layers; a bottom and top. Note that there must be four of each color, so however many number of one color are in the bottom, there will be four minus that number of the color on the top. We do casework on the color distribution of the bottom layer.
Case 1: 4, 0
In this case, there is only one possibility for the top layer - all of the other color - $\binom{4}{4}$ . Therefore there is 1 construction from this case.
Case 2: 3, 1
In this case, the top layer has four possibilities, because there are four different ways to arrange it so that it also has a 3, 1 color distribution - $\binom{4}{3}$ . Therefore there are 4 constructions from this case.
Case 3: 2, 2
In this case, the top layer has six possibilities of arrangement - $\binom{4}{2}$ . However, having adjacent colors one way can be rotated to having adjacent colors any other way, so there is only one construction for the adjacent colors subcase and similarly, only one for the diagonal color subcase. Therefore the total number of constructions for this case is 2.
The total number of constructions for the cube is thus $1+4+2=7=\boxed{7}$

Solution:
Divide the $2 \times 2 \times 2$ cube into two layers, say, front and back. Any possible construction can be rotated such that the front layer has the same or greater number of white cubes than blue cubes, so we only need to count the number of cases given that is true.
Therefore, our answer is $6+1+0=\boxed{7}$

Solution:
Since rotations of a single pattern are considered indistinguishable, we can assume that the forward upper right corner of the 2-by-2-by-2 cube is a blue cube (since we can always rotate the big cube to place a blue cube in that spot).
Once we've assigned this cube to be blue, we note that 3 1-by-1-by-1 cubes share a side with it, 3 1-by-1-by-1 cubes share a corner with it, and 1 1-by-1-by-1 cube does not touch the assigned cube at all, from the perspective of someone who can only see the cube's faces. We'll call the first 3 "adjacent", the second 3 "cornering", and the last one "opposite."
We can use a little bit of intuition to confirm that due to the rotation condition, we should treat all adjacents as indistinguishable, all cornerings as indistinguishable, and of course the opposite one is unique from all the others. Thus, we can list out like so (keeping in mind that there are 3 adjacents, 3 cornerings, and 1 opposite, and that we're choosing the positions of the remaining 3 blue cubes):
OAA, OAC, OCC, CCC, CAA, CCA, AAA.
This gives the answer to be $\boxed{7}$


Answer: 7.0

Solution:
Burnside lemma is used to counting number of orbit where the element on the same orbit can be achieved by the defined operator, naming rotation, reflection and etc.
The fact for Burnside lemma are
1. the sum of stablizer on the same orbit equals to the # of operators;
2. the sum of stablizer can be counted as $fix(g)$
3. the sum of the $fix(g)/|G|$ equals the # of orbit.
Let's start with defining the operator for a cube,
1. $\textbf{e (identity)}$
For identity, there are $\frac{8!}{4!4!} = 70$
2. ${\bf r^{1}, r^{2}, r^{3}}$ to be the rotation axis along three pair of opposite face,
each contains $r^{i}_{90}, r^{i}_{180}, r^{i}_{270}$ where $i= 1, 2, 3$
$fix(r^{i}_{90}) = fix(r^{i}_{270}) = 2\cdot1 = 2$
$fix(r^{i}_{180}) = \frac{4!}{2!\cdot2!} = 6$
therefore $fix(\bf r^{i}) = 2+2+6 = 10$ , and $fix(\bf r^{1})+fix(\bf r^{2})+fix(\bf r^{3}) = 30$
3. ${\bf r^{4}, r^{5}, r^{6}, r^{7}}$ to the rotation axis along four cube diagnals.
each contains $r^{i}_{120}, r^{i}_{240}$ where $i= 4, 5, 6, 7$
$fix(r^{i}_{120}) = fix(r^{i}_{240}) = 2\cdot1\cdot2\cdot1 = 4$
therefore $fix(\bf r^{i}) = 4+4 = 8$ , and $fix(\bf r^{4})+fix(\bf r^{5})+fix(\bf r^{6})+fix(\bf r^{7}) = 32$
4. ${\bf r^{8}, r^{9}, r^{10}, r^{11}, r^{12}, r^{13}}$ to be the rotation axis along 6 pairs of diagnally opposite sides
each contains $r^{i}_{180}$ where $i= 8, 9, 10, 11, 12, 13$
$fix(r^{i}_{180}) = \frac{4!}{2!\cdot2!} = 6$
therefore $fix(\bf r^{8})+fix(\bf r^{9})+fix(\bf r^{10})+fix(\bf r^{11})+fix(\bf r^{12})+fix(\bf r^{13}) = 36$
5. The total number of operators are
$|G| = 1 + 3\cdot3 + 4\cdot2 + 6\cdot1 = 24$
Based on 1, 2, 3, 4 the total number of stablizer is $70 + 30 + 32 + 36 = 168$
therefore the number of orbit $= \frac{168}{G=24} = \boxed{7}$


Answer: 6.0

A cube of cheese $C=\{(x, y, z)| 0 \le x, y, z \le 1\}$ is cut along the planes $x=y, y=z$ and $z=x$ . How many pieces are there?
(No cheese is moved until all three cuts are made.)
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$

Solution: The cut $x = y$ separates the cube into points with $x < y$ and points with $x > y$ , and analogous results apply for the other cuts. Thus, which piece a particular point is in depends only on the relative sizes of its coordinates $x$ $y$ , and $z$ - for example, all points with the ordering $x < y < z$ are in the same piece. Thus, as there are $3! = 6$ possible orderings, there are $6$ pieces, which is answer $\boxed{6}$


Answer: 20.0

A cube of edge $3$ cm is cut into $N$ smaller cubes, not all the same size.  If the edge of each of the smaller cubes is a whole number of centimeters, then $N=$
$\text{(A)}\ 4 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 20$

Solution:
If none of the cubes have edge length $2$ , then all of the cubes have edge length $1$ , meaning they all are the same size, a contradiction.
It is clearly impossible to split a cube of edge $3$ into two or more cubes of edge $2$ with extra unit cubes, so there is one $2\times 2\times 2$ cube and $3^3-2^3=19$ unit cubes.
The total number of cubes, $N$ , is $1+19=20\rightarrow \boxed{20}$


Answer: 330.0

A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled $A$ . The three vertices adjacent to vertex $A$ are at heights 10, 11, and 12 above the plane. The distance from vertex $A$ to the plane can be expressed as $\frac{r-\sqrt{s}}{t}$ , where $r$ $s$ , and $t$ are positive integers. Find $r+s+t$

Solution:
Set the cube at the origin with the three vertices along the axes and the plane equal to $ax+by+cz+d=0$ , where $a^2+b^2+c^2=1$ .  The distance from a point $(X,Y,Z)$ to a plane with equation $Ax+By+Cz+D=0$ is \[\frac{AX+BY+CZ+D}{\sqrt{A^2+B^2+C^2}},\] so the (directed) distance from any point $(x,y,z)$ to the plane is $ax+by+cz+d$ .  So, by looking at the three vertices, we have $10a+d=10, 10b+d=11, 10c+d=12$ , and by rearranging and summing, \[(10-d)^2+(11-d)^2+(12-d)^2= 100\cdot(a^2+b^2+c^2)=100.\]
Solving the equation is easier if we substitute $11-d=y$ , to get $3y^2+2=100$ , or $y=\sqrt {98/3}$ .  The distance from the origin to the plane is simply $d$ , which is equal to $11-\sqrt{98/3} =(33-\sqrt{294})/3$ , so $33+294+3=\boxed{330}$

Solution:
Let the vertices with distance $10,11,12$ be $B,C,D$ , respectively. An equilateral triangle $\triangle BCD$ is formed with side length $10\sqrt{2}$ . We care only about the $z$ coordinate: $B=10,C=11,D=12$ . It is well known that the centroid of a triangle is the average of the coordinates of its three vertices, so $\text{centroid}=(10+11+12)/3=11$ . Designate the midpoint of $BD$ as $M$ . Notice that median $CM$ is parallel to the plane because the $\text{centroid}$ and vertex $C$ have the same $z$ coordinate, $11$ , and the median contains $C$ and the $\text{centroid}$ . We seek the angle $\theta$ of the line: $(1)$ through the centroid $(2)$ perpendicular to the plane formed by $\triangle BCD$ $(3)$ with the plane under the cube. Since the median is parallel to the plane, this orthogonal line is also perpendicular $\textit{in slope}$ to $BD$ . Since $BD$ makes a $2-14-10\sqrt{2}$ right triangle, the orthogonal line makes the same right triangle rotated $90^\circ$ . Therefore, $\sin\theta=\frac{14}{10\sqrt{2}}=\frac{7\sqrt{2}}{10}$
It is also known that the centroid of $\triangle BCD$ is a third of the way between vertex $A$ and $H$ , the vertex farthest from the plane. Since $AH$ is a diagonal of the cube, $AH=10\sqrt{3}$ . So the distance from the $\text{centroid}$ to $A$ is $10/\sqrt{3}$ . So, the $\Delta z$ from $A$ to the centroid is $\frac{10}{\sqrt{3}}\sin\theta=\frac{10}{\sqrt{3}}\left(\frac{7\sqrt{2}}{10}\right)=\frac{7\sqrt{6}}{3}$
Thus the distance from $A$ to the plane is $11-\frac{7\sqrt{6}}{3}=\frac{33-7\sqrt{6}}{3}=\frac{33-\sqrt{294}}{3}$ , and $33+294+3=\boxed{330}$


Answer: 751.0

A cube-shaped container has vertices $A,$ $B,$ $C,$ and $D,$ where $\overline{AB}$ and $\overline{CD}$ are parallel edges of the cube, and $\overline{AC}$ and $\overline{BD}$ are diagonals of faces of the cube, as shown. Vertex $A$ of the cube is set on a horizontal plane $\mathcal{P}$ so that the plane of the rectangle $ABDC$ is perpendicular to $\mathcal{P},$ vertex $B$ is $2$ meters above $\mathcal{P},$ vertex $C$ is $8$ meters above $\mathcal{P},$ and vertex $D$ is $10$ meters above $\mathcal{P}.$ The cube contains water whose surface is parallel to $\mathcal{P}$ at a height of $7$ meters above $\mathcal{P}.$ The volume of water is $\frac{m}{n}$ cubic meters, where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution:
Let's first view the cube from a direction perpendicular to $ABDC$ , as illustrated above. Let $x$ be the cube's side length. Since $\triangle CHA \sim \triangle AGB$ , we have \[\frac{CA}{CH} = \frac{AB}{AG}.\] We know $AB = x$ $AG = \sqrt{x^2-2^2}$ $AC = \sqrt{2}x$ $CH = 8$ . Plug them into the above equation, we get \[\frac{\sqrt{2}x}{8} = \frac{x}{\sqrt{x^2-2^2}}.\] Solving this we get the cube's side length $x = 6$ , and $AC = 6\sqrt{2}.$
Let $PQ$ be the water's surface, both $P$ and $Q$ are $7$ meters from $\mathcal P$ . Notice that $C$ is $8$ meters from $\mathcal P$ , this means \[CP = \frac{1}{8}CA = \frac{3\sqrt{2}}{4}.\] Similarly, \[DQ = \frac{3}{8}CA = \frac{9\sqrt{2}}{4}.\]

Now, we realize that the 3D space inside the cube without water is a frustum, with $P$ on its smaller base and $Q$ on its larger base. To find its volume, all we need is to find the areas of both bases and the height, which is $x = 6$ . To find the smaller base, let's move our viewpoint onto the plane $ABDC$ and view the cube from a direction parallel to $ABDC$ , as shown above. The area of the smaller base is simply \[S_1 = CP^2 = \Bigl(\frac{3\sqrt{2}}{4}\Bigr)^2 = \frac{9}{8}.\] Similarly, the area of the larger base is \[S_2 = DQ^2 = \Bigl(\frac{9\sqrt{2}}{4}\Bigr)^2 = \frac{81}{8}.\]
Finally, applying the formula for a frustum's volume,
\[V = \frac{1}{3} \cdot x \cdot (S_1 + \sqrt{S_1S_2} + S_2) = \frac{1}{3} \cdot 6 \cdot \Bigl(\frac{9}{8} + \sqrt{\frac{9}{8}\cdot\frac{81}{8}} + \frac{81}{8}\Bigl) = \frac{117}{4}.\]
The water's volume is thus \[6^3 - \frac{117}{4} = \frac{747}{4},\] giving $\boxed{751}$

Solution:
Denote $h(X)$ the distance from point $X$ to $\mathcal{P},  h(A) = 0, h(B) = 2,$ $h(C) = 8, h(D) = 10, h(G) = h(I) = h(H) = 7,  AB = a, AC = a \sqrt{2}.$
Let slope $AB$ to $\mathcal{P}$ be $\alpha.$ Notation is shown in the diagram. \[\tan \alpha = \frac {\sin \alpha}{\cos \alpha} = \frac {h(B)}{AB}\cdot \frac {AC}{h(C)} = \frac {\sqrt{2}}{4} \implies a = 6.\] Let $S = GI \cap CD \implies h(S) = h(G) = 7.$ \[h(C) – h(G) = 8 - 7 = 1, h(D)- h(I) = 10 - 7 = 3.\] \[h(E) = h(F) = \frac {h(D) +h(B)}{2} = 6 \implies\] \[\frac {DI}{DE} = \frac {h(D) - h(I)}{h(D)-h(E)} = \frac {3}{4} \implies DI = DH = \frac {9}{2}.\]
Similarly $CG = \frac {3}{2} \implies SD = 9.$
Let the volume without water be $V,$ volume of the pyramid $SCGJ$ be $U.$
It is clear that $U + V = 27U = \frac {SD}{6} \cdot DI^2 = \frac {243}{8} \implies$ $V = \frac {243 \cdot 26}{8 \cdot 27 } = \frac {117}{4} = 6^3 - \frac {747}{4}$ from which $\boxed{751}.$

Solution:
We introduce a Cartesian coordinate system to the diagram.
We put the origin at $A$ . We let the $z$ -components of $B$ $C$ $D$ be positive.
We set the $x$ -axis in a direction such that $B$ is on the $x-O-z$ plane.
The coordinates of $A$ $B$ $C$ are $A = \left( 0, 0, 0 \right)$ $B = \left( x_B, 0 , 2 \right)$ $C = \left( x_C, y_C, 8 \right)$
Because $AB \perp AC$ $\overrightarrow{AB} \cdot \overrightarrow{AC} = 0$ .
Thus, \[ x_B x_C + 16 = 0 . \hspace{1cm} (1) \]
Because $AC$ is a diagonal of a face, $AC^2 = 2 AB^2$ .
Thus, \[ x_C^2 + y_C^2 + 8^2 = 2 \left( x_B^2 + 2^2 \right) . \hspace{1cm} (2) \]
Because plane $ABCD$ is perpendicular to plan $P$ $\hat z \cdot \left( \overrightarrow{AB} \times \overrightarrow{AC} \right) = 0$ .
Thus, \[ \begin{vmatrix} 0 & 0 & 1 \\ x_B & 0 & 2 \\ x_C & y_C & 8 \end{vmatrix} = 0 . \hspace{1cm} (3) \]
Jointly solving (1), (2), (3), we get one solution $x_B = 4 \sqrt{2}$ $x_C = - 2 \sqrt{2}$ $y_C = 0$ .
Thus, the side length of the cube is $|AB| = \sqrt{x_B^2 + 2^2} = 6$
Denote by $P$ and $Q$ two vertices such that $AP$ and $AQ$ are two edges, and satisfy the right-hand rule that $\widehat{AB} \times \widehat{AP} = \widehat{AQ}$ .
Now, we compute the coordinates of $P$ and $Q$
Because $|AB| = 6$ , we have $\overrightarrow{AP} \times \overrightarrow{AQ} = 6 \overrightarrow{AB}$ $\overrightarrow{AQ} \times \overrightarrow{AB} = 6 \overrightarrow{AP}$ $\overrightarrow{AB} \times \overrightarrow{AP} = 6 \overrightarrow{AQ}$
Hence, \begin{align*} \begin{bmatrix} \hat i & \hat j & \hat k \\ x_P & y_P & z_P \\ x_Q & y_Q & z_Q \end{bmatrix} & = 6 \left( 4 \sqrt{2} \hat i + 2 \hat k \right) , \\ \begin{vmatrix} \hat i & \hat j & \hat k \\ x_Q & y_Q & z_Q \\ 4 \sqrt{2} & 0 & 2 \end{vmatrix} & = 6 \left( x_P \hat i + y_P \hat j + z_P \hat k \right) , \\ \begin{vmatrix} \hat i & \hat j & \hat k \\ 4 \sqrt{2} & 0 & 2 \\ x_P & y_P & z_P \end{vmatrix} & = 6 \left( x_Q \hat i + y_Q \hat j + z_Q \hat k \right) . \end{align*}
By solving these equations, we get $y_P^2 + y_Q^2 = 36 .$
In addition, we have $\overrightarrow{AC} = \overrightarrow{AP} + \overrightarrow{AQ}$ .
Thus, $P = \left( - \sqrt{2} , 3 \sqrt{2} , 4 \right)$ $Q = \left( - \sqrt{2} , - 3 \sqrt{2} , 4 \right)$
Therefore, the volume of the water is \begin{align*} V = & 6^3 \int_{u=0}^1 \int_{v=0}^1 \int_{w=0}^1 \mathbf 1 \left\{ z_B u + z_P v + z_Q w \leq 7 \right\} dw dv du \\ & = 6^3 \int_{u=0}^1 \int_{v=0}^1 \int_{w=0}^1 \mathbf 1 \left\{ 2 u + 4 v + 4 w \leq 7 \right\} dw dv du \\ & = 6^3 - 6^3 \int_{u=0}^1 \int_{v=0}^1 \int_{w=0}^1 \mathbf 1 \left\{ 2 u + 4 v + 4 w > 7 \right\} dw dv du . \end{align*}
Define $u' = 1 - u$ $v' = 1 - v$ $w' = 1 - w$ .
Thus, \begin{align*} V & = 6^3 - 6^3 \int_{u=0}^1 \int_{v=0}^1 \int_{w=0}^1 \mathbf 1 \left\{ 2 u' + 4 v' + 4 w' < 3 \right\} dw dv du \\ & = 6^3 - 6^3 \int_{u'=0}^1 \left( \int_{v'=0}^1 \int_{w'=0}^1  \mathbf 1 \left\{ v' + w' < \frac{3}{4} - \frac{u'}{2} \right\} dw' dv' \right)  du' \\ & = 6^3 - 6^3 \int_{u'=0}^1  \frac{1}{2} \left( \frac{3}{4} - \frac{u'}{2} \right)^2 du' . \end{align*}
Define $u'' = \frac{3}{4} - \frac{u'}{2}$ .
Thus, \begin{align*} V & = 6^3 - 6^3 \int_{u'' = 1/4}^{3/4} \left( u'' \right)^2 du'' \\ & = 6^3 - 6^3 \frac{1}{3} \left( \left(\frac{3}{4}\right)^3 - \left(\frac{1}{4}\right)^3 \right) \\ & = 216 - \frac{117}{4} \\ & = \frac{747}{4} . \end{align*}
Therefore, the answer is $747 + 4 = \boxed{751}$


Answer: 86.0

A cup of boiling water ( $212^{\circ}\text{F}$ ) is placed to cool in a room whose temperature remains constant at $68^{\circ}\text{F}$ . Suppose the difference between the water temperature and the room temperature is halved every $5$ minutes. What is the water temperature, in degrees Fahrenheit, after $15$ minutes?
$\textbf{(A) } 77 \qquad \textbf{(B) } 86 \qquad \textbf{(C) } 92 \qquad \textbf{(D) } 98 \qquad \textbf{(E) } 104$

Solution:
Initially, the difference between the water temperature and the room temperature is $212-68=144$ degrees Fahrenheit.
After $5$ minutes, the difference between the temperatures is $144\div2=72$ degrees Fahrenheit.
After $10$ minutes, the difference between the temperatures is $72\div2=36$ degrees Fahrenheit.
After $15$ minutes, the difference between the temperatures is $36\div2=18$ degrees Fahrenheit. At this point, the water temperature is $68+18=\boxed{86}$ degrees Fahrenheit.


Answer: 384.0

A cylindrical barrel with radius $4$ feet and height $10$ feet is full of water. A solid cube with side length $8$ feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is $v$ cubic feet. Find $v^2$
[asy] import three; import solids; size(5cm); currentprojection=orthographic(1,-1/6,1/6);  draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,360)),white,nolight);  triple A =(8*sqrt(6)/3,0,8*sqrt(3)/3), B = (-4*sqrt(6)/3,4*sqrt(2),8*sqrt(3)/3), C = (-4*sqrt(6)/3,-4*sqrt(2),8*sqrt(3)/3), X = (0,0,-2*sqrt(2));  draw(X--X+A--X+A+B--X+A+B+C); draw(X--X+B--X+A+B); draw(X--X+C--X+A+C--X+A+B+C); draw(X+A--X+A+C); draw(X+C--X+C+B--X+A+B+C,linetype("2 4")); draw(X+B--X+C+B,linetype("2 4"));  draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,240)),white,nolight); draw((-2,-2*sqrt(3),0)..(4,0,0)..(-2,2*sqrt(3),0)); draw((-4*cos(atan(5)),-4*sin(atan(5)),0)--(-4*cos(atan(5)),-4*sin(atan(5)),-10)..(4,0,-10)..(4*cos(atan(5)),4*sin(atan(5)),-10)--(4*cos(atan(5)),4*sin(atan(5)),0)); draw((-2,-2*sqrt(3),0)..(-4,0,0)..(-2,2*sqrt(3),0),linetype("2 4")); [/asy]

Solution:
Our aim is to find the volume of the part of the cube submerged in the cylinder.
In the problem, since three edges emanate from each vertex, the boundary of the cylinder touches the cube at three points.  Because the space diagonal of the cube is vertical, by the symmetry of the cube, the three points form an equilateral triangle.  Because the radius of the circle is $4$ , by the Law of Cosines, the side length s of the equilateral triangle is
\[s^2 = 2\cdot(4^2) - 2\cdot(4^2)\cos(120^{\circ}) = 3(4^2)\]
so $s = 4\sqrt{3}$ .* Again by the symmetry of the cube, the volume we want to find is the volume of a tetrahedron with right angles on all faces at the submerged vertex, so since the lengths of the legs of the tetrahedron are $\frac{4\sqrt{3}}{\sqrt{2}} = 2\sqrt{6}$ (the three triangular faces touching the submerged vertex are all $45-45-90$ triangles) so
\[v = \frac{1}{3}(2\sqrt{6})\left(\frac{1}{2} \cdot (2\sqrt{6})^2\right) = \frac{1}{6} \cdot 48\sqrt{6} = 8\sqrt{6}\]
so
\[v^2 = 64 \cdot 6 = \boxed{384}.\]

Solution:
Visualizing the corner which is submerged in the cylinder, we can see that it's like slicing off a corner, where they slice an equal part off every edge, to make a tetrahedron where there are 3 right isosceles triangles and one equilateral triangle. With this out of the way, we can now just find the area of that equilateral triangle, using the fact that the circle of radius $4$ is the circumcircle of the equilateral triangle. Using equilateral triangle properties, you can find that the height of the triangle is $6$ , and the side length is $\frac{6}{\sqrt{3}}=2\sqrt{3} \cdot 2=4\sqrt{3}$ . As the other faces are right isosceles triangles, they are $\frac{4\sqrt{3}}{\sqrt{2}}=2\sqrt{6}$ . Therefore the volume of this tetrahedron is
\[\left(\frac{2\sqrt{6}}{2}\right)^2=12 \ \cdot \ (2\sqrt{6})=24\sqrt{6} \implies \frac{24\sqrt{6}}{3}=8\sqrt{6} \implies (8\sqrt{6})^2=\boxed{384}\]

Solution: We can use the same method as in Solution 2 to find the side length of the equilateral triangle, which is $4\sqrt3$ . From here, its area is \[\dfrac{\bigl(4\sqrt3\bigr)^2\sqrt3}4=12\sqrt3.\] The leg of the isosceles right triangle is $\dfrac{4\sqrt3}{\sqrt2}=2\sqrt6$ , and the horizontal distance from the vertex to the base of the tetrahedron is $4$ (the radius of the cylinder), so we can find the height, as shown in the diagram.  The height from the tip to the base is $2\sqrt2$ , so the volume is $\dfrac{12\sqrt3\cdot2\sqrt2}3=8\sqrt6$ , and thus the answer is $\boxed{384}$

Solution:
Since the diagonal is perpendicular to the base of the cylinder, all three edges and faces can be treated symmetrically.
The cross-section of the cube with the top face of the cylinder is an equilateral triangle inscribed in a circle with radius $4$ . This means the medians of the triangle have length $\frac{3}{2} \cdot 4=6$ , because the circumcenter is also the centroid, and the centroid divides the medians into lengths of ratio $2:1$ . Using $30-60-90$ triangles, the side length of the triangle is $4\sqrt{3}$ , and its area is $\frac{(4\sqrt{3})^2\sqrt{3}}{4}=12\sqrt{3}$
Next, consider the submerged triangular sections of the faces. Each is a $45-45-90$ triangle with leg length $x$ . The area of each is then $\frac{x^2}{2}$ . By De Gua's Theorem on the submerged pyramid (which we can apply because it has a right-angled corner), $3\left( \frac{x^2}{2} \right) ^2=(12\sqrt{3})^2$ . Solving yields $x=2\sqrt{6}$
The height of the pyramid is then $\sqrt{(2\sqrt{6})^2-4^2}=2\sqrt{2}$ , by the Pythagorean Theorem (using the slant height and circumradius). The volume is then $v=\frac{1}{3}\cdot 12\sqrt{3} \cdot 2\sqrt{2}=8\sqrt{6}$ , and the requested answer is $v^2=(8\sqrt{6})^2=\boxed{384}$ . ~bad_at_mathcounts


Answer: 216.0

A cylindrical log has diameter $12$ inches. A wedge is cut from the log by making two planar cuts that go entirely through the log. The first is perpendicular to the axis of the cylinder, and the plane of the second cut forms a $45^\circ$ angle with the plane of the first cut. The intersection of these two planes has exactly one point in common with the log. The number of cubic inches in the wedge can be expressed as $n\pi$ , where n is a positive integer. Find $n$

Solution: The volume of the wedge is half the volume of a cylinder with height $12$ and radius $6$ . (Imagine taking another identical wedge and sticking it to the existing one).  Thus, $V=\dfrac{6^2\cdot 12\pi}{2}=216\pi$ , so $n=\boxed{216}$


Answer: 36.0

A data set consists of $6$ (not distinct) positive integers: $1$ $7$ $5$ $2$ $5$ , and $X$ . The average (arithmetic mean) of the $6$ numbers equals a value in the data set. What is the sum of all possible values of $X$
$\textbf{(A) } 10 \qquad \textbf{(B) } 26 \qquad \textbf{(C) } 32 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 40$

Solution:
First, note that $1+7+5+2+5=20$ . There are $3$ possible cases:
Case 1: the mean is $5$
$X = 5 \cdot 6 - 20 = 10$
Case 2: the mean is $7$
$X = 7 \cdot 6 - 20 = 22$
Case 3: the mean is $X$
$X= \frac{20+X}{6} \Rightarrow X=4$
Therefore, the answer is $10+22+4=\boxed{36}$


Answer: 36.0

Solution:
First, note that $1+7+5+2+5=20$ . There are $3$ possible cases:
Case 1: the mean is $5$
$X = 5 \cdot 6 - 20 = 10$
Case 2: the mean is $7$
$X = 7 \cdot 6 - 20 = 22$
Case 3: the mean is $X$
$X= \frac{20+X}{6} \Rightarrow X=4$
Therefore, the answer is $10+22+4=\boxed{36}$


Answer: 12.0

A deck of cards has only red cards and black cards. The probability of a randomly chosen card being red is $\frac13$ . When $4$ black cards are added to the deck, the probability of choosing red becomes $\frac14$ . How many cards were in the deck originally?
$\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }15 \qquad \textbf{(E) }18$

Solution:
If the probability of choosing a red card is $\frac{1}{3}$ , the red and black cards are in ratio $1:2$ . This means at the beginning there are $x$ red cards and $2x$ black cards.
After $4$ black cards are added, there are $2x+4$ black cards. This time, the probability of choosing a red card is $\frac{1}{4}$ so the ratio of red to black cards is $1:3$ . This means in the new deck the number of black cards is also $3x$ for the same $x$ red cards.
So, $3x = 2x + 4$ and $x=4$ meaning there are $4$ red cards in the deck at the start and $2(4) = 8$ black cards.
So, the answer is $8+4 = 12 = \boxed{12}$

Solution: In terms of the number of cards, the original deck is $3$ times the red cards, and the final deck is $4$ times the red cards. So, the final deck is $\frac43$ times the original deck. We are given that adding $4$ cards to the original deck is the same as increasing the original deck by $\frac13$ of itself. Since $4$ cards are equal to $\frac13$ of the original deck, the original deck has $4\cdot3=\boxed{12}$ cards.

Solution:
Suppose there were $x$ cards in the deck originally. Now, the deck has $x+4$ cards, which must be a multiple of $4.$
Only $12+4=16$ is a multiple of $4,$ so the answer is $x=\boxed{12}.$


Answer: 758.0

A deck of forty cards consists of four $1$ 's, four $2$ 's,..., and four $10$ 's.  A matching pair (two cards with the same number) is removed from the deck.  Given that these cards are not returned to the deck, let $m/n$ be the probability that two randomly selected cards also form a pair, where $m$ and $n$ are relatively prime positive integers.  Find $m + n.$

Solution: There are ${38 \choose 2} = 703$ ways we can draw two cards from the reduced deck. The two cards will form a pair if both are one of the nine numbers that were not removed, which can happen in $9{4 \choose 2} = 54$ ways, or if the two cards are the remaining two cards of the number that was removed, which can happen in $1$ way. Thus, the answer is $\frac{54+1}{703} = \frac{55}{703}$ , and $m+n = \boxed{758}$


Answer: 6.0

A decorative window is made up of a rectangle with semicircles at either end. The ratio of $AD$ to $AB$ is $3:2$ . And $AB$ is 30 inches. What is the ratio of the area of the rectangle to the combined area of the semicircles?
[asy] import graph; size(5cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(8); defaultpen(dps); pen ds=black; real xmin=-4.27,xmax=14.73,ymin=-3.22,ymax=6.8; draw((0,4)--(0,0)); draw((0,0)--(2.5,0)); draw((2.5,0)--(2.5,4)); draw((2.5,4)--(0,4)); draw(shift((1.25,4))*xscale(1.25)*yscale(1.25)*arc((0,0),1,0,180)); draw(shift((1.25,0))*xscale(1.25)*yscale(1.25)*arc((0,0),1,-180,0)); dot((0,0),ds); label("$A$",(-0.26,-0.23),NE*lsf); dot((2.5,0),ds); label("$B$",(2.61,-0.26),NE*lsf); dot((0,4),ds); label("$D$",(-0.26,4.02),NE*lsf); dot((2.5,4),ds); label("$C$",(2.64,3.98),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]
$\textbf{(A)}\ 2:3\qquad\textbf{(B)}\ 3:2\qquad\textbf{(C)}\ 6:\pi\qquad\textbf{(D)}\ 9:\pi\qquad\textbf{(E)}\ 30 :\pi$

Solution:
We can set a proportion:
\[\dfrac{AD}{AB}=\dfrac{3}{2}\]
We substitute $AB$ with 30 and solve for $AD$
\[\dfrac{AD}{30}=\dfrac{3}{2}\]
\[AD=45\]
We calculate the combined area of semicircle by putting together semicircle $AB$ and $CD$ to get a circle with radius $15$ . Thus, the area is $225\pi$ . The area of the rectangle is $30\cdot 45=1350$ . We calculate the ratio:
\[\dfrac{1350}{225\pi}=\dfrac{6}{\pi}\Rightarrow\boxed{6}\]


Answer: 729.0

A dessert chef prepares the dessert for every day of a week starting with Sunday. The dessert each day is either cake, pie, ice cream, or pudding. The same dessert may not be served two days in a row. There must be cake on Friday because of a birthday. How many different dessert menus for the week are possible?
$\textbf{(A)}\ 729\qquad\textbf{(B)}\ 972\qquad\textbf{(C)}\ 1024\qquad\textbf{(D)}\ 2187\qquad\textbf{(E)}\ 2304$

Solution:
Desserts must be chosen for $7$ days: Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday.
There are $3$ choices for dessert on Saturday: pie, ice cream, or pudding, as there must be cake on Friday and the same dessert may not be served two days in a row. Likewise, there are $3$ choices for dessert on Thursday. Once dessert for Thursday is selected, there are $3$ choices for dessert on Wednesday, once Wednesday's dessert is selected there are $3$ choices for dessert on Tuesday, etc. Thus, there are $3$ choices for dessert for each of $6$ days, so the total number of possible dessert menus is $3^6$ , or $\boxed{729}$


Answer: 729.0

Solution: There are $4 \cdot 3^6$ ways for the desserts to be chosen. By symmetry, any of the desserts that are chosen on Friday share $\frac{1}{4}$ of the total arrangements. Therefore our answer is $\frac{4\cdot3^6}{4} = 3^6 = \boxed{729}.$


Answer: 729.0

Solution:
We can count the number of possible foods for each day and then multiply to enumerate the number of combinations.
On Friday, we have one possibility: cake.
On Saturday, we have three possibilities: pie, ice cream, or pudding. This is the end of the week.
On Thursday, we have three possibilities: pie, ice cream, or pudding. We can't have cake because we have to have cake the following day, which is the Friday with the birthday party.
On Wednesday, we have three possibilities: cake, plus the two things that were not eaten on Thursday.
Similarly, on Tuesday, we have three possibilities: the three things that were not eaten on Wednesday.
Likewise on Monday: three possibilities, the three things that were not eaten on Tuesday.
On Sunday, it is tempting to think there are four possibilities, but remember that cake must be served on Friday. This serves to limit the number of foods we can eat on Sunday, with the result being that there are three possibilities: The three things that were not eaten on Monday.
So the number of menus is $3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 1 \cdot 3 = 729.$ The answer is $\boxed{729}$

Solution:
We can perform casework as an understandable means of getting the answer. We can organize our counting based on the food that was served on Wednesday, because whether cake is or is not served on Wednesday seems to significantly affect the number of ways the chef can make said foods for that week.
Case 1: Cake is served on Wednesday. Here, we have three choices for food on Thursday and Saturday since cake must be served on Friday, and none of these choices are cake, which was served Wednesday. Likewise, we have three choices (pie, ice cream, and pudding) for the food served on Tuesday and thus three choices for those served on Monday and Sunday, with these three choices being whatever was not served on Tuesday and Monday, respectively. Hence, for this case, there are $3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 = 243$ possibilities.
Case 2: Cake is not served on Wednesday. Obviously, this means that pie, ice cream, and pudding are our only choices for Wednesday's food. Since cake must be served on Friday, only ice cream, pudding, and cake can be served on Thursday. However, since one of those foods was chosen for Wednesday, we only have two possibilities for Thursday's food. Like our first case, we have three possibilities for the food served on Tuesday, Monday, and Sunday: whatever was not served on Wednesday, Tuesday, and Monday, respectively. $3 \cdot 3 \cdot 3 \cdot 3 \cdot 2 \cdot 3 = 486$ possibilities thus exist for this case.
Adding the number of possibilities together yields that $243 + 486 = 729$ is the total number of menus, making our answer $\boxed{729}$

Solution:
Note that the choice of a food item on a given day is symmetric, i.e. the number of ways to create the meal plan with a cake on Friday is the same as the number of ways to create the meal plan with pudding on Friday, and the same reasoning holds for the other desserts. Since every meal plan is counted by the summation of the $4$ aforementioned plans (note that Friday's dessert has to be one of the $4$ given desserts) and that these cases are mutually exclusive (i.e. you cannot make both a cake and pudding on Friday), each case results in a quarter of the total meal plans (Since there are $4$ desserts, we multiply by $\tfrac{1}{4}$ ). The total number of plans with no restrictions can be counted with constructive counting, as follows:
We note that there are $4$ ways to choose the first dessert. Then, each dessert thereafter must be distinct from the prior one. Since there are $4$ options, and $1$ of them has been taken by the prior, each following dessert can be chosen in $(4 - 1) = 3$ ways. Thus, since there are $6$ desserts other than the first, the total number (without restrictions) is $4 \cdot 3^6$
Thus, by our symmetry argument derived prior, we know that the number of desired meal plans is $\frac{\text{\# plans w/o restrictions}}{4} = \frac{4 \cdot 3^6}{4} = 3^6 = 729$ , choice $\boxed{729}$


Answer: 9.0

A digital display shows the current date as an $8$ -digit integer consisting of a $4$ -digit year, followed by a $2$ -digit month, followed by a $2$ -digit date within the month. For example, Arbor Day this year is displayed as 20230428. For how many dates in $2023$ will each digit appear an even number of times in the 8-digital display for that date?
$\textbf{(A)}~5\qquad\textbf{(B)}~6\qquad\textbf{(C)}~7\qquad\textbf{(D)}~8\qquad\textbf{(E)}~9$

Solution:
Do careful casework by each month. In the month and the date, we need a $0$ , a $3$ , and two digits repeated (which has to be $1$ and $2$ after consideration). After the casework, we get $\boxed{9}$ .
For curious readers, the numbers (in chronological order) are:

Solution:
There is one $3$ , so we need one more (three more means that either the month or units digit of the day is $3$ ). For the same reason, we need one more $0$
If $3$ is the units digit of the month, then the $0$ can be in either of the three remaining slots. For the first case (tens digit of the month), then the last two digits must match ( $11, 22$ ). For the second (tens digit of the day), we must have the other two be $1$ , as a month can't start with $2$ or $0$ . There are $3$ successes this way.
If $3$ is the tens digit of the day, then $0$ can be either the tens digit of the month or the units digit of the day. For the first case, $1$ must go in the other slots. For the second, the other two slots must be $1$ as well. There are $2$ successes here.
If $3$ is the units digit of the day, then $0$ could go in any of the $3$ remaining slots again. If it's the tens digit of the day, then the other digits must be $1$ . If $0$ is the units digit of the day, then the other two slots must both be $1$ . If $0$ is the tens digit of the month, then the other two slots can be either both $1$ or both $2$ . In total, there are $4$ successes here.
Summing through all cases, there are $3 + 2 + 4 = \boxed{9}$ dates.

Solution: We start with $2023----$ we need an extra $0$ and an extra $3$ . So we have at least one of those extras in the days, except we can have the month $03$ . We now have $6$ possible months $01,02,03,10,11,12$ . For month $1$ we have two cases, we now have to add in another 1, and the possible days are $13,31$ . For month $2$ we need an extra $2$ so we can have the day $23$ note that we can't use $32$ because it is to large. Now for month $3$ we can have any number and multiply it by $11$ so we have the solution $11,22$ . For October we need a $1$ and a $3$ so we have $13,31$ as our choices. For November we have two choices which are $03,30$ .Now for December we have $0$ options. Summing $2+1+2+2+2$ we get $\boxed{9}$ solutions.


Answer: 9.0

Solution:
Do careful casework by each month. In the month and the date, we need a $0$ , a $3$ , and two digits repeated (which has to be $1$ and $2$ after consideration). After the casework, we get $\boxed{9}$ .
For curious readers, the numbers (in chronological order) are:

Solution:
There is one $3$ , so we need one more (three more means that either the month or units digit of the day is $3$ ). For the same reason, we need one more $0$
If $3$ is the units digit of the month, then the $0$ can be in either of the three remaining slots. For the first case (tens digit of the month), then the last two digits must match ( $11, 22$ ). For the second (tens digit of the day), we must have the other two be $1$ , as a month can't start with $2$ or $0$ . There are $3$ successes this way.
If $3$ is the tens digit of the day, then $0$ can be either the tens digit of the month or the units digit of the day. For the first case, $1$ must go in the other slots. For the second, the other two slots must be $1$ as well. There are $2$ successes here.
If $3$ is the units digit of the day, then $0$ could go in any of the $3$ remaining slots again. If it's the tens digit of the day, then the other digits must be $1$ . If $0$ is the units digit of the day, then the other two slots must both be $1$ . If $0$ is the tens digit of the month, then the other two slots can be either both $1$ or both $2$ . In total, there are $4$ successes here.
Summing through all cases, there are $3 + 2 + 4 = \boxed{9}$ dates.

Solution: We start with $2023----$ we need an extra $0$ and an extra $3$ . So we have at least one of those extras in the days, except we can have the month $03$ . We now have $6$ possible months $01,02,03,10,11,12$ . For month $1$ we have two cases, we now have to add in another 1, and the possible days are $13,31$ . For month $2$ we need an extra $2$ so we can have the day $23$ note that we can't use $32$ because it is to large. Now for month $3$ we can have any number and multiply it by $11$ so we have the solution $11,22$ . For October we need a $1$ and a $3$ so we have $13,31$ as our choices. For November we have two choices which are $03,30$ .Now for December we have $0$ options. Summing $2+1+2+2+2$ we get $\boxed{9}$ solutions.


Answer: 23.0

A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?
$\textbf{(A)}\ 17\qquad\textbf{(B)}\ 19\qquad\textbf{(C)}\ 21\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\  23$

Solution: From the greedy algorithm , we have $9$ in the hours section and $59$ in the minutes section. $9+5+9=\boxed{23}$

Solution:
With a matrix, we can see $\begin{bmatrix} 1+2&9&6&3\\ 1+1&8&5&2\\ 1+0&7&4&1 \end{bmatrix}$ The largest single digit sum we can get is $9$ .
For the minutes digits, we can combine the largest $2$ digits, which are $9,5 \Rightarrow 9+5=14$ , and finally $14+9=\boxed{23}$

Solution:
We first note that since the watch displays time in AM and PM, the value for the hours section varies from $00-12$ . Therefore, the maximum value of the digits for the hours is when the watch displays $09$ , which gives us $0+9=9$
Next, we look at the value of the minutes section, which varies from $00-59$ . Let this value be a number $ab$ . We quickly find that the maximum value for $a$ and $b$ is respectively $5$ and $9$
Adding these up, we get $9+5+9=\boxed{23}$


Answer: 23.0

Solution: From the greedy algorithm , we have $9$ in the hours section and $59$ in the minutes section. $9+5+9=\boxed{23}$

Solution:
With a matrix, we can see $\begin{bmatrix} 1+2&9&6&3\\ 1+1&8&5&2\\ 1+0&7&4&1 \end{bmatrix}$ The largest single digit sum we can get is $9$ .
For the minutes digits, we can combine the largest $2$ digits, which are $9,5 \Rightarrow 9+5=14$ , and finally $14+9=\boxed{23}$

Solution:
We first note that since the watch displays time in AM and PM, the value for the hours section varies from $00-12$ . Therefore, the maximum value of the digits for the hours is when the watch displays $09$ , which gives us $0+9=9$
Next, we look at the value of the minutes section, which varies from $00-59$ . Let this value be a number $ab$ . We quickly find that the maximum value for $a$ and $b$ is respectively $5$ and $9$
Adding these up, we get $9+5+9=\boxed{23}$


Answer: 13.0

A dilation of the plane—that is, a size transformation with a positive scale factor—sends the circle of radius $2$ centered at $A(2,2)$ to the circle of radius $3$ centered at $A’(5,6)$ . What distance does the origin $O(0,0)$ , move under this transformation?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution: The center of dilation must lie on the line $A A'$ , which can be expressed as $y = \dfrac{4x}{3} - \dfrac{2}{3}$ . Note that the center of dilation must have an $x$ -coordinate less than $2$ ; if the $x$ -coordinate were otherwise, then the circle under the transformation would not have an increased $x$ -coordinate in the coordinate plane. Also, the ratio of dilation must be equal to $\dfrac{3}{2}$ , which is the ratio of the radii of the circles. Thus, we are looking for a point $(x,y)$ such that $\dfrac{3}{2} \left( 2 - x \right) = 5 - x$ (for the $x$ -coordinates), and $\dfrac{3}{2} \left( 2 - y \right) = 6 - y$ . We do not have to include absolute value symbols because we know that the center of dilation has a lower $x$ -coordinate, and hence a lower $y$ -coordinate, from our reasoning above. Solving the two equations, we get $x = -4$ and $y = - 6$ . This means that any point $(a,b)$ on the plane will dilate to the point $\left( \dfrac{3}{2} (a + 4) - 4, \dfrac{3}{2} (b + 6) - 6 \right)$ , which means that the point $(0,0)$ dilates to $\left( 6 - 4, 9 - 6 \right) = (2,3)$ . Thus, the origin moves $\sqrt{2^2 + 3^2} = \boxed{13}$ units.

Solution:
 Using analytic geometry, we find that the center of dilation is at $(-4,-6)$ and the coefficient/factor is $1.5$ . Then, we see that the origin is $2\sqrt{13}$ from the center, and will be $1.5 \times 2\sqrt{13} = 3\sqrt{13}$ from it afterwards.
Thus, it will move $3\sqrt{13} - 2\sqrt{13} = \boxed{13}$

Solution:
Using the ratios of radii of the circles, $\frac{3}{2}$ , we find that the scale factor is $1.5$ . If the origin had not moved, this indicates that the center of the
circle would be $(3,3)$ , simply because of $(2 \cdot 1.5,  2 \cdot 1.5)$ . Since the center has moved from $(3,3)$ to $(5,6)$ , we apply the distance formula and get: $\sqrt{(6-3)^2 + (5-3)^2} = \boxed{13}$

Solution:
Before dilation, notice that the two axes are tangent to the circle with center $(2,2)$ . Using this, we can draw new axes tangent to the radius 3 circle with center $(5,6)$ , resulting in a "new origin" that is 3 units left and 3 units down from the center $(5,6)$ , or $(2,3)$ . Using the distance formula, the distance from $(0,0)$ and $(2,3)$ is $\boxed{13}$ .
~Mightyeagle

Solution: Because the dilation changes the circle of radius 2 to a circle of radius 3, the scale factor must be $\frac{3}{2}$ . The center of the circle with radius 3 is 3 units to the right and 4 units above the center of the circle with radius 2, so the center of dilation must be 6 units to the left and 8 units below the center of the radius 2 circle. So, the center of dilation is the point $(-4, -6)$ . The distance from that point to the origin is $2\sqrt{13}$ , and halving that to get the distance the origin moves gives $\boxed{13}$ . ~Meeshbot


Answer: 10.0

A disk of radius $1$ rolls all the way around the inside of a square of side length $s>4$ and sweeps out a region of area $A$ . A second disk of radius $1$ rolls all the way around the outside of the same square and sweeps out a region of area $2A$ . The value of $s$ can be written as $a+\frac{b\pi}{c}$ , where $a,b$ , and $c$ are positive integers and $b$ and $c$ are relatively prime. What is $a+b+c$
$\textbf{(A)} ~10\qquad\textbf{(B)} ~11\qquad\textbf{(C)} ~12\qquad\textbf{(D)} ~13\qquad\textbf{(E)} ~14$

Solution: The side length of the inner square traced out by the disk with radius $1$ is $s-4.$ However, there is a piece at each corner (bounded by two line segments and one $90^\circ$ arc) where the disk never sweeps out. The combined area of these four pieces is $(1+1)^2-\pi\cdot1^2=4-\pi.$ As a result, we have \[A=s^2-(s-4)^2-(4-\pi)=8s-20+\pi.\] Now, we consider the second disk. The part it sweeps is comprised of four quarter circles with radius $2$ and four rectangles with side lengths of $2$ and $s.$ When we add it all together, we have $2A=8s+4\pi,$ or \[A=4s+2\pi.\] We equate the expressions for $A,$ and then solve for $s:$ \[8s-20+\pi=4s+2\pi.\] We get $s=5+\frac{\pi}{4},$ so the answer is $5+1+4=\boxed{10}.$