Answer: 20.0

Each of the $12$ edges of a cube is labeled $0$ or $1$ . Two labelings are considered different even if one can be obtained from the other by a sequence of one or more rotations and/or reflections. For how many such labelings is the sum of the labels on the edges of each of the $6$ faces of the cube equal to $2$
$\textbf{(A) } 8 \qquad\textbf{(B) } 10 \qquad\textbf{(C) } 12 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 20$

Solution:
For simplicity, we will name this cube $ABCDEFGH$ by vertices, as shown below.  Note that for each face of this cube, two edges are labeled $0,$ and two edges are labeled $1.$ For all twelve edges of this cube, we conclude that six edges are labeled $0,$ and six edges are labeled $1.$
We apply casework to face $ABCD.$ Recall that there are $\binom42=6$ ways to label its edges:
Therefore, we have $4+16=\boxed{20}$ such labelings in total.

Solution:
Since we want the sum of the edges of each face to be $2$ , we need there to be two $1$ s and two $0$ s on each face. Through experimentation, we find that either $2, 4,$ or all of them have $1$ s adjacent to $1$ s and $0$ s adjacent to $0$ on each face. WLOG, let the first face (counterclockwise) be $0,0,1,1$ . In this case we are trying to have all of them be adjacent to each other. First face: $0,0,1,1$ . Second face: $2$ choices: $1,0,0,1$ or $0,0,1,1$ . After that, it is basically forced and everything will fall in to place. Since we assumed WLOG, we need to multiply $2$ by $4$ to get a total of $8$ different arrangements.
Secondly, $4$ of the faces have all of them adjacent and $2$ of the faces do not: WLOG counting counterclockwise, we have $0,0,1,1$ . Then, we choose the other face next to it. There are two cases, which are $0,1,0,1$ and $1,0,1,0$ . Therefore, this subcase has $4$ different arrangements. Then, we can choose the face at front to be $1,0,1,0$ . This has $4$ cases. The sides can either be $0,1,1,0$ or $1,1,0,0$ . Therefore, we have another $8$ cases.
Summing these up, we have $8+4+8 = 20$ . Therefore, our answer is $\boxed{20}$

Solution:
We see that if the $3$ edges connecting to $A$ has two $0$ 's, and one $1$ , it would have the same solutions as if it had two $1$ 's, and one $0$ . The solutions would just be inverted. As case 2.1 and case 2.2.2 are inverses, and case 2.2.1 has case 1 as an inverse, there would not be any additional solutions.
Similarly, if the $3$ edges connecting to $A$ has three $0$ 's, it would be the same as the inverse of case 1, or case 2.2.1, resulting in no new solutions.
Putting all the cases together, we have $4+6+4+6=\boxed{20}$ solutions.

Solution:
The problem states the sum of the labels on the edges of each of the $6$ faces of the cube equal to $2$ . That is, the sum of the labels on the $4$ edges of a face is equal to $2$ . The labels can only be $0$ or $1$ , meaning $2$ edges are labeled $1$ , the other $2$ are labeled $0$
This problem can be approached by Graph Coloring of Graph Theory . Note that each face of the cube connects to $4$ other faces, each with a shared edge. We use the following graph to represent the problem. Each vertex represents a face, each edge represent the cube's edge. Each vertex has $4$ edges connecting to $4$ other vertices. The edges can be colored red or blue, with red as label $0$ , and blue as label $1$ . Each vertex must have $2$ red edges and $2$ blue edges.
$\textbf{Case 1}$ $2$ adjacent red edges from vertex A. There are $4$ ways to choose $2$ red edges adjacent to each other and connect to $2$ vertices with an edge between them as shown below.
$\textbf{Case 1.1}$ $2$ adjacent red edges from vertex $A$ form a closed loop with a third red edge. There is only $1$ case as shown below.
$\textbf{Case 1.2}$ $2$ adjacent red edges from vertex $A$ does not form a closed loop with a third red edge. There are $3$ cases as shown below.
In case $1$ , there are total $4 \cdot (1 + 3) = 16$ ways.
$\textbf{Case 2}$ $2$ red edges from vertex $A$ with $1$ blue edge in between. There are $2$ ways to choose $2$ red edges with $1$ blue edge in between.
There are only $2$ cases as shown below.
In case $2$ , there are total $2 \cdot 2 = 4$ ways.
From both case $1$ and case $2$ , there are $16 + 4 = \boxed{20}$ ways in total.


Answer: 16.0

Each of the $20$ balls is tossed independently and at random into one of the $5$ bins. Let $p$ be the probability that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Let $q$ be the probability that every bin ends up with $4$ balls. What is $\frac{p}{q}$
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\  4 \qquad\textbf{(C)}\  8 \qquad\textbf{(D)}\  12 \qquad\textbf{(E)}\ 16$

Solution:
For simplicity purposes, we assume that the balls and the bins are both distinguishable.
Recall that there are $5^{20}$ ways to distribute $20$ balls into $5$ bins. We have \[p=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{5^{20}} \text{ and } q=\frac{\binom{20}{4,4,4,4,4}}{5^{20}}.\] Therefore, the answer is \[\frac pq=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{\binom{20}{4,4,4,4,4}}=\frac{5\cdot4\cdot\frac{20!}{3!\cdot5!\cdot4!\cdot4!\cdot4!}}{\frac{20!}{4!\cdot4!\cdot4!\cdot4!\cdot4!}}=\frac{5\cdot4\cdot(4!\cdot4!\cdot4!\cdot4!\cdot4!)}{3!\cdot5!\cdot4!\cdot4!\cdot4!}=\frac{5\cdot4\cdot4}{5}=\boxed{16}.\] ~MRENTHUSIASM ~Jesshuang

Solution:
For simplicity purposes, we assume that the balls and the bins are both distinguishable.
Let $q=\frac{x}{a},$ where $a$ is the total number of combinations and $x$ is the number of cases where every bin ends up with $4$ balls.
We can take $1$ ball from one bin and place it in another bin so that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Note that one configuration of $4{-}4{-}4{-}4{-}4$ corresponds to $5\cdot4\cdot4=80$ configurations of $3{-}5{-}4{-}4{-}4.$ On the other hand, one configuration of $3{-}5{-}4{-}4{-}4$ corresponds to $5$ configurations of $4{-}4{-}4{-}4{-}4.$
Therefore, we have \[p = \frac{80}{5}\cdot\frac{x}{a} = 16\cdot\frac{x}{a},\] from which $\frac{p}{q} = \boxed{16}.$

Solution: Since both of the cases will have $3$ bins with $4$ balls in them, we can leave those out. There are $2 \cdot \binom {5}{2} = 20$ ways to choose where to place the $3$ and the $5$ . After that, there are $\binom {8}{3} = 56$ ways to put the $3$ and $5$ balls being put into the bins. For the $4,4,4,4,4$ case, after we canceled the $4,4,4$ out, we have $\binom {8}{4} = 70$ ways to put the $4$ balls inside the bins. Therefore, we have $\frac {56\cdot 20}{70}$ which is equal to $8 \cdot 2 = \boxed{16}$

Solution:
Construct the set $A$ consisting of all possible $3{-}5{-}4{-}4{-}4$ bin configurations, and construct set $B$ consisting of all possible $4{-}4{-}4{-}4{-}4$ configurations. If we let $N$ be the total number of configurations possible, it's clear we want to solve for $\frac{p}{q} = \frac{\frac{|A|}{N}}{\frac{|B|}{N}} = \frac{|A|}{|B|}$
Consider drawing an edge between an element in $A$ and an element in $B$ if it is possible to reach one configuration from the other by moving a single ball (Note this process is reversible.). Let us consider the total number of edges drawn.
For any element in $A$ , we may choose one of the $5$ balls in the $5$ -bin and move it to the $3$ -bin to get a valid element in $B$ . This implies the number of edges is $5|A|$
On the other hand, for any element in $B$ , we may choose one of the $20$ balls and move it to one of the other $4$ -bins to get a valid element in $A$ . This implies the number of edges is $80|B|$
We equate the expressions to get $5|A| = 80|B|$ , from which $\frac{|A|}{|B|} = \frac{80}{5} = \boxed{16}$


Answer: 16.0

Solution:
For simplicity purposes, we assume that the balls and the bins are both distinguishable.
Recall that there are $5^{20}$ ways to distribute $20$ balls into $5$ bins. We have \[p=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{5^{20}} \text{ and } q=\frac{\binom{20}{4,4,4,4,4}}{5^{20}}.\] Therefore, the answer is \[\frac pq=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{\binom{20}{4,4,4,4,4}}=\frac{5\cdot4\cdot\frac{20!}{3!\cdot5!\cdot4!\cdot4!\cdot4!}}{\frac{20!}{4!\cdot4!\cdot4!\cdot4!\cdot4!}}=\frac{5\cdot4\cdot(4!\cdot4!\cdot4!\cdot4!\cdot4!)}{3!\cdot5!\cdot4!\cdot4!\cdot4!}=\frac{5\cdot4\cdot4}{5}=\boxed{16}.\] ~MRENTHUSIASM ~Jesshuang

Solution:
For simplicity purposes, we assume that the balls and the bins are both distinguishable.
Let $q=\frac{x}{a},$ where $a$ is the total number of combinations and $x$ is the number of cases where every bin ends up with $4$ balls.
We can take $1$ ball from one bin and place it in another bin so that some bin ends up with $3$ balls, another with $5$ balls, and the other three with $4$ balls each. Note that one configuration of $4{-}4{-}4{-}4{-}4$ corresponds to $5\cdot4\cdot4=80$ configurations of $3{-}5{-}4{-}4{-}4.$ On the other hand, one configuration of $3{-}5{-}4{-}4{-}4$ corresponds to $5$ configurations of $4{-}4{-}4{-}4{-}4.$
Therefore, we have \[p = \frac{80}{5}\cdot\frac{x}{a} = 16\cdot\frac{x}{a},\] from which $\frac{p}{q} = \boxed{16}.$

Solution: Since both of the cases will have $3$ bins with $4$ balls in them, we can leave those out. There are $2 \cdot \binom {5}{2} = 20$ ways to choose where to place the $3$ and the $5$ . After that, there are $\binom {8}{3} = 56$ ways to put the $3$ and $5$ balls being put into the bins. For the $4,4,4,4,4$ case, after we canceled the $4,4,4$ out, we have $\binom {8}{4} = 70$ ways to put the $4$ balls inside the bins. Therefore, we have $\frac {56\cdot 20}{70}$ which is equal to $8 \cdot 2 = \boxed{16}$

Solution:
Construct the set $A$ consisting of all possible $3{-}5{-}4{-}4{-}4$ bin configurations, and construct set $B$ consisting of all possible $4{-}4{-}4{-}4{-}4$ configurations. If we let $N$ be the total number of configurations possible, it's clear we want to solve for $\frac{p}{q} = \frac{\frac{|A|}{N}}{\frac{|B|}{N}} = \frac{|A|}{|B|}$
Consider drawing an edge between an element in $A$ and an element in $B$ if it is possible to reach one configuration from the other by moving a single ball (Note this process is reversible.). Let us consider the total number of edges drawn.
For any element in $A$ , we may choose one of the $5$ balls in the $5$ -bin and move it to the $3$ -bin to get a valid element in $B$ . This implies the number of edges is $5|A|$
On the other hand, for any element in $B$ , we may choose one of the $20$ balls and move it to one of the other $4$ -bins to get a valid element in $A$ . This implies the number of edges is $80|B|$
We equate the expressions to get $5|A| = 80|B|$ , from which $\frac{|A|}{|B|} = \frac{80}{5} = \boxed{16}$


Answer: 298.0

Each of the $2001$ students at a high school studies either Spanish or French, and some study both. The number who study Spanish is between $80$ percent and $85$ percent of the school population, and the number who study French is between $30$ percent and $40$ percent. Let $m$ be the smallest number of students who could study both languages, and let $M$ be the largest number of students who could study both languages. Find $M-m$

Solution:
Let $S$ be the percent of people who study Spanish, $F$ be the number of people who study French, and let $S \cap F$ be the number of students who study both. Then $\left\lceil 80\% \cdot 2001 \right\rceil = 1601 \le S \le \left\lfloor 85\% \cdot 2001 \right\rfloor = 1700$ , and $\left\lceil 30\% \cdot 2001 \right\rceil = 601 \le F \le \left\lfloor 40\% \cdot 2001 \right\rfloor = 800$ . By the Principle of Inclusion-Exclusion
\[S+F- S \cap F = S \cup F = 2001\]
For $m = S \cap F$ to be smallest, $S$ and $F$ must be minimized.
\[1601 + 601 - m = 2001 \Longrightarrow m = 201\]
For $M = S \cap F$ to be largest, $S$ and $F$ must be maximized.
\[1700 + 800 - M = 2001 \Longrightarrow M = 499\]
Therefore, the answer is $M - m = 499 - 201 = \boxed{298}$


Answer: 7.0

Each of the $39$ students in the eighth grade at Lincoln Middle School has one dog or one cat or both a dog and a cat. Twenty students have a dog and $26$ students have a cat. How many students have both a dog and a cat?
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 13\qquad \textbf{(C)}\ 19\qquad \textbf{(D)}\ 39\qquad \textbf{(E)}\ 46$

Solution: The union of two sets is equal to the sum of each set minus their intersection. The number of students that have both a dog and a cat is $20+26-39 = \boxed{7}$

Solution:
We create a diagram:
Let $x$ be the number of students with both a dog and a cat.
Therefore, we have
\[26+20-x = 39\] \[46-x = 39\] \[x = \boxed{7}\]


Answer: 87431.0

Each of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 is used only once to make two five-digit numbers so that they have the largest possible sum. Which of the following could be one of the numbers?
$\textbf{(A)}\hspace{.05in}76531\qquad\textbf{(B)}\hspace{.05in}86724\qquad\textbf{(C)}\hspace{.05in}87431\qquad\textbf{(D)}\hspace{.05in}96240\qquad\textbf{(E)}\hspace{.05in}97403$

Solution:
In order to maximize the sum of the numbers, the numbers must have their digits ordered in decreasing value. There are only two numbers from the answer choices with this property: $76531$ and $87431$ . To determine the answer we will have to use estimation and the first two digits of the numbers.
For $76531$ the number that would maximize the sum would start with $98$ . The first two digits of $76531$ (when rounded) are $77$ . Adding $98$ and $77$ , we find that the first three digits of the sum of the two numbers would be $175$
For $87431$ the number that would maximize the sum would start with $96$ . The first two digits of $87431$ (when rounded) are $87$ . Adding $96$ and $87$ , we find that the first three digits of the sum of the two numbers would be $183$
From the estimations, we can say that the answer to this problem is $\boxed{87431}$

Solution:
In order to determine the largest number possible, we have to evenly distribute the digits when adding. The two numbers that show an example of this are $97531$ and $86420$ . The digits can be interchangeable between numbers because we only care about the actual digits.
The first digit must be either $9$ or $8$ . This immediately knocks out $\textbf{(A)}\ 76531$
The second digit must be either $7$ or $6$ . This doesn't cancel any choices.
The third digit must be either $5$ or $4$ . This knocks out $\textbf{(B)}\ 86724$ and $\textbf{(D)}\ 96240$
The fourth digit must be $3$ or $2$ . This cancels out $\textbf{(E)}\ 97403$
This leaves us with $\boxed{87431}$

Solution:
If we use intuition, we know that the digits will be IN ORDER, to maximze the number. This eliminates $\textbf{(B)}$ $\textbf{(D)}$ ,and $\textbf{(E)}$ . Additionally, the 2 two numbers must have 9 and 8 for the first digit to maximize the sum, eliminating $\textbf{(A)}$ . This leaves $\boxed{87431}$ .
-written by litttle_master purely, not copied from anywhere


Answer: 3.0

Each of the equations $3x^2-2=25, (2x-1)^2=(x-1)^2, \sqrt{x^2-7}=\sqrt{x-1}$ has:
$\textbf{(A)}\ \text{two integral roots}\qquad\textbf{(B)}\ \text{no root greater than 3}\qquad\textbf{(C)}\ \text{no root zero}\\ \textbf{(D)}\ \text{only one root}\qquad\textbf{(E)}\ \text{one negative root and one positive root}$
Solution

Solution:
Since the question asks us about the unifying characteristic of all three equations' roots, we have to first determine them.
$3x^2-2 = 25$ can be rewritten as $3x^2 - 27 = 0$ , which gives the following roots $+3$ and $-3$
$(2x-1)^2 = (x-1)^2$ can be expanded to $4x^2-4x+1=x^2-2x+1$ , which in turn leads to $3x^2-2x=0$ . The roots here are $0$ and $\frac{2}{3}$
$\sqrt{x^2-7}=\sqrt{x-1}$ , when squared, also turns into a quadratic equation: $x^2 - x - 6 = 0$ . Binomial factoring gives us the roots $-2$ and $3$
We can clearly see that, between all of the equations, there is $\boxed{3}$


Answer: 24.0

Each of the five numbers 1, 4, 7, 10, and 13 is placed in one of the five squares so that the sum of the three numbers in the horizontal row equals the sum of the three numbers in the vertical column. The largest possible value for the horizontal or vertical sum is
[asy] draw((0,0)--(3,0)--(3,1)--(0,1)--cycle); draw((1,-1)--(2,-1)--(2,2)--(1,2)--cycle); [/asy]
$\text{(A)}\ 20 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 22 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 30$

Solution: The largest sum occurs when $13$ is placed in the center. This sum is $13 + 10 + 1 = 13 + 7 + 4 = \boxed{24}$ . Note: Two other common sums, $18$ and $21$ , are also possible.

Solution:
Since the horizontal sum equals the vertical sum, twice this sum will be the sum
of the five numbers plus the number in the center. When the center number is $13$ , the sum is the largest, \[[10 + 4 + 1 + 7 + 2(13)]=2S\\ 48=2S\\ S=\boxed{24}\] The other
four numbers are divided into two pairs with equal sums.


Answer: 25.0

Each of the following four large congruent squares is subdivided into combinations of congruent triangles or rectangles and is partially bolded . What percent of the total area is partially bolded? [asy] import graph; size(7.01cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.42,xmax=14.59,ymin=-10.08,ymax=5.26;  pair A=(0,0), B=(4,0), C=(0,4), D=(4,4), F=(2,0), G=(3,0), H=(1,4), I=(2,4), J=(3,4), K=(0,-2), L=(4,-2), M=(0,-6), O=(0,-4), P=(4,-4), Q=(2,-2), R=(2,-6), T=(6,4), U=(10,0), V=(10,4), Z=(10,2), A_1=(8,4), B_1=(8,0), C_1=(6,-2), D_1=(10,-2), E_1=(6,-6), F_1=(10,-6), G_1=(6,-4), H_1=(10,-4), I_1=(8,-2), J_1=(8,-6), K_1=(8,-4);  draw(C--H--(1,0)--A--cycle,linewidth(1.6)); draw(M--O--Q--R--cycle,linewidth(1.6)); draw(A_1--V--Z--cycle,linewidth(1.6)); draw(G_1--K_1--J_1--E_1--cycle,linewidth(1.6));  draw(C--D); draw(D--B); draw(B--A); draw(A--C); draw(H--(1,0)); draw(I--F); draw(J--G); draw(C--H,linewidth(1.6)); draw(H--(1,0),linewidth(1.6)); draw((1,0)--A,linewidth(1.6)); draw(A--C,linewidth(1.6)); draw(K--L); draw((4,-6)--L); draw((4,-6)--M); draw(M--K); draw(O--P); draw(Q--R); draw(O--Q); draw(M--O,linewidth(1.6)); draw(O--Q,linewidth(1.6)); draw(Q--R,linewidth(1.6)); draw(R--M,linewidth(1.6)); draw(T--V); draw(V--U); draw(U--(6,0)); draw((6,0)--T); draw((6,2)--Z); draw(A_1--B_1); draw(A_1--Z); draw(A_1--V,linewidth(1.6)); draw(V--Z,linewidth(1.6)); draw(Z--A_1,linewidth(1.6)); draw(C_1--D_1); draw(D_1--F_1); draw(F_1--E_1); draw(E_1--C_1); draw(G_1--H_1); draw(I_1--J_1); draw(G_1--K_1,linewidth(1.6)); draw(K_1--J_1,linewidth(1.6)); draw(J_1--E_1,linewidth(1.6)); draw(E_1--G_1,linewidth(1.6));  dot(A,linewidth(1pt)+ds); dot(B,linewidth(1pt)+ds); dot(C,linewidth(1pt)+ds); dot(D,linewidth(1pt)+ds); dot((1,0),linewidth(1pt)+ds); dot(F,linewidth(1pt)+ds); dot(G,linewidth(1pt)+ds); dot(H,linewidth(1pt)+ds); dot(I,linewidth(1pt)+ds); dot(J,linewidth(1pt)+ds); dot(K,linewidth(1pt)+ds); dot(L,linewidth(1pt)+ds); dot(M,linewidth(1pt)+ds); dot((4,-6),linewidth(1pt)+ds); dot(O,linewidth(1pt)+ds); dot(P,linewidth(1pt)+ds); dot(Q,linewidth(1pt)+ds); dot(R,linewidth(1pt)+ds); dot((6,0),linewidth(1pt)+ds); dot(T,linewidth(1pt)+ds); dot(U,linewidth(1pt)+ds); dot(V,linewidth(1pt)+ds); dot((6,2),linewidth(1pt)+ds); dot(Z,linewidth(1pt)+ds); dot(A_1,linewidth(1pt)+ds); dot(B_1,linewidth(1pt)+ds); dot(C_1,linewidth(1pt)+ds); dot(D_1,linewidth(1pt)+ds); dot(E_1,linewidth(1pt)+ds); dot(F_1,linewidth(1pt)+ds); dot(G_1,linewidth(1pt)+ds); dot(H_1,linewidth(1pt)+ds); dot(I_1,linewidth(1pt)+ds); dot(J_1,linewidth(1pt)+ds); dot(K_1,linewidth(1pt)+ds);  clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy] $\textbf{(A) }12\frac{1}{2} \qquad\textbf{(B) }20 \qquad\textbf{(C) }25 \qquad\textbf{(D) }33\frac{1}{3} \qquad\textbf{(E) }37\frac{1}{2}$

Solution: Assume that the area of each square is $1$ .  Then, the area of the bolded region in the top left square is $\dfrac{1}{4}$ .  The area of the top right bolded region is $\dfrac{1}{8}$ .  The area of the bottom left bolded region is $\dfrac{3}{8}$ .  And the area of the bottom right bolded region is $\dfrac{1}{4}$ .  Add the four fractions: $\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{3}{8} + \dfrac{1}{4} = 1$ .  The four squares together have an area of $4$ , so the percentage bolded is $\dfrac{1}{4} \cdot 100 = \boxed{25}$


Answer: 7.0

Each of the letters $\text{W}$ $\text{X}$ $\text{Y}$ , and $\text{Z}$ represents a different integer in the set $\{ 1,2,3,4\}$ , but not necessarily in that order.  If $\dfrac{\text{W}}{\text{X}} - \dfrac{\text{Y}}{\text{Z}}=1$ , then the sum of $\text{W}$ and $\text{Y}$ is
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 7$

Solution:
There are different ways to approach this problem, and I'll start with the different factor of the numbers of the set $\{ 1,2,3,4\}$
$1$ has factor $1$
$2$ has factors $1$ and $2$
$3$ has factors $1$ and $3$
$4$ has factors $1$ $2$ , and $4$
From here, we note that even though all numbers have the factor $1$ , only $4$ has another factor other than $1$ in the set (ie. $2$
We could therefore have one fraction be $\frac{4}{2}$ and another $\frac{3}{1}$
The sum of the numerators is $4+3=7=\boxed{7}$


Answer: 5.0

Each of the points $A,B,C,D,E,$ and $F$ in the figure below represents a different digit from $1$ to $6.$ Each of the five lines shown passes through some of these points. The digits along each line are added to produce five sums, one for each line. The total of the five sums is $47.$ What is the digit represented by $B?$
[asy] size(200); dotfactor = 10;  pair p1 = (-28,0); pair p2 = (-111,213); draw(p1--p2,linewidth(1));  pair p3 = (-160,0); pair p4 = (-244,213); draw(p3--p4,linewidth(1));  pair p5 = (-316,0); pair p6 = (-67,213); draw(p5--p6,linewidth(1));  pair p7 = (0, 68); pair p8 = (-350,10); draw(p7--p8,linewidth(1));  pair p9 = (0, 150); pair p10 = (-350, 62); draw(p9--p10,linewidth(1));  pair A = intersectionpoint(p1--p2, p5--p6); dot("$A$", A, 2*W);  pair B = intersectionpoint(p5--p6, p3--p4); dot("$B$", B, 2*WNW);  pair C = intersectionpoint(p7--p8, p5--p6); dot("$C$", C, 1.5*NW);  pair D = intersectionpoint(p3--p4, p7--p8); dot("$D$", D, 2*NNE);  pair EE = intersectionpoint(p1--p2, p7--p8); dot("$E$", EE, 2*NNE);  pair F = intersectionpoint(p1--p2, p9--p10); dot("$F$", F, 2*NNE); [/asy]
$\textbf{(A) }1 \qquad \textbf{(B) }2 \qquad \textbf{(C) }3 \qquad \textbf{(D) }4 \qquad \textbf{(E) }5$

Solution:
We can form the following expressions for the sum along each line: \[\begin{dcases}A+B+C\\A+E+F\\C+D+E\\B+D\\B+F\end{dcases}\] Adding these together, we must have $2A+3B+2C+2D+2E+2F=47$ , i.e. $2(A+B+C+D+E+F)+B=47$ . Since $A,B,C,D,E,F$ are unique integers between $1$ and $6$ , we obtain $A+B+C+D+E+F=1+2+3+4+5+6=21$ (where the order doesn't matter as addition is commutative), so our equation simplifies to $42 + B = 47$ . This means $B = \boxed{5}$ .
~RJ5303707

Solution: Following the first few steps of Solution 1, we have $2(A+C+D+E+F)+3B=47$ . Because an even number ( $2(A+C+D+E+F)$ ) subtracted from an odd number (47) is always odd, we know that $3B$ is odd, showing that $B$ is odd. Now we know that $B$ is either 1, 3, or 5. If we try $B=1$ , we get $43=47$ which is not true. Testing $B=3$ , we get $45=47$ , which is also not true. Therefore, we have $B = \boxed{5}$


Answer: 4.0

Each of the sides of a square $S_1$ with area $16$ is bisected, and a smaller square $S_2$ is constructed using the bisection points as vertices. The same process is carried out on $S_2$ to construct an even smaller square $S_3$ . What is the area of $S_3$
$\mathrm{(A)}\ \frac{1}{2}\qquad\mathrm{(B)}\ 1\qquad\mathrm{(C)}\ 2\qquad\mathrm{(D)}\ 3\qquad\mathrm{(E)}\ 4$

Solution: Since the length ratio is $\frac{1}{\sqrt{2}}$ , then the area ratio is $\frac{1}{2}$ (since the area ratio between two similar 2-dimensional objects is equal to the side ratio of those objects squared). This means that $S_2 = 8$ and $S_3 = \boxed{4}$


Answer: 2.0

Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.
[asy] unitsize(.3cm); path c=Circle((0,2),1); filldraw(Circle((0,0),3),grey,black); filldraw(Circle((0,0),1),white,black); filldraw(c,white,black); filldraw(rotate(60)*c,white,black); filldraw(rotate(120)*c,white,black); filldraw(rotate(180)*c,white,black); filldraw(rotate(240)*c,white,black); filldraw(rotate(300)*c,white,black); [/asy]
$\textbf{(A)}\ \pi \qquad \textbf{(B)}\ 1.5\pi \qquad \textbf{(C)}\ 2\pi \qquad \textbf{(D)}\ 3\pi \qquad \textbf{(E)}\ 3.5\pi$

Solution: The outer circle has radius $1+1+1=3$ , and thus area $9\pi$ . The little circles have area $\pi$ each; since there are 7, their total area is $7\pi$ . Thus, our answer is $9\pi-7\pi=\boxed{2}$


Answer: 2.0

Solution: The outer circle has radius $1+1+1=3$ , and thus area $9\pi$ . The little circles have area $\pi$ each; since there are 7, their total area is $7\pi$ . Thus, our answer is $9\pi-7\pi=\boxed{2}$


Answer: 6.0

Each of the twenty dots on the graph below represents one of Sarah's classmates.  Classmates who are friends are connected with a line segment.  For her birthday party, Sarah is inviting only the following:  all of her friends and all of those classmates who are friends with at least one of her friends.  How many classmates will not be invited to Sarah's party? [asy]/* AMC8 2003 #18 Problem */ pair a=(102,256), b=(68,131), c=(162,101), d=(134,150); pair e=(269,105), f=(359,104), g=(303,12), h=(579,211); pair i=(534, 342), j=(442,432), k=(374,484), l=(278,501); pair m=(282,411), n=(147,451), o=(103,437), p=(31,373); pair q=(419,175), r=(462,209), s=(477,288), t=(443,358); pair oval=(282,303); draw(l--m--n--cycle); draw(p--oval); draw(o--oval); draw(b--d--oval); draw(c--d--e--oval); draw(e--f--g--h--i--j--oval); draw(k--oval); draw(q--oval); draw(s--oval); draw(r--s--t--oval); dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); dot(g); dot(h); dot(i); dot(j); dot(k); dot(l); dot(m); dot(n); dot(o); dot(p); dot(q); dot(r); dot(s); dot(t); filldraw(yscale(.5)*Circle((282,606),80),white,black); label(scale(0.75)*"Sarah", oval);[/asy]
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$

Solution: There are $3$ people who are friends with only each other who won't be invited, plus $1$ person who has no friends, and $2$ people who are friends of friends of friends who won’t be invited. So the answer is $\boxed{6}$


Answer: 26.0

Each of two boxes contains both black and white marbles, and the total number of marbles in the two boxes is $25.$ One marble is taken out of each box randomly. The probability that both marbles are black is $27/50,$ and the probability that both marbles are white is $m/n,$ where $m$ and $n$ are relatively prime positive integers. What is $m + n$

Solution:
If we work with the problem for a little bit, we quickly see that there is no direct combinatorics way to calculate $m/n$ . The Principle of Inclusion-Exclusion still requires us to find the individual probability of each box.
Let $a, b$ represent the number of marbles in each box, and without loss of generality let $a>b$ . Then, $a + b = 25$ , and since the $ab$ may be reduced to form $50$ on the denominator of $\frac{27}{50}$ $50|ab$ . It follows that $5|a,b$ , so there are 2 pairs of $a$ and $b: (20,5),(15,10)$
Thus, $m + n = \boxed{026}$

Solution:
Let $w_1, w_2, b_1,$ and $b_2$ represent the white and black marbles in boxes 1 and 2.
Since there are $25$ marbles in the box:
$w_1 + w_2 + b_1 + b_2 = 25$
From the fact that there is a $\frac{27}{50}$ chance of drawing one black marble from each box:
$\frac{b_1 \cdot b_2}{(b_1 + w_1)(b_2 + w_2)} = \frac{27}{50} = \frac{54}{100} = \frac{81}{150}$
Thinking of the numerator and denominator separately, if $\frac{27}{50}$ was not a reduced fraction when calculating out the probability, then $b_1 \cdot b_2 = 27$ .  Since $b_1 < 25$ , this forces the variables to be $3$ and $9$ in some permutation.  Without loss of generality, let $b_1 = 3$ and $b_2 = 9$
The denominator becomes: $(3 + w_1)(9 + w_2) = 50$
Since there have been $12$ black marbles used, there must be $13$ white marbles.  Substituting that in:
$(3 + w_1)(9 + (13 - w_1)) = 50$
$(3 + w_1)(22 - w_1) = 50$
Since the factors of $50$ that are greater than $3$ are $5, 10, 25,$ and $50$ , the quantity $3 + w_1$ must equal one of those.  However, since $w_1 < 13$ , testing $2$ and $7$ for $w_1$ does not give a correct product.  Thus, $\frac{27}{50}$ must be a reduced form of the actual fraction.
First assume that the fraction was reduced from $\frac{54}{100}$ , yielding the equations $b_1\cdot b_2 = 54$ and $(b_1 + w_1)(b_2 + w_2) = 100$ .
Factoring $b_1 \cdot b_2 = 54$ and saying WLOG that $b_1 <  b_2 < 25$ gives $(b_1, b_2) = (3, 18)$ or $(6, 9)$ .  Trying the first pair and setting the denominator equal to 100 gives: $(3 + w_1)(18 + w_2) = 100$
Since $w_1 + w_2 = 4$ , the pairs $(w_1, w_2) = (1, 3), (2,2),$ and $(3,1)$ can be tried, since each box must contain at least one white marble.  Plugging in $w_1 = w_2 = 2$ gives the true equation $(3 + 2)(18 + 2) =100$ , so the number of marbles are $(w_1, w_2, b_1, b_2) = (2, 2, 3, 18)$
Thus, the chance of drawing 2 white marbles is $\frac{w_1 \cdot w_2 }{(w_1+ b_1)(w_2 + b_2)} = \frac{4}{100} = \frac{1}{25}$ in lowest terms, and the answer to the problem is $1 + 25 = \boxed{026}.$

Solution: We know that $\frac{27}{50} = \frac{b_1}{t_1} \cdot \frac{b_2}{t_2}$ , where $b_1$ and $b_2$ are the number of black marbles in the first and the second box respectively, and $t_1$ and $t_2$ is the total number of marbles in the first and the second boxes respectively. So, $t_1 + t_2 = 25$ . Then, we can realize that $\frac{27}{50} = \frac{9}{10} \cdot \frac{3}{5} = \frac{9}{10} \cdot \frac{9}{15}$ , which means that having 9 black marbles out of 10 total in the first box and 9 marbles out of 15 total the second box is valid. Then there is 1 white marble in the first box and 6 in the second box. So, the probability of drawing two white marbles becomes $\frac{1}{10} \cdot \frac{6}{15} = \frac{1}{25}$ . The answer is $1 + 25 = \boxed{026}$


Answer: 21.0

Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either $12$ pieces of red candy, $14$ pieces of green candy, $15$ pieces of blue candy, or $n$ pieces of purple candy. A piece of purple candy costs $20$ cents. What is the smallest possible value of $n$
$\textbf{(A) } 18 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 24\qquad \textbf{(D) } 25 \qquad \textbf{(E) } 28$

Solution: If he has enough money to buy $12$ pieces of red candy, $14$ pieces of green candy, and $15$ pieces of blue candy, then the smallest amount of money he could have is $\text{lcm}{(12,14,15)} = 420$ cents. Since a piece of purple candy costs $20$ cents, the smallest possible value of $n$ is $\frac{420}{20} = \boxed{21}$

Solution: We simply need to find a value of $20n$ that is divisible by $12$ $14$ , and $15$ . Observe that $20 \cdot 18$ is divisible by $12$ and $15$ , but not $14$ $20 \cdot 21$ is divisible by $12$ $14$ , and $15$ , meaning that we have exact change (in this case, $420$ cents) to buy each type of candy, so the minimum value of $n$ is $\boxed{21}$

Solution: We can notice that the number of purple candy times $20$ has to be divisible by $7$ , because of the $14$ green candies, and $3$ , because of the $12$ red candies. $7\cdot3=21$ , so the answer has to be $\boxed{21}$


Answer: 21.0

Solution: If he has enough money to buy $12$ pieces of red candy, $14$ pieces of green candy, and $15$ pieces of blue candy, then the smallest amount of money he could have is $\text{lcm}{(12,14,15)} = 420$ cents. Since a piece of purple candy costs $20$ cents, the smallest possible value of $n$ is $\frac{420}{20} = \boxed{21}$

Solution: We simply need to find a value of $20n$ that is divisible by $12$ $14$ , and $15$ . Observe that $20 \cdot 18$ is divisible by $12$ and $15$ , but not $14$ $20 \cdot 21$ is divisible by $12$ $14$ , and $15$ , meaning that we have exact change (in this case, $420$ cents) to buy each type of candy, so the minimum value of $n$ is $\boxed{21}$

Solution: We can notice that the number of purple candy times $20$ has to be divisible by $7$ , because of the $14$ green candies, and $3$ , because of the $12$ red candies. $7\cdot3=21$ , so the answer has to be $\boxed{21}$


Answer: 4.0

Each principal of Lincoln High School serves exactly one $3$ -year term. What is the maximum number of principals this school could have during an $8$ -year period?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 8$

Solution:
If the first year of the $8$ -year period was the final year of a principal's
term, then in the next six years two more principals would serve, and the last year of the
period would be the first year of the fourth principal's term. Therefore, the maximum
number of principals who can serve during an $8$ -year period is $4$ , so the answer is $\boxed{4}$ if the terms are divided $1\ |\ 2\ 3\ 4\ |\ 5\ 6\ 7\ |\ 8$


Answer: 363.0

Each row of the Misty Moon Amphitheater has $33$ seats. Rows $12$ through $22$ are reserved for a youth club. How many seats are reserved for this club?
$\mathrm{(A) \ } 297 \qquad \mathrm{(B) \ } 330\qquad \mathrm{(C) \ } 363\qquad \mathrm{(D) \ } 396\qquad \mathrm{(E) \ } 726$

Solution: There are $22-12+1=11$ rows of $33$ seats, giving $11\times 33=\boxed{363}$ seats.


Answer: 63.0

Each side of $\triangle ABC$ is $12$ units. $D$ is the foot of the perpendicular dropped from $A$ on $BC$ ,
and $E$ is the midpoint of $AD$ . The length of $BE$ , in the same unit, is:
$\textbf{(A)}\ \sqrt{18} \qquad \textbf{(B)}\ \sqrt{28} \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ \sqrt{63} \qquad \textbf{(E)}\ \sqrt{98}$

Solution: Note that $\triangle ABC$ is an equilateral triangle .  From the Pythagorean Theorem (or by using 30-60-90 triangles), $AD = 6\sqrt{3}$ .  That means $DE = 3\sqrt{3}$ .  Using the Pythagorean Theorem again, $BE = \sqrt{63}$ , which is answer choice $\boxed{63}$


Answer: 72.0

Solution:
Let a "tile" denote a $1\times1$ square and "square" refer to $2\times2$
We first have $4!=24$ possible ways to fill out the top left square. We then fill out the bottom right tile. In the bottom right square, we already have one corner filled out (from our initial coloring), and we now have $3$ options left to pick from.
We then look at the right middle tile. It is part of two squares: the top right and top left. Among these squares, $3$ colors have already been used, so we only have one more option for it. Similarly, every other square only has one more option, so we have a total of $3\cdot4!=\boxed{72}$ ways.

Solution:
 We can split this problem into $2$ cases as shown above. We can swap a set of equal colors for another set of equal colors to create a new square.
Square 1:
The first square can be rotated to create another square so we have to multiply the number of arrangements by $2$ . We have $4! = 24$ arrangements without rotating and $24\cdot 2 = 48$ arrangements in total for the first square.
Square 2:
There are $4! = 24$ ways to arrange the colors.
In total, we have $48 + 24 = \boxed{72}$ arrangements.

Solution:
Let’s call the top-right corner color A, the top-middle color B, the top-right color C, and so on, with color D being the middle row, and right corner square, and color G being the bottom-left square’s color. WLOG A=Red, B=White, D=Blue, and E=Green. We will now consider squares C and F’s colors.
Case 1 : C=Red and F=Blue
In this case, we get that G and H have to be Red and White in some order, and the same for H and I. We can color this in 2 ways.
Case 2 : C=Blue and F=Red
In this case, one of G and H needs to be White and Red, and  H and I needs to be White and Blue. There is 1 way to color this.
In total, we get 24*(2+1)=72 ways to color the grid. $\boxed{72}$

Solution:
We will choose colors step-by-step:
1. There are $4$ ways to choose a color in the center.
2. Then we select any corner and there would be $3$ ways to choose a color as we can't use the same color as the one in the center.
3. Consider the $2\times 2$ square that contains the center and the corner we have selected. For the other $2$ squares, there are $2$ ways to choose colors.
4. Now, consider how many configurations it makes sense to construct the $2\times 2$ square opposite to the corner we have selected using the $2$ other $2\times 2$ squares, and we get $3$ configurations.
Finally, the answer is $4 \cdot 3 \cdot 2 \cdot 3 = \boxed{72}$

Solution:
Note that there can be no overlap between colors in each square.
Then, we can choose $1$ color to be in the center. ${4 \choose 1}$ = 4
Now, we have some casework:
Case 1: 1 color is placed in 4 corners and then others are placed on opposite edges. $232$ $414$ $232$ There's $3!=6$ ways to do this.
Case 2: 2 colors are placed with 2 in adjacent corners and 1 edge opposite them. The final color is placed in the remaining 2 edges. $232$ $414$ $323$ The orientation of the 2 colors has 2 possibilities, and there are $3!$ color permutations. $2*3!=12$
There can't be any more ways to do this, as we have combined all cases such that each color is used once and only once per $2*2$ square.
We multiply the start with the sum of the 2 cases: $4(6+12)=\boxed{72}$

Solution: Note that we could fill the 3 by 3 square with numbers of choices, rather than letters or color names. The top-left corner receives a 3 because there are 3 total choices to choose from: R, G and B. The squares bordering them has values of 2 and 1, regardless of order. 2 indicates that the small square can have any color excluding the one existing color, 1 indicates the remaining color of the 2 by 2 square. Finally, the middle square receives 3, since the first 2 by 2 square has RGB already, and it does not matter what color it has. Moving onto the next 2 by 2 square, we see that there are already 2 decided colors, so the other squares must be 2 and 1, again, regardless of the order. The same applys to the third 2 by 2 square, and finally the last square has the value of one, as it is the only square left. Multiplying the numbers, $2\times2\times2\times3\time3\times3$ $\boxed{72}$


Answer: 22.0

Each square in a $5 \times 5$ grid is either filled or empty, and has up to eight adjacent neighboring squares, where neighboring squares share either a side or a corner. The grid is transformed by the following rules:
A sample transformation is shown in the figure below. [asy]         import geometry;         unitsize(0.6cm);          void ds(pair x) {             filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle,mediumgray,invisible);         }          ds((1,1));         ds((2,1));         ds((3,1));         ds((1,3));          for (int i = 0; i <= 5; ++i) {             draw((0,i)--(5,i));             draw((i,0)--(i,5));         }          label("Initial", (2.5,-1));         draw((6,2.5)--(8,2.5),Arrow);          ds((10,2));         ds((11,1));         ds((11,0));          for (int i = 0; i <= 5; ++i) {             draw((9,i)--(14,i));             draw((i+9,0)--(i+9,5));         }          label("Transformed", (11.5,-1)); [/asy] Suppose the $5 \times 5$ grid has a border of empty squares surrounding a $3 \times 3$ subgrid. How many initial configurations will lead to a transformed grid consisting of a single filled square in the center after a single transformation? (Rotations and reflections of the same configuration are considered different.) [asy]         import geometry;         unitsize(0.6cm);          void ds(pair x) {             filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle,mediumgray,invisible);         }          for (int i = 1; i < 4; ++ i) {             for (int j = 1; j < 4; ++j) {                 label("?",(i + 0.5, j + 0.5));             }         }          for (int i = 0; i <= 5; ++i) {             draw((0,i)--(5,i));             draw((i,0)--(i,5));         }          label("Initial", (2.5,-1));         draw((6,2.5)--(8,2.5),Arrow);          ds((11,2));          for (int i = 0; i <= 5; ++i) {             draw((9,i)--(14,i));             draw((i+9,0)--(i+9,5));         }          label("Transformed", (11.5,-1)); [/asy] $\textbf{(A)}\ 14 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 22 \qquad\textbf{(D)}\ 26 \qquad\textbf{(E)}\ 30$

Solution:
There are two cases for the initial configuration:
Together, the answer is $2+20=\boxed{22}.$


Answer: 22.0

Solution:
There are two cases for the initial configuration:
Together, the answer is $2+20=\boxed{22}.$


Answer: 64.0

Each third-grade classroom at Pearl Creek Elementary has $18$ students and $2$ pet rabbits. How many more students than rabbits are there in all $4$ of the third-grade classrooms?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 72\qquad\textbf{(E)}\ 80$

Solution: In each class, there are $18-2=16$ more students than rabbits. So for all classrooms, the difference between students and rabbits is $16 \times 4 = \boxed{64}$


Answer: 64.0

Solution: Multiplying $18$ and $2$ by $4$ we get $72$ students and $8$ rabbits. We then subtract: $72 - 8 = \boxed{64}.$

Solution: In each class, there are $18-2=16$ more students than rabbits. So for all classrooms, the difference between students and rabbits is $16 \times 4 = \boxed{64}$


Answer: 929.0

Each unit square of a 3-by-3 unit-square grid is to be colored either blue or red. For each square, either color is equally likely to be used. The probability of obtaining a grid that does not have a 2-by-2 red square is $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution:
We can use complementary counting , counting all of the colorings that have at least one red $2\times 2$ square.
By the Principle of Inclusion-Exclusion , there are (alternatively subtracting and adding) $128-40+8-1=95$ ways to have at least one red $2 \times 2$ square.
There are $2^9=512$ ways to paint the $3 \times 3$ square with no restrictions, so there are $512-95=417$ ways to paint the square with the restriction. Therefore, the probability of obtaining a grid that does not have a $2 \times 2$ red square is $\frac{417}{512}$ , and $417+512=\boxed{929}$

Solution:
We consider how many ways we can have 2*2 grid
$(1)$ : All the grids are red-- $1$ case
$(2)$ : One unit square is blue--The blue lies on the center of the bigger square, makes no 2*2 grid $9-1=8$ cases
$(3)$ : Two unit squares are blue--one of the squares lies in the center of the bigger square, makes no 2*2 grid, $8$ cases.
Or, two squares lie on second column, first row, second column third row; second row first column, second row third column, 2 extra cases. $\binom 9 2-8-2=26$ cases
$(4)$ Three unit squares are blue. We find that if a 2*2 square is formed, there are 5 extra unit squares can be painted. But cases that three squares in the same column or same row is overcomunted. So in this case, there are $4\cdot (\binom 5 3)-4=36$
$(5)$ Four unit squares are blue, no overcomunted case will be considered. there are $4\cdot \binom 5 4=20$
$(6)$ Five unit squares are blue, $4$ cases in all
Sum up those cases, there are $1+8+26+36+20+4=95$ cases that a 2*2 grid can be formed.
In all, there are $2^9=512$ possible ways to paint the big square, so the answer is $1-\frac{95}{512}=\frac{417}{512}$ leads to $\boxed{929}$

Solution:
\[\begin{array}{|c|c|c|} \hline C_{11} & C_{12} & C_{13}\\ \hline C_{21} & C_{22} & C_{23}\\ \hline C_{31} & C_{32} & C_{33}\\ \hline \end{array}\]
Case 1: The 3-by-3 unit-square grid has exactly $1$ 2-by-2 red square
Assume the 2-by-2 red square is at $C_{11}, C_{12}, C_{21}, C_{22}$ . To make sure there are no more 2-by-2 red squares, $C_{31} \text{and} C_{32}$ can't both be red and $C_{13} \text{and} C_{23}$ can't both be red. Meaning that there are $2^2-1=3$ coloring methods for $C_{31} \text{and} C_{32}$ and $C_{13} \text{and} C_{23}$ $C_{33}$ can be colored with either colors. However, the coloring method where $C_{23}, C_{32}, C_{33}$ are all red needs to be removed. For exactly one 2-by-2 red square at $C_{11}, C_{12}, C_{21}, C_{22}$ , there are $3 \cdot 3 \cdot 2 -1=17$ coloring methods. As there are $4$ locations for the 2-by-2 red square on the 3-by-3 unit-square grid, there are $17 \cdot 4 = 68$ coloring methods.
Case 2: The 3-by-3 unit-square grid has exactly $2$ 2-by-2 red squares
Case 2.1: $2$ 2-by-2 red squares take up $6$ unit grids
Assume the $2$ 2-by-2 red squares are at $C_{11}, C_{12}, C_{21}, C_{22}, C_{31}, C_{32}$ . To make sure there are no more 2-by-2 red squares, $C_{13} \text{and} C_{23}$ can't both be red. Meaning that there are $2^2-1=3$ coloring methods for $C_{13} \text{and} C_{23}$ $C_{33}$ can be colored with either colors. However, the coloring method where $C_{13}, C_{23}, C_{33}$ are all red needs to be removed. For exactly $2$ 2-by-2 red squares at $C_{11}, C_{12}, C_{21}, C_{22}, C_{31}, C_{32}$ , there are $3 \cdot 2 -1=5$ coloring methods. As there are $4$ locations for the $2$ 2-by-2 red squares on the 3-by-3 unit-square grid, there are $5 \cdot 4 = 20$ coloring methods.
Case 2.2: $2$ 2-by-2 red squares take up $7$ unit grids
Assume the $2$ 2-by-2 red squares are at $C_{11}, C_{12}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}$ . To make sure there are no more 2-by-2 red squares, $C_{13}$ and $C_{31}$ can't be red. Meaning that there is $1$ coloring method for $C_{13}$ and $C_{31}$ . For exactly $2$ 2-by-2 red squares at $C_{11}, C_{12}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}$ , there is $1$ coloring method. As there are $2$ locations for the $2$ 2-by-2 red squares on the 3-by-3 unit-square grid, there are $1 \cdot 2 = 2$ coloring methods.
Hence, if the 3-by-3 unit-square grid has exactly $2$ 2-by-2 red squares, there are $20+2 = 22$ coloring methods.
Case 3: The 3-by-3 unit-square grid has exactly $3$ 2-by-2 red squares
Assume the $3$ 2-by-2 red squares are at $C_{11}, C_{12}, C_{13}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}$ . To make sure there are no more 2-by-2 red squares, $C_{33}$ can't be red. Meaning that there is $1$ coloring method for $C_{33}$ . For exactly $3$ 2-by-2 red squares at $C_{11}, C_{12}, C_{13}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}$ , there is $1$ coloring method. As there are $4$ locations for the $3$ 2-by-2 red squares on the 3-by-3 unit-square grid, there are $1 \cdot 4 = 4$ coloring methods.
Case 4: The 3-by-3 unit-square grid has exactly $4$ 2-by-2 red squares
If the 3-by-3 unit-square grid has exactly $4$ 2-by-2 red squares, all $9$ unit grids are red and there is $1$ coloring method.
In total, there are $68+22+4+1=95$ coloring methods with 2-by-2 red squares. \[\frac{m}{n}=1-\frac{95}{2^9}=\frac{417}{512}\]
\[m+n=417+512=\boxed{929}\]


Answer: 1.2

Each valve $A$ $B$ , and $C$ , when open, releases water into a tank at its own constant rate. With all three valves open, the tank fills in 1 hour, with only valves $A$ and $C$ open it takes 1.5 hours, and with only valves $B$ and $C$ open it takes 2 hours. The number of hours required with only valves $A$ and $B$ open is
$\textbf{(A)}\ 1.1\qquad\textbf{(B)}\ 1.15\qquad\textbf{(C)}\ 1.2\qquad\textbf{(D)}\ 1.25\qquad\textbf{(E)}\ 1.75$

Solution: Let the rate of water flowing through valve $A$ be $a$ , the rate of water flowing through valve $B$ be $b$ , and the rate of water flowing through valve $C$ be $c$ WLOG , let the volume of the tank be 1 liter, and let the units for the rates be liters per hour.  With this information, we can write three equations. \[\frac{1}{a+b+c} = 1\] \[\frac{1}{a+c} = \frac{3}{2}\] \[\frac{1}{b+c} = 2\] Manipulate each equation to get \[1 = a+b+c\] \[\frac{2}{3} = a+c\] \[\frac{1}{2} = b+c\] Solving for $a$ yields $a = \tfrac{1}{2}$ , and solving for $b$ yields $b = \tfrac{1}{3}$ .  The number of hours to fill the tub with only valves $A$ and $B$ on is $\frac{1}{\frac{1}{2} + \frac{1}{3}} = \frac{6}{5} = \boxed{1.2}$


Answer: 6.0

Each vertex of a cube is to be labeled with an integer $1$ through $8$ , with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?
$\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24$

Solution:
Note that the sum of the numbers on each face must be 18, because $\frac{1+2+\cdots+8}{2}=18$
So now consider the opposite edges (two edges which are parallel but not on same face of the cube);
they must have the same sum value too.
Now think about the points $1$ and $8$ . If they are not on the same edge, they must be endpoints of opposite edges, and we should have $1+X=8+Y$ . However, this scenario would yield no solution for $[7,2]$ , which is a contradiction. (Try drawing out the cube if it doesn't make sense to you.)
The points $1$ and $8$ are therefore on the same side and all edges parallel must also sum to $9$
Now we have $4$ parallel sides $1-8, 2-7, 3-6, 4-5$ .
Thinking about $4$ endpoints, we realize they need to sum to $18$ .
It is easy to notice only $1-7-6-4$ and $8-2-3-5$ would work.
So if we fix one direction $1-8 ($ or $8-1)$ all other $3$ parallel sides must lay in one particular direction. $(1-8,7-2,6-3,4-5)$ or $(8-1,2-7,3-6,5-4)$
Now, the problem is the same as arranging $4$ points in a two-dimensional square, which is $\frac{4!}{4}=\boxed{6.}$

Solution:
Again, all faces sum to $18.$ If $x,y,z$ are the vertices next to $1$ , then the remaining vertices are $17-x-y, 17-y-z, 17-x-z, x+y+z-16.$ Now it remains to test possibilities. Note that we must have $x+y+z>17.$ Without loss of generality, let $x<y<z.$ \[3,7,8\text{: Does not work.}\] \[4,6,8\text{: Works.}\] \[4,7,8\text{: Works.}\] \[5,6,7\text{: Does not work.}\] \[5,6,8\text{: Does not work.}\] \[5,7,8\text{: Does not work.}\] \[6,7,8\text{: Works.}\]
Keeping in mind that a) solutions that can be obtained by rotating each other count as one solution and b) a cyclic sequence is always asymmetrical, we can see that there are two solutions (one with $[x, y, z]$ and one with $[z, y, x]$ ) for each combination of $x$ $y$ , and $z$ from above. So, our answer is $3\cdot 2=\boxed{6.}$

Solution:
We know the sum of each face is $18.$ If we look at an edge of the cube whose numbers sum to $x$ , it must be possible to achieve the sum $18-x$ in two distinct ways, looking at the two faces which contain the edge. If $8$ and $6$ were on the same edge, it is possible to achieve the desired sum only with the numbers $1$ and $3$ since the values must be distinct. Similarly, if $8$ and $7$ were on the same edge, the only way to get the sum is with $1$ and $2$ . This means that $6$ and $7$ are not on the same edge as $8$ , or in other words they are diagonally across from it on the same face, or on the other end of the cube.
Now we look at three cases, each yielding two solutions which are reflections of each other:
1) $6$ and $7$ are diagonally opposite $8$ on the same face.
2) $6$ is diagonally across the cube from $8$ , while $7$ is diagonally across from $8$ on the same face.
3) $7$ is diagonally across the cube from $8$ , while $6$ is diagonally across from $8$ on the same face.
This means the answer is $3\cdot 2=\boxed{6.}$


Answer: 6.0

Solution:
Note that the sum of the numbers on each face must be 18, because $\frac{1+2+\cdots+8}{2}=18$
So now consider the opposite edges (two edges which are parallel but not on same face of the cube);
they must have the same sum value too.
Now think about the points $1$ and $8$ . If they are not on the same edge, they must be endpoints of opposite edges, and we should have $1+X=8+Y$ . However, this scenario would yield no solution for $[7,2]$ , which is a contradiction. (Try drawing out the cube if it doesn't make sense to you.)
The points $1$ and $8$ are therefore on the same side and all edges parallel must also sum to $9$
Now we have $4$ parallel sides $1-8, 2-7, 3-6, 4-5$ .
Thinking about $4$ endpoints, we realize they need to sum to $18$ .
It is easy to notice only $1-7-6-4$ and $8-2-3-5$ would work.
So if we fix one direction $1-8 ($ or $8-1)$ all other $3$ parallel sides must lay in one particular direction. $(1-8,7-2,6-3,4-5)$ or $(8-1,2-7,3-6,5-4)$
Now, the problem is the same as arranging $4$ points in a two-dimensional square, which is $\frac{4!}{4}=\boxed{6.}$

Solution:
Again, all faces sum to $18.$ If $x,y,z$ are the vertices next to $1$ , then the remaining vertices are $17-x-y, 17-y-z, 17-x-z, x+y+z-16.$ Now it remains to test possibilities. Note that we must have $x+y+z>17.$ Without loss of generality, let $x<y<z.$ \[3,7,8\text{: Does not work.}\] \[4,6,8\text{: Works.}\] \[4,7,8\text{: Works.}\] \[5,6,7\text{: Does not work.}\] \[5,6,8\text{: Does not work.}\] \[5,7,8\text{: Does not work.}\] \[6,7,8\text{: Works.}\]
Keeping in mind that a) solutions that can be obtained by rotating each other count as one solution and b) a cyclic sequence is always asymmetrical, we can see that there are two solutions (one with $[x, y, z]$ and one with $[z, y, x]$ ) for each combination of $x$ $y$ , and $z$ from above. So, our answer is $3\cdot 2=\boxed{6.}$

Solution:
We know the sum of each face is $18.$ If we look at an edge of the cube whose numbers sum to $x$ , it must be possible to achieve the sum $18-x$ in two distinct ways, looking at the two faces which contain the edge. If $8$ and $6$ were on the same edge, it is possible to achieve the desired sum only with the numbers $1$ and $3$ since the values must be distinct. Similarly, if $8$ and $7$ were on the same edge, the only way to get the sum is with $1$ and $2$ . This means that $6$ and $7$ are not on the same edge as $8$ , or in other words they are diagonally across from it on the same face, or on the other end of the cube.
Now we look at three cases, each yielding two solutions which are reflections of each other:
1) $6$ and $7$ are diagonally opposite $8$ on the same face.
2) $6$ is diagonally across the cube from $8$ , while $7$ is diagonally across from $8$ on the same face.
3) $7$ is diagonally across the cube from $8$ , while $6$ is diagonally across from $8$ on the same face.
This means the answer is $3\cdot 2=\boxed{6.}$


Answer: 64.0

Each vertex of a regular dodecagon ( $12$ -gon) is to be colored either red or blue, and thus there are $2^{12}$ possible colorings. Find the number of these colorings with the property that no four vertices colored the same color are the four vertices of a rectangle.

Solution:
\[\text{First, we notice that a rectangle is made from two pairs of vertices 1/2 turn away from each other.}\]
\[\textit{Note: The image is }\frac{\textit{280}}{\textit{841}}\approx\frac{\textit{1}}{\textit{3}}\textit{ size.}\]
\[\text{For there to be no rectangles, there can be at most one same-colored pair for each color and the rest need to have one red and blue.}\]
\[\textit{\underline{Case 1: \textbf{No pairs}}}\]
\[\text{Each pair has two ways to color: One red or the other red. Thus, there are }2^6=\boxed{64}\]


Answer: 928.0

Solution:
Note that the condition is equivalent to stating that there are no 2 pairs of oppositely spaced vertices with the same color.
Case 1: There are no pairs. This yields $2$ options for each vertices 1-6, and the remaining vertices 7-12 are set, yielding $2^6=64$ cases.
Case 2: There is one pair. Again start with 2 options for each vertex in 1-6, but now multiply by 6 since there are 6 possibilities for which pair can have the same color assigned instead of the opposite. Thus, the cases are: $2^6*6=384$
case 3: There are two pairs, but oppositely colored. Start with $2^6$ for assigning 1-6, then multiply by 6C2=15 for assigning which have repeated colors. Divide by 2 due to half the cases having the same colored opposites. $2^6*15/2=480$
It is apparent that no other cases exist, as more pairs would force there to be 2 pairs of same colored oppositely spaced vertices with the same color. Thus, the answer is: $64+384+480=\boxed{928}$

Solution:
First, we identify the geometric condition for the sentence ``four vertices colored the
same color are the four vertices of a rectangle Consider any four vertices on the dodecagon, $A$ $B$ $C$ $D$ .
Denote by $O$ the center of the dodecagon.
Because $OA = OB = OC$ $\angle OAB = \angle OBA$ and $\angle OBC = \angle OCB$
Thus, \begin{align*} \angle ABC & = \angle OBA + \angle OBC \\ & = \frac{\angle OBA + \angle OAB}{2} + \frac{\angle OBC + \angle OCB}{2} \\ & = \frac{180^\circ - \angle AOB}{2} +  \frac{180^\circ - \angle COB}{2} \\ & = 180^\circ - \frac{\angle AOB + \angle COB}{2} \\ & = 180^\circ - \frac{\angle AOC}{2} . \end{align*}
Hence, $\angle ABC = 90^\circ$ if and only if $\angle AOC = 180^\circ$ .
Similarly, $\angle ADC = 90^\circ$ if and only if $\angle AOC = 180^\circ$ , and $\angle BCD = 90^\circ$ (or $\angle DAB = 90^\circ$ ) if and only if $\angle BOD = 180^\circ$
Therefore, $ABCD$ is a rectangle if and only if two diagonals both pass through $O$
Now, we categorize 12 vertices into 6 groups. Each group contains 2 diagonal vertices.
Next, we compute the number of coloring configurations such that the above same-color rectangles do not exist.
Case 1: Two vertices in each group has distinct colors.
For each group, we only need to determine which vertex is red. The other one must be blue.
Therefore, the number of configurations in this case is $2^6$
Case 2: There is one group who vertices have the same color. All other groups are with vertices that have distinct colors.
We construct such an instance in the following steps.
Step 1: We determine which group has two vertices that have the same color.
The number of ways is 6.
Step 2: For the selected group, we choose a color for its two vertices.
The number of ways is 2.
Step 3: For each unselected group, we determine which vertex is red.
The number of ways is $2^5$
Following from the rule of product, the total number of configurations in this case is $6 \cdot 2 \cdot 2^5 = 6 \cdot 2^6$
Case 3: One group has two red vertices, one group has two blue vertices, and each of the other four groups has vertices with distinct colors.
We construct such an instance in the following steps.
Step 1: We determine which group has two vertices that have both red color.
The number of ways is 6.
Step 2: We determine which group has two vertices that have both blue color.
The number of ways is 5.
Step 3: For each unselected group, we determine which vertex is red.
The number of ways is $2^4$
Following from the rule of product, the total number of configurations in this case is $6 \cdot 5 \cdot 2^4 = 30 \cdot 2^4$
Putting all cases together, the total number of feasible configurations is \[ 2^6 + 6 \cdot 2^6 + 30 \cdot 2^4 = \boxed{928}. \]

Solution:
Note that a rectangle is formed if and only if it's diagonals pass through the center of the dodecagon and the diagonal's endpoints are the same color.
Consider the $6$ diagonals that pass through the center. A rectangle is formed if the endpoints of some pair of them are all the same color. We can now perform casework on the diagonals:
Case $1:$ The endpoints of all the diagonals are all different (e.g. for all diagonals, the endpoints are red and blue in some order).
The amount of ways to do this is $2^6=64,$ since there are $2$ ways to order which vertice is red and which is blue for each diagonal.
Case $2:$ There exists one diagonal such that the endpoints are the same color.
There are $6$ ways to choose this diagonal, $2$ ways to choose which color it is, and $2^5$ ways to color the rest of the diagonals. Therefore, the total for this case is $6 \cdot 2 \cdot 2^5 = 384.$
Case $3:$ There exists two diagonals such that the endpoints are the same color, but they are different colors
Then, there are $6 \cdot 5=30$ ways to choose these two diagonals, and $2^4=16$ ways to color the rest, so the total for this case is $30 \cdot 16 = 480.$
Summing all of the cases, we have $64+384+480=\boxed{928}.$


Answer: 3120.0

Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
$\textbf{(A)}\ 2520 \qquad \textbf{(B)}\ 2880 \qquad \textbf{(C)}\ 3120 \qquad \textbf{(D)}\ 3250 \qquad \textbf{(E)}\ 3750$

Solution:
This solution essentially explains other ways of thinking about the cases stated in Solution 2.
Case 1:
${6\choose5} \cdot 5!$
5 colors need to be chosen from the group of 6. Those 5 colors have 5! distinct arrangements on the pentagon's vertices.
Case 2:
${6\choose4} \cdot4\cdot5\cdot3!$
4 colors need to be chosen from the group of 6. Out of those 4 colors, one needs to be chosen to form a pair of 2 identical colors. Then, for arranging this layout onto the pentagon, one of the five sides (same thing as pair of adjacent vertices) of the pentagon needs to be established for the pair. The remaining 3 unique colors can be arranged 3! different ways on the remaining 3 vertices.
Case 3:
${6\choose3} \cdot {3\choose2} \cdot5\cdot2$
3 colors need to be chosen from the group of 6. Out of those 3 colors, 2 need to be chosen to be the pairs. Then, for arranging this layout onto the pentagon, start out by thinking about the 1 color that is not part of a pair, as it makes things easier. It can be any one of the 5 vertices of the pentagon. The remaining 2 pairs of colors can only be arranged 2 ways on the remaining 4 vertices.
Solving each case and adding them up gets you 3120. $\boxed{3120}$


Answer: 3120.0

Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
$\textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750$

Solution:
Let vertex $A$ be any vertex, then vertex $B$ be one of the diagonal vertices to $A$ $C$ be one of the diagonal vertices to $B$ , and so on. We consider cases for this problem.
In the case that $C$ has the same color as $A$ $D$ has a different color from $A$ and so $E$ has a different color from $A$ and $D$ . In this case, $A$ has $6$ choices, $B$ has $5$ choices (any color but the color of $A$ ), $C$ has $1$ choice, $D$ has $5$ choices, and $E$ has $4$ choices, resulting in a possible of $6 \cdot 5 \cdot 1 \cdot 5 \cdot 4 = 600$ combinations.
In the case that $C$ has a different color from $A$ and $D$ has a different color from $A$ $A$ has $6$ choices, $B$ has $5$ choices, $C$ has $4$ choices (since $A$ and $B$ necessarily have different colors), $D$ has $4$ choices, and $E$ has $4$ choices, resulting in a possible of $6 \cdot 5 \cdot 4 \cdot 4 \cdot 4 = 1920$ combinations.
In the case that $C$ has a different color from $A$ and $D$ has the same color as $A$ $A$ has $6$ choices, $B$ has $5$ choices, $C$ has $4$ choices, $D$ has $1$ choice, and $E$ has $5$ choices, resulting in a possible of $6 \cdot 5 \cdot 4 \cdot 1 \cdot 5 = 600$ combinations.
Adding all those combinations up, we get $600+1920+600=\boxed{3120}$

Solution:
First, notice that there can be $3$ cases. One with all vertices painted different colors, one with one pair of adjacent vertices painted the same color and the final one with two pairs of adjacent vertices painted two different colors.
Case $1$ : There are $6!$ ways of assigning each vertex a different color. $6! = 720$
Case $2$ : There are $\frac {6!}{2!} * 5$ ways. After picking four colors, we can rotate our pentagon in $5$ ways to get different outcomes. $\frac {6!}{2} * 5 = 1800$
Case $3$ : There are $\frac {\frac {6!}{3!} * 10}{2!}$ ways of arranging the final case. We can pick $3$ colors for our pentagon. There are $5$ spots for the first pair of colors. Then, there are $2$ possible ways we can put the final pair in the last $3$ spaces. But because the two pairs are indistinguishable, we divide by $2!$ $\frac {\frac {6!}{3!} * 10}{2!} = 600$
Adding all the possibilities up, we get $720+1800+600=\boxed{3120}$

Solution:
There are $6$ ways to assign a color to $A$ . WLOG, give vertex $A$ a color; we can multiply by $6$ at the end. Since vertices $A$ and $C$ cannot have the same color, there are $5$ ways to assign colors to vertex $C$ . Using this same logic, there are $5$ ways to assign a color to vertices $E$ $B$ , and $D$ , giving a total of $5^4=625$ ways. However, vertex $D$ cannot be the same color as vertex $A$ . To use complementary counting, we need to find the amount of ways for $D$ and $A$ to have the same color. Notice that this is equivalent to the amount of ways to arrange colors among the vertices of a square. Using the same logic as above, there are $5^3=125$ ways, except we must subtract the number of ways for a triangle. Each time, there is $1$ less vertex, so $5$ times less ways to color. This process stops when there are only $2$ vertices left; in this case there are simply $5$ ways to color this figure.
So in conclusion, there are $6(5^4-(5^3-(5^2-(5))))=6(5^4-5^3+5^2-5)=\boxed{3120}$ ways.

Solution:
This problem is a direct application of the chromatic polynomial of a graph, represented by $P_k(G)$ , which returns a polynomial representing the number of ways to color a graph $G$ with $k$ colors such that no two adjacent vertices share the same color. In other words, it gives us a polynomial in terms of $k$ where we can plug in $k=6$ to get our answer.
Well, if we want to find $P_k$ for a graph $G$ , we should probably first draw the graph, right? We can first draw five vertices (without any edges) in the shape of a pentagon. After connecting the diagonals, we get a star. We attempt to shift the vertices around to simplify the graph, which we quickly realize is isomorphic (can be turned into) the five-cycle, or the pentagon.
Finding $P_k$ on a graph with a cycle of length at most 3 is straightforward -- we first pick a vertex v. It has 0 colored (visited) neighbors, so we can color it in k ways. We then move on to the vertices adjacent to v, etc, and at the end we multiply all these together. For example, the chromatic polynomial of a triangle is $k(k-1)(k-2)$ . However, cycles with length > 3 introduce ambiguity, and thus the above technique fails. Thus, we need to use the recursive formula $P_k(G) = P_k(G-e)-P_k(G*e)$ , where $G-e$ represents removing e and $G*e$ represents contracting e, or collapsing the two endpoints of e into one. When we hit a graph where the longest cycle has length 3, we can use the first technique to quickly find $P_k$
After about four iterations of the algorithm, we get that the chromatic polynomial is $k(k-1)^4 - k(k-1)^3 + k(k-1)(k-2)$ . Plugging in 6 gives us $\boxed{3120}$

Solution:
In pentagon $ABCDE$ , fix any vertex $A$ . Now draw diagonal $AC$ . There are six choices for vertex $A$ and $5$ choices for vertex $C$
Now draw diagonal $CE$ . Since $E$ cannot be the same color as vertex $C$ , we have $5$ choices for $E$ . Again, we have five choices for vertex $D$ (draw diagonal $AD$ ).
Thus there are $6 \cdot 5$ choices for vertices $A$ and $C$ and $5 \cdot 5$ combinations for $D$ and $E$
To determine the final count, we consider two cases for the final $25$ combinations of $D$ and $E$ , which uniquely determines $B$ . Then, we multiply by $30$ since the choices of $A$ and $C$ are independent from these two cases.
Case $1$ $D$ and $E$ are the same color. There are $4$ possible pairs (This is because $D$ and $E$ are not chosen from the same 5 colors. $D$ cannot be $A$ as they are on a diagonal, but $E$ can be. ~ primegn), and thus we have $5$ choices for $B$ . There are $20$ cases here.
Case $2$ $D$ and $E$ are different. There are $25-4 = 21$ possible combinations and we have $4$ choices for $B$ (not the color of $D$ nor $E$ ). In this case we have $21 \cdot 4 = 84$ cases.
Our final count is $30(84+20) = \boxed{3120}$

Solution:
WLOG, draw $ABCDE$ such that point $A$ is at the top, and write the letters in counterclockwise order. WLOG, fill in $A$ first. There are $6$ ways to do so. From here we proceed with casework on the color of $B$
Summing up the cases, $600+600+1920=3120 \Longrightarrow \boxed{3120}$

Solution:
Notice that a minimum of, $3$ and a maximum of $5$ colours can be used.
This is straight forward and there are $6P5 = 720$ ways.
If $x$ and $y$ have same colours, they have to be adjacent. Let $x$ be directly right of $y$ $4$ consecutive points from $x$ clockwise have different colours. They can be coloured in $6P4=360$ ways. And $x$ can be chosen in $5$ ways, bringing the total to $360 \times 5=1800$
Note that $3$ points can't have the same colour as at least $1$ of the lines joining them will be a diagonal. Then the arrangement is $2-2-1$ . The point with distinct colour can be chosen in $5$ ways and can be coloured in $6$ ways. The next two points clockwise can be coloured in $5$ ways and the next two in $4$ ways. The total is $5 \times 6 \times 5 \times 4=600$
Finally, adding all cases, the answer is $600+1800+720=3120 \Longrightarrow \boxed{3120}$

Solution:
Name the pentagon $ADBEC$ . Then $A,B;$ $B,C;$ $C,D;$ $D,E$ and $E,A$ have to be of different colours.
$A$ can be coloured in $6$ ways, $B$ in $5$ ways (Without the one used for $A$ ), $C$ in $5$ ways (Without the one used for $B$ ), $D$ in $5$ ways (Without the one used for $C$ ).
Now we find the expected value of how many ways can $E$ be coloured.
If $A$ and $C$ are of the same colour, $A$ and $D$ are always different. This happens $\frac{1}{5}$ times and leaves $4$ options for $E$
If $A$ and $C$ are of different colours $\left(\frac{4}{5} \ \text{of the time} \right)$ then there is a $\frac{1}{5}$ chance that $A$ and $D$ are of same colour and $\frac{4}{5}$ chance that they are different. The first one leaves $5$ ways to colour $E$ and the second one leaves $4$
Calculating all these, we find the expected value for the number of ways $E$ can be coloured is $\frac{1}{5} \times 4 + \frac{4}{5} \left( \frac{1}{5} \times 5 + \frac{4}{5} \times 4 \right) = 4.16$
Therefore total number or ways $= 6 \times 5 \times 5 \times 5 \times 4.16 = \boxed{3120}$

Solution:
First, we know that there are $6$ ways to assign a color to vertex $A$ . Then, there are $6$ ways to assign a color to vertex $B$ , because it doesn't depend on vertex $A$ . There are also $5$ ways to assign a color to vertex $C$ , because it can't be the same color as vertex $A$ . Notice that both vertices $E$ and $D$ have 5 or 4 choices, depending on whether $A$ $B$ , and $C$ are the same color. This means that the we know that the amount of choices must be bigger than $6 \times 6 \times 5 \times 4 \times 4 = 2880$ , which eliminates options $A$ and $B$
Then, notice that the amount of choices must be smaller than $6 \times 6 \times 5 \times 4.5 \times 4.5 = 3645$ , because the expected amount of choices $E$ and $D$ have are less than 4.5. (There is a greater probability that $A$ $B$ , and $B$ $C$ are both different). This eliminates option $E$ , which leaves us with options $C$ and $D$
We know that the amount of choices must be a multiple of $6$ , so $D$ is eliminated, leaving us with option $\boxed{3120}$


Answer: 40.0

Ed and Ann both have lemonade with their lunch. Ed orders the regular size. Ann gets the large lemonade, which is 50% more than the regular. After both consume $\frac{3}{4}$ of their drinks, Ann gives Ed a third of what she has left, and 2 additional ounces. When they finish their lemonades they realize that they both drank the same amount. How many ounces of lemonade did they drink together?
$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 50$

Solution:
Let the size of Ed's drink equal $x$ ounces, and let the size of Ann's drink equal $\frac{3}{2}x$ ounces.  After both consume $\frac{3}{4}$ of their drinks, Ed and Ann have $\frac{x}{4}$ and $\frac{3x}{8}$ ounces of their drinks remaining.  Ann gives away $\frac{x}{8} + 2$ ounces to Ed.
In the end, Ed drank everything in his original lemonade plus what Ann gave him, and Ann drank everything in her original lemonade minus what she gave Ed. Thus we have \[x + \frac{x}{8} + 2 = \frac{3x}{2} - \frac{x}{8} - 2\] \[x = 16\] The total amount the two of them drank is simply \[x + \frac{3}{2}x = 16 + 24 = \boxed{40}\]


Answer: 314.0

Ed and Sue bike at equal and constant rates.  Similarly, they jog at equal and constant rates, and they swim at equal and constant rates.  Ed covers $74$ kilometers after biking for $2$ hours, jogging for $3$ hours, and swimming for $4$ hours, while Sue covers $91$ kilometers after jogging for $2$ hours, swimming for $3$ hours, and biking for $4$ hours.  Their biking, jogging, and swimming rates are all whole numbers of kilometers per hour.  Find the sum of the squares of Ed's biking, jogging, and swimming rates.

Solution:
Let the biking rate be $b$ , swimming rate be $s$ , jogging rate be $j$ , all in km/h.
We have $2b + 3j + 4s = 74,2j + 3s + 4b = 91$ . Subtracting the second from twice the first gives $4j + 5s = 57$ . Mod 4, we need $s\equiv1\pmod{4}$ . Thus, $(j,s) = (13,1),(8,5),(3,9)$
$(13,1)$ and $(3,9)$ give non-integral $b$ , but $(8,5)$ gives $b = 15$ . Thus, our answer is $15^{2} + 8^{2} + 5^{2} = \boxed{314}$

Solution: Let $b$ $j$ , and $s$ be the biking, jogging, and swimming rates of the two people.  Hence, $2b + 3j + 4s = 74$ and $4b + 2j + 3s = 91$ .  Subtracting gives us that $2b - j - s = 17$ .  Adding three times this to the first equation gives that $8b + s = 125\implies b\le 15$ .  Adding four times the previous equation to the first given one gives us that $10b - j = 142\implies b > 14\implies b\ge 15$ .  This gives us that $b = 15$ , and then $j = 8$ and $s = 5$ .  Therefore, $b^2 + s^2 + j^2 = 225 + 64 + 25 = \boxed{314}$

Solution:
Creating two systems, we get $2x+3y+4z=74$ , and $2y+3z+4x=91$ . Subtracting the two expressions we get $y+z-2x=-17$ . Note that $-17$ is odd, so one of $x,y,z$ is odd. We see from our second expression that $z$ must be odd, because $91$ is also odd and $2y$ and $4x$ are odd. Thus, with this information, we can test cases quickly:
When readdressing the first equation, we see that if $2x+3y$ will be a multiple of $6$ $4z \equiv 2 \pmod{6} = 5$ , we get that $x=15$ and $y=8$ , which works because of integer values. Therefore, $225+64+25=\boxed{314}$

Solution: Building on top of Solution 3, we can add $j+s-2b=17$ and $2b+3j+4s=74$ (sorry, I used different variables) to get $4j+5s=57$ . Logically speaking, most athletic people swim a lot faster than 1 km/h (0.62 mph), so we test out the next case that works, which is 5 km/h (3.1 mph). This seems much more logical, so we plug it into the equation to get $4j+25=57\implies j=8$ . This seems reasonable, as people usually jog at around 8 to 16 km/h. Plugging these values into $2b+3j+4s=74$ , we get $b=15$ . 15 km/h is a little slow (most people bike at 20 km/h), but is still reasonable. So, we get $5^2+8^2+15^2=\boxed{314}$


Answer: 3.0

Ed has five identical green marbles, and a large supply of identical red marbles. He arranges the green marbles and some of the red ones in a row and finds that the number of marbles whose right hand neighbor is the same color as themselves is equal to the number of marbles whose right hand neighbor is the other color. An example of such an arrangement is GGRRRGGRG. Let $m$ be the maximum number of red marbles for which such an arrangement is possible, and let $N$ be the number of ways he can arrange the $m+5$ marbles to satisfy the requirement. Find the remainder when $N$ is divided by $1000$

Solution: We are limited by the number of marbles whose right hand neighbor is not the same color as the marble.  By surrounding every green marble with red marbles - RGRGRGRGRGR.  That's 10 "not the same colors" and 0 "same colors."  Now, for every red marble we add, we will add one "same color" pair and keep all 10 "not the same color" pairs.  It follows that we can add 10 more red marbles for a total of $m = 16$ .  We can place those ten marbles  in any of 6 "boxes": To the left of the first green marble, to the right of the first but left of the second, etc. up until to the right of the last.  This is a stars-and-bars problem, the solution of which can be found as $\binom{n+k}{k}$ where n is the number of stars and k is the number of bars.  There are 10 stars (The unassigned Rs, since each "box" must contain at least one, are not counted here) and 5 "bars," the green marbles.  So the answer is $\binom{15}{5} = 3003$ , take the remainder when divided by 1000 to get the answer: $\boxed{003}$

Solution: Begin with the above solution's reasoning to find that the ideal sequence is RGRGRGRGRGR + 10 Rs. To count the number of arrangements where the Gs are separate, group a G and an R together (GR) which will be rearranged/counted as one. However this also introduces the case of having GRGR.... which results in too few number of marbles whose right-hand neighbor is different, so we fix an R in the beginning of our sequence which also removes it from our counting calculation. Finally we are looking at the arrangement of 5 (GR)s + 10 Rs, all indistinguishable, which is $\frac{15!}{5!10!}=3003\Longrightarrow\boxed{003}$


Answer: 18.0

Eight $1\times 1$ square tiles are arranged as shown so their outside edges form a polygon with a perimeter of $14$ units.  Two additional tiles of the same size are added to the figure so that at least one side of each tile is shared with a side of one of the squares in the original figure.  Which of the following could be the perimeter of the new figure?
[asy] for (int a=1; a <= 4; ++a) {     draw((a,0)--(a,2)); } draw((0,0)--(4,0)); draw((0,1)--(5,1)); draw((1,2)--(5,2)); draw((0,0)--(0,1)); draw((5,1)--(5,2)); [/asy]
$\text{(A)}\ 15 \qquad \text{(B)}\ 17 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 19 \qquad \text{(E)}\ 20$

Solution:
One such figure would be

The perimeter of this figure is $\boxed{18}$


Answer: 65.0

Eight circles of diameter 1 are packed in the first quadrant of the coordinate plane as shown. Let region $\mathcal{R}$ be the union of the eight circular regions. Line $l,$ with slope 3, divides $\mathcal{R}$ into two regions of equal area. Line $l$ 's equation can be expressed in the form $ax=by+c,$ where $a, b,$ and $c$ are positive integers whose greatest common divisor is 1. Find $a^2+b^2+c^2.$ [asy] size(150);defaultpen(linewidth(0.7)); draw((6.5,0)--origin--(0,6.5), Arrows(5)); int[] array={3,3,2}; int i,j; for(i=0; i<3; i=i+1) { for(j=0; j<array[i]; j=j+1) { draw(Circle((1+2*i,1+2*j),1)); }} label("x", (7,0)); label("y", (0,7));[/asy]

Solution: The line passing through the tangency point of the bottom left circle and the one to its right and through the tangency of the top circle in the middle column and the one beneath it is the line we are looking for: a line passing through the tangency of two circles cuts congruent areas, so our line cuts through the four aforementioned circles splitting into congruent areas, and there are an additional two circles on each side. The line passes through $\left(1,\frac 12\right)$ and $\left(\frac 32,2\right)$ , which can be easily solved to be $6x = 2y + 5$ . Thus, $a^2 + b^2 + c^2 = \boxed{065}$

Solution: This problem looks daunting at a first glance, but we can make geometric inequality inferences by drawing lines that simplify the problem by removing sections of the total area. To begin, we can eliminate the possibility of the line intersecting the circle on the top left (call it circle A), or the circle on the bottom right (call it circle B). This is can be seen visually by drawing a line with slope 3 that is tangent to either of these circles. The area is clearly larger on one side; this can be proven by counting full circles. We can go on with the same mindset and eliminate the circle below circle A and the circle above circle B. By removing pairs of circles and proving the line will never intersect with them, we can safely work with whatever is remaining. By now you should have 4 circles making an L shape (waluigi style). Now the two biggest contenders for this method are the two circles on the bottom row. Using the same strategy, we can see that a line that goes through the tangent point of these two circles also goes through the tangent point of the other two circles. This clearly will cut the 4 circles into two regions of equal area. Using super advanced linear algebra we get: $6x = 2y + 5$ . The answer is then $6^2 + 2^2 + 5^2 = \boxed{065}$ . This solution is an alternate explanation to solution 1.


Answer: 197.0

Eight circles of radius $34$ are sequentially tangent, and two of the circles are tangent to $AB$ and $BC$ of triangle $ABC$ , respectively. $2024$ circles of radius $1$ can be arranged in the same manner. The inradius of triangle $ABC$ can be expressed as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] pair A = (2,1); pair B = (0,0); pair C = (3,0); dot(A^^B^^C); label("$A$", A, N); label("$B$", B, S); label("$C$", C, S); draw(A--B--C--cycle); for(real i=0.62; i<2.7; i+=0.29){ draw(circle((i,0.145), 0.145)); } [/asy]

Solution: Draw an altitude from both end circles of the diagram with the circles of radius one, and call the lengths you get drawing the altitudes of the circles down to $BC$ $a$ and $b$ . Now we have the length of side $BC$ of being $(2)(2022)+1+1+a+b$ . However, the side $BC$ can also be written as $(6)(68)+34+34+34a+34b$ , due to similar triangles from the second diagram. If we set the equations equal, we have $\frac{1190}{11} = a+b$ . Call the radius of the incircle $r$ , then we have the side BC to be $r(a+b)$ . We find $r$ as $\frac{4046+\frac{1190}{11}}{\frac{1190}{11}}$ , which simplifies to $\frac{10+((34)(11))}{10}$ ,so we have $\frac{192}{5}$ , which sums to $\boxed{197}$

Solution:
Assume that $ABC$ is isosceles with $AB=AC$
If we let $P_1$ be the intersection of $BC$ and the leftmost of the eight circles of radius $34$ $N_1$ the center of the leftmost circle, and $M_1$ the intersection of the leftmost circle and $AB$ , and we do the same for the $2024$ circles of radius $1$ , naming the points $P_2$ $N_2$ , and $M_2$ , respectively, then we see that $BP_1N_1M_1\sim BP_2N_2M_2$ . The same goes for vertex $C$ , and the corresponding quadrilaterals are congruent.
Let $x=BP_2$ . We see that $BP_1=34x$ by similarity ratios (due to the radii). The corresponding figures on vertex $C$ are also these values. If we combine the distances of the figures, we see that $BC=2x+4046$ and $BC=68x+476$ , and solving this system, we find that $x=\frac{595}{11}$
If we consider that the incircle of $\triangle ABC$ is essentially the case of $1$ circle with $r$ radius (the inradius of $\triangle ABC$ , we can find that $BC=2rx$ . From $BC=2x+4046$ , we have:
$r=1+\frac{2023}{x}$
$=1+\frac{11\cdot2023}{595}$
$=1+\frac{187}{5}$
$=\frac{192}{5}$
Thus the answer is $192+5=\boxed{197}$

Solution:
Let $x = \cot{\frac{B}{2}} + \cot{\frac{C}{2}}$ . By representing $BC$ in two ways, we have the following: \[34x + 7\cdot 34\cdot 2 = BC\] \[x + 2023 \cdot 2 = BC\]
Solving we find $x = \frac{1190}{11}$ .
Now draw the inradius, let it be $r$ . We find that $rx =BC$ , hence \[xr = x + 4046 \implies r-1 = \frac{11}{1190}\cdot 4046 = \frac{187}{5}.\] Thus \[r = \frac{192}{5} \implies \boxed{197}.\] ~AtharvNaphade

Solution:
First, let the circle tangent to $AB$ and $BC$ be $O$ and the other circle that is tangent to $AC$ and $BC$ be $R$ . Let $x$ be the distance from the tangency point on line segment $BC$ of the circle $O$ to $B$ . Also, let $y$ be the distance of the tangency point of circle $R$ on the line segment $BC$ to point $C$ . Realize that we can let $n$ be the number of circles tangent to line segment $BC$ and $r$ be the corresponding radius of each of the circles. Also, the circles that are tangent to $BC$ are similar. So, we can build the equation $BC = (x+y+2(n-1)) \times r$ . Looking at the given information, we see that when $n=8$ $r=34$ , and when $n=2024$ $r=1$ , and we also want to find the radius $r$ in the case where $n=1$ . Using these facts, we can write the following equations:
$BC = (x+y+2(8-1)) \times 34 = (x+y+2(2024-1)) \times 1 = (x+y+2(1-1)) \times r$
We can find that $x+y = \frac{1190}{11}$ . Now, let $(x+y+2(2024-1)) \times 1 = (x+y+2(1-1)) \times r$
Substituting $x+y = \frac{1190}{11}$ in, we find that \[r = \frac{192}{5} \implies \boxed{197}.\]

Solution:
Define $I, x_1, x_8, y_1, y_{2024}$ to be the incenter and centers of the first and last circles of the $8$ and $2024$ tangent circles to $BC,$ and define $r$ to be the inradius of triangle $\bigtriangleup ABC.$ We calculate $\overline{x_1x_8} = 34 \cdot 14$ and $\overline{y_1y_{2024}} = 1 \cdot 4046$ because connecting the center of the circles voids two extra radii.
We can easily see that $B, x_1, x_8,$ and $I$ are collinear, and the same follows for $C, y_1, y_2024,$ and $I$ (think angle bisectors).
We observe that triangles $\bigtriangleup I x_1 x_8$ and $\bigtriangleup I y_1 y_{2024}$ are similar, and therefore the ratio of the altitude to the base is the same, so we note
\[\frac{\text{altitude}}{\text{base}} = \frac{r-34}{34\cdot 14} = \frac{r-1}{1\cdot 4046}.\]
Solving yields $r = \frac{192}{5},$ so the answer is $192+5 = \boxed{197}.$


Answer: 1680.0

Eight congruent equilateral triangles , each of a different color, are used to construct a regular octahedron . How many distinguishable ways are there to construct the octahedron? (Two colored octahedrons are distinguishable if neither can be rotated to look just like the other.)
$\textbf {(A)}\ 210 \qquad \textbf {(B)}\ 560 \qquad \textbf {(C)}\ 840 \qquad \textbf {(D)}\ 1260 \qquad \textbf {(E)}\ 1680$

Solution:
This problem can be approached by Graph Theory . Note that each face of the octahedron is connected to 3 other faces. We use the above graph to represent the problem. Each vertex represents a face of the octahedron, each edge represent the octahedron's edge.
Now the problem becomes how many distinguishable ways to color the $8$ vertices such that two colored graphs are distinguishable if neither can be rotated and reflected to become the other.
Notice that once the outer 4 vertices are colored, no matter how the inner 4 vertices are colored, the resulting graphs are distinguishable graphs.
There are $8$ colors and $4$ outer vertices, therefore there are $\binom{8}{4}$ ways to color outer 4 vertices. Combination is used because the coloring has to be distinguishable when rotated and reflected. There are $4$ colors left, therefore there are $4!$ ways to color inner 4 vertices. Permutation is used because the coloring of the inner vertices have no restrictions. In total that is $\binom{8}{4} \cdot 4! = \boxed{1680}$

Solution:
Let the colors be $1$ to $8$ inclusive, then rotate the octahedron such that color $1$ is on top. You have $7$ choices of what color is on the bottom, WLOG $2$ . Then, there's two rings of each $3$ colors on the top and bottom. For the top ring, you can choose any $3$ out of the $6$ remaining colors, and there's two ways to orient them. The octahedron is now fixed in place, so you can have $3!$ ways to put the three remaining colors in three spaces.
In total this is $7 \cdot \binom{6}{3} \cdot 2 \cdot 3!=\boxed{1680}$


Answer: 140.0

Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill?
$\textbf{(A)}\ \text{\textdollar}120\qquad\textbf{(B)}\ \text{\textdollar}128\qquad\textbf{(C)}\ \text{\textdollar}140\qquad\textbf{(D)}\ \text{\textdollar}144\qquad\textbf{(E)}\ \text{\textdollar}160$

Solution: Since Judi's 7 friends had to pay $2.50 extra each to cover the total amount that Judi should have paid, we multiply $2.50\cdot7=17.50$ is the bill Judi would have paid if she had money. Hence, to calculate the total amount, we multiply $17.50\cdot8=\boxed{140}$ to find the total the 8 friends paid.

Solution: Let $m$ be the total bill the 8 friends split up; each person was supposed to pay $$\frac{m}{8}$ . Since Judi forgot her money, we have to add $$\frac{m}{8}+2.50$ to get the amount of money each of the $7$ friends have to pay. After multipling by 7, we have $7(\frac{m}{8}+2.50)=\frac{7m}{8} + 17.5$ . This is equal to $m$ because both sides represent the total bill. Solving for $m$ \[\frac{7m}{8} + 17.5=m\] \[\frac{m}{8}=17.5\] \[m=\boxed{140}.\]


Answer: 28.0

Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point inside the circle. How many triangles with all three vertices in the interior of the circle are created?
$\textbf{(A)}\ 28 \qquad \textbf{(B)}\ 56 \qquad \textbf{(C)}\ 70 \qquad \textbf{(D)}\ 84 \qquad \textbf{(E)}\ 140$

Solution: To choose a chord, we know that two points must be chosen. This implies that for three chords to create a triangle and not intersect at a single point, six points need to be chosen. We also know that for any six points we pick, there is only $1$ way to connect the points such that a triangle is formed in the circle's interior (this is because we want no two chords to be nonintersecting ~Williamgolly). Therefore, the answer is ${{8}\choose{6}}$ , which is equivalent to $\boxed{28}$

Solution: To make a triangle, where the $3$ points are arranged on a circle, you just need to choose $3$ points because no $3$ points are arranged in a straight line on a circle, meaning that to count the number of triangles we get $\binom{8}{3}=56$ . However, this is incorrect because of the restriction that no three chords intersect in a single point in the circle meaning that the answer is most likely less than $56$ , and the only answer choice that satisfies this condition is $\boxed{28}$ ~Batmanstark


Answer: 27.0

Eight points are spaced around at intervals of one unit around a $2 \times 2$ square, as shown. Two of the $8$ points are chosen at random. What is the probability that the two points are one unit apart? [asy] size((50)); dot((5,0)); dot((5,5)); dot((0,5)); dot((-5,5)); dot((-5,0)); dot((-5,-5)); dot((0,-5)); dot((5,-5)); [/asy] $\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{2}{7}\qquad\textbf{(C)}\ \frac{4}{11}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{4}{7}$

Solution:
The two points are one unit apart at $8$ places around the edge of the square. There are $8 \choose 2$ $= 28$ ways to choose two points. The probability is
\[\frac{8}{28} = \boxed{27}\]

Solution: Arbitrarily pick a point in the grid. Clearly, we see two options for the other point to be placed, so the answer is $\boxed{27}$


Answer: 152.0

Eight spheres of radius 100 are placed on a flat surface so that each sphere is tangent to two others and their centers are the vertices of a regular octagon .  A ninth sphere is placed on the flat surface so that it is tangent to each of the other eight spheres.  The radius of this last sphere is $a +b\sqrt {c},$ where $a, b,$ and $c$ are positive integers , and $c$ is not divisible by the square of any prime .  Find $a + b + c$

Solution:
The key is to realize the significance that the figures are spheres, not circles . The 2D analogue of the diagram onto the flat surface will not contain 8 circles tangent to a ninth one; instead the circles will overlap since the middle sphere has a larger radius and will sort of “bulge” out.
1998 AIME-10a.png
Let us examine the relation between one of the outside 8 spheres and the center one (with radius $r$ ):
1998 AIME-10b.png
If we draw the segment containing the centers and the radii perpendicular to the flat surface, we get a trapezoid ; if we draw the segment parallel to the surface that connects the center of the smaller sphere to the radii of the larger, we get a right triangle. Call that segment $x$ . Then by the Pythagorean Theorem
\[x^2 + (r-100)^2 = (r+100)^2 \Longrightarrow x = 20\sqrt{r}\]
1998 AIME-10c.png
$x$ is the distance from one of the vertices of the octagon to the center, so the diagonal of the octagon is of length $2x =40\sqrt{r}$ . We can draw another right triangle as shown above. One leg has a length of $200$ . The other can be found by partitioning the leg into three sections and using $45-45-90 \triangle$ s to see that the leg is $100\sqrt{2} + 200 + 100\sqrt{2} = 200(\sqrt{2} + 1)$ . Pythagorean Theorem:
\begin{eqnarray*} (40\sqrt{r})^2 &=& 200^2 + [200(\sqrt{2}+1)]^2\\ 1600r &=& 200^2[(1 + \sqrt{2})^2 + 1] \\ r &=& 100 + 50\sqrt{2} \end{eqnarray*}
Thus $a + b + c = 100 + 50 + 2 = \boxed{152}$

Solution:
Isolate a triangle, with base length $200$ (a side of the octagon). This triangle is obviously isoceles. Denote the other side length as $x$ . Since the interior angle is $45$ degrees (due to the shape being an octagon), then we can apply Law of Cosines to this triangle. We get: \begin{eqnarray*} 200^2 &=& 2x^2 - 2x^2*cos(45^\circ) \\ &=& 2x^2 - 2x^2*\frac{\sqrt{2}}{2} \\ &=& (2-\sqrt{2})x^2 \end{eqnarray*}
And thus \[x = \frac{200}{\sqrt{2-\sqrt{2}}}\]
From the above, $x = 20\sqrt{r}$ , so we get
\begin{eqnarray*} r &=& (\frac{200}{20(\sqrt{2-\sqrt{2}})})^2 \\ &=& (\frac{10}{\sqrt{2-\sqrt{2}})})^2 \cdot \frac{2+\sqrt{2}}{2+\sqrt{2}} \\ &=& \frac{200 + 100\sqrt{2}}{2} \\ &=& 100 + 50\sqrt{2} \end{eqnarray*}
And hence the answer is $100 + 50 + 2 \Rightarrow \boxed{152}$


Answer: 3.0

Eleven members of the Middle School Math Club each paid the same amount for a guest speaker to talk about problem solving at their math club meeting. They paid their guest speaker $\textdollar\underline{1} \underline{A} \underline{2}$ . What is the missing digit $A$ of this $3$ -digit number?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad \textbf{(E) }4$

Solution: Since all the eleven members paid the same amount, that means that the total must be divisible by $11$ . We can do some trial-and-error to get $A=3$ , so our answer is $\boxed{3}$ ~SparklyFlowers

Solution: We know that a number is divisible by $11$ if the odd digits added together minus the even digits added together (or vice versa) is a multiple of $11$ . Thus, we have $1+2-A$ = a multiple of $11$ . The only multiple that works here is $0$ , as $11 \cdot 0 = 0$ . Thus, $A = \boxed{3}$ ~fn106068


Answer: 247.0

Ellina has twelve blocks, two each of red ( $\textbf{R}$ ), blue ( $\textbf{B}$ ), yellow ( $\textbf{Y}$ ), green ( $\textbf{G}$ ), orange ( $\textbf{O}$ ), and purple ( $\textbf{P}$ ). Call an arrangement of blocks $\textit{even}$ if there is an even number of blocks between each pair of blocks of the same color. For example, the arrangement \[\textbf{R B B Y G G Y R O P P O}\] is even. Ellina arranges her blocks in a row in random order. The probability that her arrangement is even is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: Consider this position chart: \[\textbf{1 2 3 4 5 6 7 8 9 10 11 12}\] Since there has to be an even number of spaces between each pair of the same color, spots $1$ $3$ $5$ $7$ $9$ , and $11$ contain some permutation of all $6$ colored balls. Likewise, so do the even spots, so the number of even configurations is $6! \cdot 6!$ (after putting every pair of colored balls in opposite parity positions, the configuration can be shown to be even). This is out of $\frac{12!}{(2!)^6}$ possible arrangements, so the probability is: \[\frac{6!\cdot6!}{\frac{12!}{(2!)^6}} = \frac{6!\cdot2^6}{7\cdot8\cdot9\cdot10\cdot11\cdot12} = \frac{2^4}{7\cdot11\cdot3} = \frac{16}{231},\] which is in simplest form. So, $m + n = 16 + 231 = \boxed{247}$

Solution: We can simply use constructive counting. First, let us place the red blocks; choose the first slot in $12$ ways, and the second in $6$ ways, because the number is cut in half due to the condition in the problem. This gives $12 \cdot 6$ ways to place the red blocks. Similarly, there are $10 \cdot 5$ ways to place the blue blocks, and so on, until there are $2 \cdot 1$ ways to place the purple blocks. Thus, the probability is \[\frac{12 \cdot 6 \cdot 10 \cdot 5 \cdot 8 \cdot 4 \cdot 6 \cdot 3 \cdot 4 \cdot 2 \cdot 2 \cdot 1}{12!}=\frac{16}{231},\] and the desired answer is $16+231=\boxed{247}$


Answer: 8.0

Elmer the emu takes $44$ equal strides to walk between consecutive telephone poles on a rural road. Oscar the ostrich can cover the same distance in $12$ equal leaps. The telephone poles are evenly spaced, and the $41$ st pole along this road is exactly one mile ( $5280$ feet) from the first pole. How much longer, in feet, is Oscar's leap than Elmer's stride?
$\textbf{(A) }6\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }15$

Solution:
There are $41-1=40$ gaps between the $41$ telephone poles, so the distance of each gap is $5280\div40=132$ feet.
Each of Oscar's leaps covers $132\div12=11$ feet, and each of Elmer's strides covers $132\div44=3$ feet.
Therefore, Oscar's leap is $11-3=\boxed{8}$ feet longer than Elmer's stride.

Solution: There are $41-1=40$ gaps between the $41$ telephone poles, so Elmer takes $44 \cdot 40 = 1760$ strides in total, and Oscar takes $12 \cdot 40 = 480$ leaps in total. Therefore, the answer is $(5280 \div 480) - (5280 \div 1760) = 11-3=\boxed{8}$


Answer: 8.0

Solution:
There are $41-1=40$ gaps between the $41$ telephone poles, so the distance of each gap is $5280\div40=132$ feet.
Each of Oscar's leaps covers $132\div12=11$ feet, and each of Elmer's strides covers $132\div44=3$ feet.
Therefore, Oscar's leap is $11-3=\boxed{8}$ feet longer than Elmer's stride.

Solution: There are $41-1=40$ gaps between the $41$ telephone poles, so Elmer takes $44 \cdot 40 = 1760$ strides in total, and Oscar takes $12 \cdot 40 = 480$ leaps in total. Therefore, the answer is $(5280 \div 480) - (5280 \div 1760) = 11-3=\boxed{8}$