Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $A,B,C$ be angles of a triangle with \begin{align*} \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\ \cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9} \end{align*} There are positive integers $p$ $q$ $r$ , and $s$ for which \[\cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac{p-q\sqrt{r}}{s},\] where $p+q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime.  Find $p+q+r+s$

Solution: Let \begin{align*} \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{     ------ (1)}\\ \cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9} \text{     ------ (2)}\\ \cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B &= x \text{                        ------ (3)}\\ \end{align*}
Adding (1) and (3) we get: \[2 \cos^2 A + \cos^2 B + \cos^2 C + 2 \sin A( \sin B \cos C +  \sin C \cos B)  = \frac{15}{8} + x\] or \[2 \cos^2 A + \cos^2 B + \cos^2 C + 2 \sin A \sin (B+C)  = \frac{15}{8} + x\] or \[2 \cos^2 A + \cos^2 B + \cos^2 C + 2 \sin ^2 A = \frac{15}{8} + x\] or \[\cos^2 B + \cos^2 C =  x - \frac{1}{8}     \text{     ------ (4)}\]
Similarly adding (2) and (3) we get: \[\cos^2 A + \cos^2 B =  x - \frac{4}{9}     \text{     ------ (5)}\] Similarly adding (1) and (2) we get: \[\cos^2 A + \cos^2 C =  \frac{14}{9}  - \frac{1}{8}     \text{     ------ (6)}\]
And (4) - (5) gives: \[\cos^2 C - \cos^2 A =  \frac{4}{9}  - \frac{1}{8}    \text{     ------ (7)}\]
Now (6) - (7) gives: $\cos^2 A  =  \frac{5}{9}$ or $\cos A = \sqrt{\dfrac{5}{9}}$ and $\sin A = \frac{2}{3}$ so $\cos C$ is $\sqrt{\dfrac{7}{8}}$ and therefore $\sin C$ is $\sqrt{\dfrac{1}{8}}$
Now $\sin B = \sin(A+C)$ can be computed first and then $\cos^2 B$ is easily found.
Thus $\cos^2 B$ and $\cos^2 C$ can be plugged into (4) above to give x = $\dfrac{111-4\sqrt{35}}{72}$
Hence the answer is = $111+4+35+72 = \boxed{222}$

Answer:  222
Problem: Let $A,B,C$ be angles of a triangle with \begin{align*} \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\ \cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9} \end{align*} There are positive integers $p$ $q$ $r$ , and $s$ for which \[\cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac{p-q\sqrt{r}}{s},\] where $p+q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime.  Find $p+q+r+s$

Solution: Let's take the first equation $\cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C = \frac{15}{8}$ . Substituting $180 - A - B$ for C, given A, B, and C form a triangle, and that $\cos C = \cos(A + B)$ , gives us:
$\cos^2 A + \cos^2 B + 2 \sin A \sin B \cos (A+B) = \frac{15}{8}$
Expanding out gives us $\cos^2 A + \cos^2 B + 2 \sin^2 A \sin^2 B - 2 \sin A \sin B \cos A \cos B = \frac{15}{8}$
Using the double angle formula $\cos^2 k = \frac{\cos (2k) + 1}{2}$ , we can substitute for each of the squares $\cos^2 A$ and $\cos^2 B$ . Next we can use the Pythagorean identity on the $\sin^2 A$ and $\sin^2 B$ terms. Lastly we can use the sine double angle to simplify.
$\cos^2 A + \cos^2 B + 2(1 - \cos^2 A)(1 - \cos^2 B) - \frac{1}{2} \cdot \sin 2A \sin 2B = \frac{15}{8}$
Expanding and canceling yields, and again using double angle substitution,
$1 + 2 \cdot \frac{\cos (2A) + 1}{2} \cdot \frac{\cos (2B) + 1}{2} - \frac{1}{2} \cdot \sin 2A \sin 2B = \frac{15}{8}$
Further simplifying yields:
$\frac{3}{2} + \frac{\cos 2A \cos 2B - \sin 2A \sin 2B}{2} = \frac{15}{8}$
Using cosine angle addition formula and simplifying further yields, and applying the same logic to Equation $2$ yields:
$\cos (2A + 2B) = \frac{3}{4}$ and $\cos (2B + 2C) = \frac{1}{9}$
Substituting the identity $\cos (2A + 2B) = \cos(2C)$ , we get:
$\cos (2C) = \frac{3}{4}$ and $\cos (2A) = \frac{1}{9}$
Since the third expression simplifies to the expression $\frac{3}{2} + \frac{\cos (2A + 2C)}{2}$ , taking inverse cosine and using the angles in angle addition formula yields the answer, $\frac{111 - 4\sqrt{35}}{72}$ , giving us the answer $\boxed{222}$

Answer:  222
Problem: Let $A,B,C$ be angles of a triangle with \begin{align*} \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\ \cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9} \end{align*} There are positive integers $p$ $q$ $r$ , and $s$ for which \[\cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac{p-q\sqrt{r}}{s},\] where $p+q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime.  Find $p+q+r+s$

Solution: We will use the sum to product formula to simply these equations. Recall \[2\sin{\frac{\alpha+\beta}{2}}\sin{\frac{\alpha-\beta}{2}} = \cos{\alpha}+\cos{\beta}.\] Using this, let's rewrite the first equation: \[\cos^2(A) + \cos^2(B) + 2 \sin(A) \sin(B) \cos(C) = \frac{15}{8}\] \[\cos^2(A) + \cos^2(B) + (\cos(A+B)+\cos(A-B))\cos(C).\] Now, note that $\cos(C)=-\cos(A+B)$ \[\cos^2(A) + \cos^2(B) + (\cos(A+B)+\cos(A-B))(-\cos(A+B))\] \[\cos^2(A) + \cos^2(B) - \cos(A+B)\cos(A-B)+cos^2(A+B)=\frac{15}{8}.\] We apply the sum to product formula again. \[\cos^2(A) + \cos^2(B) - \frac{\cos(2A)+\cos(2B)}{2}+cos^2(A+B)=\frac{15}{8}.\] Now, recall that $\cos(2\alpha)=2\cos^2(\alpha)-1$ . We apply this and simplify our expression to get: \[\cos^2(A+B)=\frac{7}{8}\] \[\cos^2(C)=\frac{7}{8}.\] Analogously, \[\cos^2(A)=\frac{5}{9}.\] \[\cos^2(A+C)=\frac{p-q\sqrt{r}}{s}-1.\] We can find this value easily by angle sum formula. After a few calculations, we get $\frac{111 - 4\sqrt{35}}{72}$ , giving us the answer $\boxed{222}$ .
~superagh

Answer:  222
Problem: Let $A,B,C$ be angles of a triangle with \begin{align*} \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\ \cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9} \end{align*} There are positive integers $p$ $q$ $r$ , and $s$ for which \[\cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac{p-q\sqrt{r}}{s},\] where $p+q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime.  Find $p+q+r+s$

Solution: According to LOC $a^2+b^2-2ab\cos{\angle{c}}=c^2$ , we can write it into $\sin^2{\angle{A}}+\sin^2{\angle{B}}-2\sin{\angle{A}}\sin{\angle{B}}\cos{\angle{C}}=\sin^2{\angle{C}}$ $\sin^2{\angle{A}}+\sin^2{\angle{B}}-2\sin{\angle{A}}\sin{\angle{B}}\cos{\angle{C}}+cos^2A+cos^2B+2sinAsinBcosC=\frac{15}{8}+sin^2C$ We can simplify to $2=sin^2C+\frac{15}{8}$ . Similarly, we can generalize $2=sin^2A+\frac{14}{9}$ . After solving, we can get that $sinA=\frac{2}{3}; cosA=\frac{\sqrt{5}}{3}; sinC=\frac{\sqrt{2}}{4}; cosC=\frac{\sqrt{14}}{4}$ Assume the value we are looking for is $x$ , we get $sin^2B+x=2$ , while $sinB=sin(180^{\circ}-A-C)=sin(A+C)$ which is $\frac{2\sqrt{14}+\sqrt{10}}{12}$ , so $x=\frac{111 - 4\sqrt{35}}{72}$ , giving us the answer $\boxed{222}$ .~bluesoul

Answer:  222
Problem: Find the number of positive integers with three not necessarily distinct digits, $abc$ , with $a \neq 0$ and $c \neq 0$ such that both $abc$ and $cba$ are multiples of $4$

Solution: A positive integer is divisible by $4$ if and only if its last two digits are divisible by $4.$ For any value of $b$ , there are two possible values for $a$ and $c$ , since we find that if $b$ is even, $a$ and $c$ must be either $4$ or $8$ , and if $b$ is odd, $a$ and $c$ must be either $2$ or $6$ . There are thus $2 \cdot 2 = 4$ ways to choose $a$ and $c$ for each $b,$ and $10$ ways to choose $b$ since $b$ can be any digit. The final answer is then $4 \cdot 10 = \boxed{040}$

Answer:  040
Problem: Find the number of positive integers with three not necessarily distinct digits, $abc$ , with $a \neq 0$ and $c \neq 0$ such that both $abc$ and $cba$ are multiples of $4$

Solution: A number is divisible by four if its last two digits are divisible by 4. Thus, we require that $10b + a$ and $10b + c$ are both divisible by $4$ . If $b$ is odd, then $a$ and $c$ must both be $2 \pmod 4$ meaning that $a$ and $c$ are $2$ or $6$ . If $b$ is even, then $a$ and $c$ must be $0 \pmod 4$ meaning that $a$ and $c$ are $4$ or $8$ . For each choice of $b$ there are $2$ choices for $a$ and $2$ for $c$ for a total of $10 \cdot 2 \cdot 2 = \boxed{040}$ numbers.

Answer:  040
Problem: Find the number of positive integers with three not necessarily distinct digits, $abc$ , with $a \neq 0$ and $c \neq 0$ such that both $abc$ and $cba$ are multiples of $4$

Solution: For this number to fit the requirements $bc$ and $ba$ must be divisible by 4. So $bc = 00, 04, 08, 12, 16, ... , 92, 96$ and so must $ba$ for each two digits of $bc$ . There are two possibilities for $ba$ if $b$ is odd and three possibilities if $b$ is even. So there are $2^{2} \cdot 5+3^{2} \cdot 4 = 65$ possibilities but this overcounts when $a$ or $c = 0$ . So when $bc = 00, 20, 40, 60, 80$ and the corresponding $ba$ should be removed, so $65 - 5 \cdot 3 = 50$ . But we are still overcounting when $b$ is even because then $a$ can be 0. So the answer is $50 - 10 \cdot 1 = \boxed{040}$

Answer:  040
Problem: The terms of an arithmetic sequence add to $715$ . The first term of the sequence is increased by $1$ , the second term is increased by $3$ , the third term is increased by $5$ , and in general, the $k$ th term is increased by the $k$ th odd positive integer. The terms of the new sequence add to $836$ . Find the sum of the first, last, and middle terms of the original sequence.

Solution: If the sum of the original sequence is $\sum_{i=1}^{n} a_i$ then the sum of the new sequence can be expressed as $\sum_{i=1}^{n} a_i + (2i - 1) = n^2 + \sum_{i=1}^{n} a_i.$ Therefore, $836 = n^2 + 715 \rightarrow n=11.$ Now the middle term of the original sequence is simply the average of all the terms, or $\frac{715}{11} = 65,$ and the first and last terms average to this middle term, so the desired sum is simply three times the middle term, or $\boxed{195}.$

Answer:  195
Problem: The terms of an arithmetic sequence add to $715$ . The first term of the sequence is increased by $1$ , the second term is increased by $3$ , the third term is increased by $5$ , and in general, the $k$ th term is increased by the $k$ th odd positive integer. The terms of the new sequence add to $836$ . Find the sum of the first, last, and middle terms of the original sequence.

Solution: After the adding of the odd numbers, the total of the sequence increases by $836 - 715 = 121 = 11^2$ . Since the sum of the first $n$ positive odd numbers is $n^2$ , there must be $11$ terms in the sequence, so the mean of the sequence is $\dfrac{715}{11} = 65$ . Since the first, last, and middle terms are centered around the mean, our final answer is $65 \cdot 3 = \boxed{195}$

Answer:  195
Problem: The terms of an arithmetic sequence add to $715$ . The first term of the sequence is increased by $1$ , the second term is increased by $3$ , the third term is increased by $5$ , and in general, the $k$ th term is increased by the $k$ th odd positive integer. The terms of the new sequence add to $836$ . Find the sum of the first, last, and middle terms of the original sequence.

Solution: Proceed as in Solution 2 until it is noted that there are 11 terms in the sequence. Since the sum of the terms of the original arithmetic sequence is 715, we note that $\frac{2a_1 + 10d}{2} \cdot 11 = 715$ or $2a_1 + 10d = 130$ for all sets of first terms and common differences that fit the conditions given in the problem. Assume WLOG that $a_1 = 60$ and $d = 1$ . Then the first term of the corresponding arithmetic sequence will be $60$ , the sixth (middle) term will be $65$ , and the eleventh (largest) term will be $70$ . Thus, our final answer is $60 + 65 + 70 = \boxed{195}$

Answer:  195
Problem: Nine people sit down for dinner where there are three choices of meals. Three people order the beef meal, three order the chicken meal, and three order the fish meal. The waiter serves the nine meals in random order. Find the number of ways in which the waiter could serve the meal types to the nine people so that exactly one person receives the type of meal ordered by that person.

Solution: Call a beef meal $B,$ a chicken meal $C,$ and a fish meal $F.$ Now say the nine people order meals $\text{BBBCCCFFF}$ respectively and say that the person who receives the correct meal is the first person. We will solve for this case and then multiply by $9$ to account for the $9$ different ways in which the person to receive the correct meal could be picked. Note, this implies that the dishes are indistinguishable, though the people aren't. For example, two people who order chicken are separate, though if they receive fish, there is only 1 way to order them.
The problem we must solve is to distribute meals $\text{BBCCCFFF}$ to orders $\text{BBCCCFFF}$ with no matches. The two people who ordered $B$ 's can either both get $C$ 's, both get $F$ 's, or get one $C$ and one $F.$ We proceed with casework.
Summing across the cases we see there are $24$ possibilities, so the answer is $9 \cdot 24 = \boxed{216.}$

Answer:  216.
Problem: Nine people sit down for dinner where there are three choices of meals. Three people order the beef meal, three order the chicken meal, and three order the fish meal. The waiter serves the nine meals in random order. Find the number of ways in which the waiter could serve the meal types to the nine people so that exactly one person receives the type of meal ordered by that person.

Solution: We only need to figure out the number of ways to order the string $BBBCCCFFF$ , where exactly one $B$ is in the first three positions, one $C$ is in the $4^{th}$ to $6^{th}$ positions, and one $F$ is in the last three positions. There are $3^3=27$ ways to place the first $3$ meals. Then for the other two people, there are $2$ ways to serve their meals. Thus, there are $(3\cdot2)^3=\boxed{216}$ ways to serve their meals.

Answer:  216
Problem: Butch and Sundance need to get out of Dodge. To travel as quickly as possible, each alternates walking and riding their only horse, Sparky, as follows. Butch begins by walking while Sundance rides. When Sundance reaches the first of the hitching posts that are conveniently located at one-mile intervals along their route, he ties Sparky to the post and begins walking. When Butch reaches Sparky, he rides until he passes Sundance, then leaves Sparky at the next hitching post and resumes walking, and they continue in this manner. Sparky, Butch, and Sundance walk at $6,$ $4,$ and $2.5$ miles per hour, respectively. The first time Butch and Sundance meet at a milepost, they are $n$ miles from Dodge, and they have been traveling for $t$ minutes. Find $n + t$

Solution: When they meet at the milepost, Sparky has been ridden for $n$ miles total. Assume Butch rides Sparky for $a$ miles, and Sundance rides for $n-a$ miles. Thus, we can set up an equation, given that Sparky takes $\frac{1}{6}$ hours per mile, Butch takes $\frac{1}{4}$ hours per mile, and Sundance takes $\frac{2}{5}$ hours per mile:
\[\frac{a}{6} + \frac{n-a}{4} = \frac{n-a}{6} + \frac{2a}{5} \rightarrow  a = \frac{5}{19}n.\]
The smallest possible integral value of $n$ is $19$ , so we plug in $n = 19$ and $a = 5$ and get $t = \frac{13}{3}$ hours, or $260$ minutes. So our answer is $19 + 260 = \boxed{279}$

Answer:  279
Problem: Let $B$ be the set of all binary integers that can be written using exactly $5$ zeros and $8$ ones where leading zeros are allowed. If all possible subtractions are performed in which one element of $B$ is subtracted from another, find the number of times the answer $1$ is obtained.

Solution: When $1$ is subtracted from a binary number, the number of digits will remain constant if and only if the original number ended in $10.$ Therefore, every subtraction involving two numbers from $B$ will necessarily involve exactly one number ending in $10.$ To solve the problem, then, we can simply count the instances of such numbers. With the $10$ in place, the seven remaining $1$ 's can be distributed in any of the remaining $11$ spaces, so the answer is ${11 \choose 7} = \boxed{330}$

Answer:  330
Problem: The complex numbers $z$ and $w$ satisfy $z^{13} = w,$ $w^{11} = z,$ and the imaginary part of $z$ is $\sin{\frac{m\pi}{n}}$ , for relatively prime positive integers $m$ and $n$ with $m<n.$ Find $n.$

Solution: Substituting the first equation into the second, we find that $(z^{13})^{11} = z$ and thus $z^{143} = z.$ We know that $z \neq 0,$ because we are given the imaginary part of $z,$ so we can divide by $z$ to get $z^{142} = 1.$ So, $z$ must be a $142$ nd root of unity, and thus, by De Moivre's theorem, the imaginary part of $z$ will be of the form $\sin{\frac{2k\pi}{142}} = \sin{\frac{k\pi}{71}},$ where $k \in \{1, 2, \ldots, 70\}.$ Note that $71$ is prime and $k<71$ by the conditions of the problem, so the denominator in the argument of this value will always be $71.$ Thus, $n = \boxed{071}.$

Answer:  071
Problem: At each of the sixteen circles in the network below stands a student. A total of $3360$ coins are distributed among the sixteen students. All at once, all students give away all their coins by passing an equal number of coins to each of their neighbors in the network. After the trade, all students have the same number of coins as they started with. Find the number of coins the student standing at the center circle had originally.

Solution: Say the student in the center starts out with $a$ coins, the students neighboring the center student each start with $b$ coins, and all other students start out with $c$ coins. Then the $a$ -coin student has five neighbors, all the $b$ -coin students have three neighbors, and all the $c$ -coin students have four neighbors.
Now in order for each student's number of coins to remain equal after the trade, the number of coins given by each student must be equal to the number received, and thus
\begin{align*} a &= 5 \cdot \frac{b}{3}\\ b &= \frac{a}{5} + 2 \cdot \frac{c}{4}\\ c &= 2 \cdot \frac{c}{4} + 2 \cdot \frac{b}{3}. \end{align*}
Solving these equations, we see that $\frac{a}{5} = \frac{b}{3} = \frac{c}{4}.$ Also, the total number of coins is $a + 5b + 10c = 3360,$ so $a + 5 \cdot \frac{3a}{5} + 10 \cdot \frac{4a}{5} = 3360 \rightarrow a = \frac{3360}{12} = \boxed{280}.$

Answer:  280
Problem: At each of the sixteen circles in the network below stands a student. A total of $3360$ coins are distributed among the sixteen students. All at once, all students give away all their coins by passing an equal number of coins to each of their neighbors in the network. After the trade, all students have the same number of coins as they started with. Find the number of coins the student standing at the center circle had originally.

Solution: Since the students give the same number of gifts of coins as they receive and still end up the same number of coins, we can assume that every gift of coins has the same number of coins. Let $x$ be the number of coins in each gift of coins. There $10$ people who give $4$ gifts of coins, $5$ people who give $3$ gifts of coins, and $1$ person who gives $5$ gifts of coins. Thus,
\begin{align*} 10(4x)+5(3x)+5x &= 3360\\ 40x+15x+5x &= 3360\\ 60x &= 3360\\ x &= 56 \end{align*} Therefore the answer is $5(56) = \boxed{280}.$

Answer:  280
Problem: At each of the sixteen circles in the network below stands a student. A total of $3360$ coins are distributed among the sixteen students. All at once, all students give away all their coins by passing an equal number of coins to each of their neighbors in the network. After the trade, all students have the same number of coins as they started with. Find the number of coins the student standing at the center circle had originally.

Solution: Mark the number of coins from inside to outside as $a$ $b_1$ $b_2$ $b_3$ $b_4$ $b_5$ $c_1$ $c_2$ $c_3$ $c_4$ $c_5$ $d_1$ $d_2$ $d_3$ $d_4$ $d_5$ .
Then, we obtain \begin{align*} d_1 &= \frac{d_5}{4} + \frac{d_2}{4} + \frac{c_1}{4} + \frac{c_2}{4}\\ d_2 &= \frac{d_1}{4} + \frac{d_3}{4} + \frac{c_2}{4} + \frac{c_3}{4}\\ d_3 &= \frac{d_2}{4} + \frac{d_4}{4} + \frac{c_3}{4} + \frac{c_4}{4}\\ d_4 &= \frac{d_3}{4} + \frac{d_5}{4} + \frac{c_4}{4} + \frac{c_5}{4}\\ d_5 &= \frac{d_4}{4} + \frac{d_1}{4} + \frac{c_5}{4} + \frac{c_1}{4}\\ \end{align*} Letting $D = d_1 + d_2 + d_3 + d_4 + d_5$ $C = c_1 + c_2 + c_3 + c_4 + c_5$ gets us $D = \frac{D}{4} + \frac{D}{4} + \frac{C}{4} + \frac{C}{4}$ and $D = C$ .
In the same way, $C = \frac{D}{4} + \frac{D}{4} + \frac{B}{3} + \frac{B}{3}$ $B = \frac{3D}{4}$ $B = \frac{C}{4} + \frac{C}{4} + a$ $a = \frac{D}{4}$ .
Then, with $a + B + C + D = 3360$ $D = 1120$ $a = \boxed{280}$

Answer:  280
Problem: At each of the sixteen circles in the network below stands a student. A total of $3360$ coins are distributed among the sixteen students. All at once, all students give away all their coins by passing an equal number of coins to each of their neighbors in the network. After the trade, all students have the same number of coins as they started with. Find the number of coins the student standing at the center circle had originally.

Solution: Define $x$ as the number of coins the student in the middle has. Since this student connects to $5$ other students, each of those students must have passed $\dfrac15 x$ coins to the center to maintain the same number of coins.
Each of these students connect to $3$ other students, passing $\dfrac15 x$ coins to each, so they must have $\dfrac35 x$ coins. These students must then recieve $\dfrac35 x$ coins, $\dfrac 15 x$ of which were given to by the center student. Thus, they must also have received $\dfrac25 x$ coins from the outer layer, and since this figure has symmetry, these must be the same.
Each of the next layer of students must have given $\dfrac15 x$ coins. Since they gave $4$ people coins, they must have started with $\dfrac45 x$ coins. They received $\dfrac25 x$ of them from the inner layer, making the two other connections have given them the same amount. By a similar argument, they recieved $\dfrac15 x$ from each of them.
By a similar argument, the outermost hexagon of students must have had $\dfrac45 x$ coins each. Summing this all up, we get the number of total coins passed out as a function of $x$ . This ends up to be \begin{align*}  \dfrac45 x + \dfrac45 x + \dfrac45 x + \dfrac45 x + \dfrac45 x + \dfrac45 x + \dfrac45 x + \dfrac45 x + \dfrac45 x + \dfrac45 x + \dfrac35 x + \dfrac35 x + \dfrac35 x + \dfrac35 x + \dfrac35 x + x &= 10 \cdot \dfrac45 x  + 5 \cdot \dfrac35 x + x \\  &= 8x + 3x + x \\  &= 12 x.  \end{align*}
Since this all sums up to $3360$ , which is given, we find that \begin{align*} 12x &= 3360 \\ x &= 280 \end{align*}
We defined $x$ to be the number of coins that the center person has, so the answer is $x$ , which is $\boxed{280}$

Answer:  280
Problem: Cube $ABCDEFGH,$ labeled as shown below, has edge length $1$ and is cut by a plane passing through vertex $D$ and the midpoints $M$ and $N$ of $\overline{AB}$ and $\overline{CG}$ respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form $\tfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$

Solution: Define a coordinate system with $D$ at the origin and $C,$ $A,$ and $H$ on the $x$ $y$ , and $z$ axes respectively. Then $D=(0,0,0),$ $M=(.5,1,0),$ and $N=(1,0,.5).$ It follows that the plane going through $D,$ $M,$ and $N$ has equation $2x-y-4z=0.$ Let $Q = (1,1,.25)$ be the intersection of this plane and edge $BF$ and let $P = (1,2,0).$ Now since $2(1) - 1(2) - 4(0) = 0,$ $P$ is on the plane. Also, $P$ lies on the extensions of segments $DM,$ $NQ,$ and $CB$ so the solid $DPCN = DMBCQN + MPBQ$ is a right triangular pyramid. Note also that pyramid $MPBQ$ is similar to $DPCN$ with scale factor $.5$ and thus the volume of solid $DMBCQN,$ which is one of the solids bounded by the cube and the plane, is $[DPCN] - [MPBQ] = [DPCN] - \left(\frac{1}{2}\right)^3[DPCN] = \frac{7}{8}[DPCN].$ But the volume of $DPCN$ is simply the volume of a pyramid with base $1$ and height $.5$ which is $\frac{1}{3} \cdot 1 \cdot .5 = \frac{1}{6}.$ So $[DMBCQN] = \frac{7}{8} \cdot \frac{1}{6} = \frac{7}{48}.$ Note, however, that this volume is less than $.5$ and thus this solid is the smaller of the two solids. The desired volume is then $[ABCDEFGH] - [DMBCQN] = 1 - \frac{7}{48} = \frac{41}{48} \rightarrow p+q = \boxed{089.}$

Answer:  089.