Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, and 8 exactly once?
$\textbf{(A) }48000\qquad\textbf{(B) }49999.5\qquad\textbf{(C) }53332.8\qquad\textbf{(D) }55555\qquad\textbf{(E) }56432.8$

Solution: We can first solve for the mean for the digits $1, 3, 5, 7,$ and $9$ since each is $2$ away from each other. The mean of the numbers than can be solved using these digits is $55555$ . The total amount of numbers that can be formed using these digits is $5! =120$ . The sum of these numbers is $55555(120) = 6666600$ . Now we can find out the total value that was gained by replacing the $8$ with a $9$ . We can start how be calculating the gain when the $8$ was in the ones digit. Since there are $4! = 24$ numbers with the $8$ in the ones digit and $1$ was gain from each of them, $24$ is the number gained. Then, we repeat this with the tens, hundreds, thousands, and ten thousands place, leading to a total of $24+240+2400+24000+240000=266664$ as the total amount that was gained. Subtract this amount from the sum of the digits using the $9$ instead of the $8$ to get $6666600-266664=6399936$ . Finally, we divide this by $120$ to get the average. $\frac{6399936}{120}= \boxed{53332.8}$

Answer: C 53332.8
Problem: What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, and 8 exactly once?
$\textbf{(A) }48000\qquad\textbf{(B) }49999.5\qquad\textbf{(C) }53332.8\qquad\textbf{(D) }55555\qquad\textbf{(E) }56432.8$

Solution: The average value of the digits is $(1 + 3 + 5 + 7 + 8)/5 = 4.8$ .
Values occur in every place so $4.8 * 11111 = \boxed{53332.8}$

Answer: C 53332.8
Problem: Forty slips are placed into a hat, each bearing a number $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ , or $10$ , with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let $p$ be the probability that all four slips bear the same number. Let $q$ be the probability that two of the slips bear a number $a$ and the other two bear a number $b \neq a$ . What is the value of $q/p$
$\textbf{(A) } 162 \qquad \textbf{(B) } 180 \qquad \textbf{(C) } 324 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 720$

Solution: There are $10$ ways to determine which number to pick. There are $4!$ ways to then draw those four slips with that number, and $40 \cdot 39 \cdot 38 \cdot 37$ total ways to draw four slips. Thus $p = \frac{10\cdot 4!}{40 \cdot 39 \cdot 38 \cdot 37}$
There are ${10 \choose 2} = 45$ ways to determine which two numbers to pick for the second probability. There are ${4 \choose 2} = 6$ ways to arrange the order which we draw the non-equal slips, and in each order there are $4 \times 3 \times 4 \times 3$ ways to pick the slips, so $q = \frac{45 \cdot 6 \cdot 4^2 \cdot 3^2}{40 \cdot 39 \cdot 38 \cdot 37}$
Hence, the answer is $\frac{q}{p} = \frac{45 \cdot 6 \cdot 4^2 \cdot 3^2}{10\cdot 4!} = \boxed{162}$

Answer: A 162
Problem: Forty slips are placed into a hat, each bearing a number $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ , or $10$ , with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let $p$ be the probability that all four slips bear the same number. Let $q$ be the probability that two of the slips bear a number $a$ and the other two bear a number $b \neq a$ . What is the value of $q/p$
$\textbf{(A) } 162 \qquad \textbf{(B) } 180 \qquad \textbf{(C) } 324 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 720$

Solution: For probability $p$ , there are $\binom{10}{1}=10$ ways to choose the number you want to show up $4$ times.
Hence, the probability is $\frac{10}{\binom{40}{4}}$
For probability $q$ , there are $\binom{10}{2}=45$ ways to choose the $2$ numbers you want to show up twice. There are $\binom{4}{2}\cdot\binom{4}{2}$ ways to pick which slips you want out of the $4$ of each.
Hence, the probability is $\frac{45\cdot6\cdot6}{\binom{40}{4}}$
Hence, $\frac{q}{p}=\frac{45\cdot6\cdot6}{10}=9\cdot6\cdot3=\boxed{162}$

Answer: A 162
Problem: For how many positive integers $n$ less than or equal to $24$ is $n!$ evenly divisible by $1 + 2 + \cdots + n?$
$\textbf{(A) } 8 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 21$

Solution: Since $1 + 2 + \cdots + n = \frac{n(n+1)}{2}$ , the condition is equivalent to having an integer value for $\frac{n!} {\frac{n(n+1)}{2}}$ . This reduces, when $n\ge 1$ , to having an integer value for $\frac{2(n-1)!}{n+1}$ . This fraction is an 
integer unless $n+1$ is an odd prime. There are $8$ odd primes less than or equal to $24$ , so there
are $24 - 8 =  \boxed{16}$ numbers less than or equal to $24$ that satisfy the condition.

Answer: C 16
Problem: In trapezoid $ABCD$ we have $\overline{AB}$ parallel to $\overline{DC}$ $E$ as the midpoint of $\overline{BC}$ , and $F$ as the midpoint of $\overline{DA}$ . The area of $ABEF$ is twice the area of $FECD$ . What is $AB/DC$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 8$

Solution: Since the heights of both trapezoids are equal, and the area of $ABEF$ is twice the area of $FECD$
$\frac{AB+EF}{2}=2\left(\frac{DC+EF}{2}\right)$
$\frac{AB+EF}{2}=DC+EF$ , so
$AB+EF=2DC+2EF$
$EF$ is exactly halfway between $AB$ and $DC$ , so $EF=\frac{AB+DC}{2}$
$AB+\frac{AB+DC}{2}=2DC+AB+DC$ , so
$\frac{3}{2}AB+\frac{1}{2}DC=3DC+AB$ , and
$\frac{1}{2}AB=\frac{5}{2}DC$
$\frac{AB}{DC} = \boxed{5}$

Answer: C 5
Problem: In trapezoid $ABCD$ we have $\overline{AB}$ parallel to $\overline{DC}$ $E$ as the midpoint of $\overline{BC}$ , and $F$ as the midpoint of $\overline{DA}$ . The area of $ABEF$ is twice the area of $FECD$ . What is $AB/DC$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 8$

Solution: Mark $DC=z$ $AB=x$ , and $FE=y.$ Note that the heights of trapezoids $ABEF$ $FECD$ are the same. Mark the height to be $h$
Then, we have that $\tfrac{x+y}{2}\cdot h=2(\tfrac{y+z}{2} \cdot h)$
From this, we get that $x=2z+y$
We also get that $\tfrac{x+z}{2} \cdot 2h= 3(\tfrac{y+z}{2} \cdot h)$
Simplifying, we get that $2x=z+3y$
Notice that we want $\tfrac{AB}{DC}=\tfrac{x}{z}$
Dividing the first equation by $z$ , we get that $\tfrac{x}{z}=2+\tfrac{y}{z}\implies 3(\tfrac{x}{z})=6+3(\tfrac{y}{z})$
Dividing the second equation by $z$ , we get that $2(\tfrac{x}{z})=1+3(\tfrac{y}{z})$
Now, when we subtract the top equation from the bottom, we get that $\tfrac{x}{z}=\boxed{5}$

Answer: C 5
Problem: In trapezoid $ABCD$ we have $\overline{AB}$ parallel to $\overline{DC}$ $E$ as the midpoint of $\overline{BC}$ , and $F$ as the midpoint of $\overline{DA}$ . The area of $ABEF$ is twice the area of $FECD$ . What is $AB/DC$
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 8$

Solution: Since the bases of the trapezoids along with the height are the same, the only thing that matters is the second base. Denote the length of the bigger trapezoid $x$ . The area of the smaller trapezoid is $A$ $h \cdot \frac {b_1 + b_2}{2}$ . The area of the larger trapezoid is $A$ $h \cdot \frac {b_1 + x}{2}$ . Since this problem asks for proportions, assume that $b_1 = 1$ and $b_2 = 2$
The smaller trapezoid has area $h$ while the larger trapezoid must have area $5h$ . We have the equation $\frac {x}{2} = 5$ $x$ = 10, and our answer is $\boxed{5}$

Answer: C 5
Problem: Let $x$ and $y$ be two-digit integers such that $y$ is obtained by reversing the digits
of $x$ . The integers $x$ and $y$ satisfy $x^2 - y^2 = m^2$ for some positive integer $m$ .
What is $x + y + m$
$\textbf{(A) } 88 \qquad \textbf{(B) } 112 \qquad \textbf{(C) } 116 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 154$

Solution: Let $x = 10a+b, y = 10b+a$ . The given conditions imply $x>y$ , which implies $a>b$ , and they also imply that both $a$ and $b$ are nonzero.
Then, $x^2 - y^2 = (x-y)(x+y) = (9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2$
Since this must be a perfect square, all the exponents in its prime factorization must be even. $99$ factorizes into $3^2 \cdot 11$ , so $11|(a-b)(a+b)$ . However, the maximum value of $a-b$ is $9-1=8$ , so $11|a+b$ . The maximum value of $a+b$ is $9+8=17$ , so $a+b=11$
Then, we have $33^2(a-b) = m^2$ , so $a-b$ is a perfect square, but the only perfect squares that are within our bound on $a-b$ are $1$ and $4$ . We know $a+b=11$ , and, for $a-b=1$ , adding equations to eliminate $b$ gives us $2a=12 \Longrightarrow a=6, b=5$ . Testing $a-b=4$ gives us $2a=15 \Longrightarrow a=\frac{15}{2}, b=\frac{7}{2}$ , which is impossible, as $a$ and $b$ must be digits. Therefore, $(a,b) = (6,5)$ , and $x+y+m=\boxed{154}$

Answer: E 154
Problem: Let $x$ and $y$ be two-digit integers such that $y$ is obtained by reversing the digits
of $x$ . The integers $x$ and $y$ satisfy $x^2 - y^2 = m^2$ for some positive integer $m$ .
What is $x + y + m$
$\textbf{(A) } 88 \qquad \textbf{(B) } 112 \qquad \textbf{(C) } 116 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 154$

Solution: The first steps are the same as Solution 1.  Let $x = 10a+b, y = 10b+a$ , where we know that a and b are digits (whole numbers less than $10$ ).
Like Solution 1, we end up getting $(9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2$ .  This is where the solution diverges.
We know that the left side of the equation is a perfect square because $m$ is an integer.  If we factor $99$ into its prime factors, we get $3^2\cdot 11$ .  In order to get a perfect square on the left side, $(a-b)(a+b)$ must make both prime exponents even.  Because the a and b are digits, a simple guess would be that $(a+b)$ (the bigger number) equals $11$ while $(a-b)$ is a factor of nine (1 or 9).  The correct guesses are $a = 6, b = 5$ causing $x = 65, y = 56,$ and $m = 33$ .  The sum of the numbers is $\boxed{154}$

Answer: E 154
Problem: Let $x$ and $y$ be two-digit integers such that $y$ is obtained by reversing the digits
of $x$ . The integers $x$ and $y$ satisfy $x^2 - y^2 = m^2$ for some positive integer $m$ .
What is $x + y + m$
$\textbf{(A) } 88 \qquad \textbf{(B) } 112 \qquad \textbf{(C) } 116 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 154$

Solution: Once again, the solution is quite similar as the above solutions. Since $x$ and $y$ are two digit integers, we can write $x = 10a+b, y = 10b+a$ and because $x^2 - y^2 = (x-y)(x+y)$ , substituting and factoring, we get $99(a+b)(a-b) = m^2$
Therefore, $(a+b)(a-b) = \frac{m^2}{99}$ and $\frac{m^2}{99}$ must be an integer. A quick strategy is to find the smallest such integer $m$ such that $\frac{m^2}{99}$ is an integer. We notice that 99 has a prime factorization of $3^2 \cdot 11.$
Let $m^2 = n.$ Since we need a perfect square and 3 is already squared, we just need to square 11. So $3^2 \cdot 11^2$ gives us 1089 as $n$ and $m = \sqrt{1089} = 33.$ We now get the equation $(x-y)(x+y) = 33^2$ , which we can also write as $(x-y)(x+y) = 11^2 \cdot 3^2$
A very simple guess assumes that $x-y=3^2$ and $x+y=11^2$ since $x$ and $y$ are positive. Finally, we come to the conclusion that $x=65$ and $y=56$ , so $x+y+m$ $=$ $\boxed{154}$

Answer: E 154
Problem: Let $x$ and $y$ be two-digit integers such that $y$ is obtained by reversing the digits
of $x$ . The integers $x$ and $y$ satisfy $x^2 - y^2 = m^2$ for some positive integer $m$ .
What is $x + y + m$
$\textbf{(A) } 88 \qquad \textbf{(B) } 112 \qquad \textbf{(C) } 116 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 154$

Solution: Continue the same as Solution $3$ until we get $33$ . Knowing that $33^2 = 1089$ , we have narrowed down our Pythagorean triples. We know that the $2$ other squares should be larger than $33^2$ , so we can start testing.
If we start testing the $40$ s, it is fruitless since the closest to $33^2$ would be $33 - 45 - 54$ which is not a Pythagorean triple. We can start by testing out the $50$ s, and it turns our that $33 - 56 - 65$ is a Pythagorean triple. Therefore, our answer is $33+56+65$ $\boxed{154}$

Answer: E 154
Problem: subset $B$ of the set of integers from $1$ to $100$ , inclusive, has the property that no two elements of $B$ sum to $125$ . What is the maximum possible number of elements in $B$
$\textbf{(A) } 50 \qquad \textbf{(B) } 51 \qquad \textbf{(C) } 62 \qquad \textbf{(D) } 65 \qquad \textbf{(E) } 68$

Solution: The question asks for the maximum possible number of elements. The integers from $1$ to $24$ can be included because you cannot make $125$ with integers from $1$ to $24$ without the other number being greater than $100$ . The integers from $25$ to $100$ are left. They can be paired so the sum is $125$ $25+100$ $26+99$ $27+98$ $\ldots$ $62+63$ . That is $38$ pairs, and at most one number from each pair can be included in the set. The total is $24 + 38 = \boxed{62}$

Answer: C 62
Problem: subset $B$ of the set of integers from $1$ to $100$ , inclusive, has the property that no two elements of $B$ sum to $125$ . What is the maximum possible number of elements in $B$
$\textbf{(A) } 50 \qquad \textbf{(B) } 51 \qquad \textbf{(C) } 62 \qquad \textbf{(D) } 65 \qquad \textbf{(E) } 68$

Solution: "Cut" $125$ into half. The maximum integer value in the smaller half is $62$ . Thus the answer is $\boxed{62}$

Answer: C 62
Problem: subset $B$ of the set of integers from $1$ to $100$ , inclusive, has the property that no two elements of $B$ sum to $125$ . What is the maximum possible number of elements in $B$
$\textbf{(A) } 50 \qquad \textbf{(B) } 51 \qquad \textbf{(C) } 62 \qquad \textbf{(D) } 65 \qquad \textbf{(E) } 68$

Solution: The maximum possible number of elements includes the smallest numbers. So, subset $B = \{1,2,3....n-1,n\}$ where n is the maximum number of elements in subset $B$ . So, we have to find two consecutive numbers, $n$ and $n+1$ , whose sum is $125$ . Setting up our equation, we have $n+(n+1) = 2n+1 = 125$ . When we solve for $n$ , we get $n =\boxed{62}$

Answer: C 62
Problem: subset $B$ of the set of integers from $1$ to $100$ , inclusive, has the property that no two elements of $B$ sum to $125$ . What is the maximum possible number of elements in $B$
$\textbf{(A) } 50 \qquad \textbf{(B) } 51 \qquad \textbf{(C) } 62 \qquad \textbf{(D) } 65 \qquad \textbf{(E) } 68$

Solution: We can put all odd numbers into subset B, or we can put all even numbers into subset B, so now there are 50 numbers in the set. I will use all even numbers in this solution. Now, we need to add other odd(or even!) numbers possible in this subset, which is all odd(or even) numbers that can be added so that the sum with 100(or 99) plus the biggest possible odd number(or even) to get 123. This will get us the numbers 1,3,5...21,23(or numbers 2,4,6...22,24), which gives us 12 more numbers, and adding that to the 50 original numbers, we get $B =\boxed{62}$

Answer: C 62
Problem: You and five friends need to raise $1500$ dollars in donations for a charity, dividing the fundraising equally.  How many dollars will each of you need to raise?
$\mathrm{(A) \ } 250\qquad \mathrm{(B) \ } 300 \qquad \mathrm{(C) \ } 1500 \qquad \mathrm{(D) \ } 7500 \qquad \mathrm{(E) \ } 9000$

Solution: There are $6$ people to split the $1500$ dollars among, so each person must raise $\frac{1500}6=250$ dollars. $\Rightarrow\boxed{250}$

Answer: A 250
Problem: Alicia earns 20 dollars per hour, of which $1.45\%$ is deducted to pay local taxes.  How many cents per hour of Alicia's wages are used to pay local taxes?
$\mathrm{(A) \ } 0.0029 \qquad \mathrm{(B) \ } 0.029 \qquad \mathrm{(C) \ } 0.29 \qquad \mathrm{(D) \ } 2.9 \qquad \mathrm{(E) \ } 29$

Solution: $20$ dollars is the same as $2000$ cents, and $1.45\%$ of $2000$ is $0.0145\times2000=29$ cents. $\Rightarrow\boxed{29}$

Answer: E 29
Problem: Alicia earns 20 dollars per hour, of which $1.45\%$ is deducted to pay local taxes.  How many cents per hour of Alicia's wages are used to pay local taxes?
$\mathrm{(A) \ } 0.0029 \qquad \mathrm{(B) \ } 0.029 \qquad \mathrm{(C) \ } 0.29 \qquad \mathrm{(D) \ } 2.9 \qquad \mathrm{(E) \ } 29$

Solution: Since there can't be decimal values of cents, the answer must be $\Rightarrow\boxed{29}$

Answer: E 29
Problem: What is the value of $x$ if $|x-1|=|x-2|$
$\mathrm{(A) \ } -\frac12 \qquad \mathrm{(B) \ } \frac12 \qquad \mathrm{(C) \ } 1 \qquad \mathrm{(D) \ } \frac32 \qquad \mathrm{(E) \ } 2$

Solution: $|x-1|$ is the distance between $x$ and $1$ $|x-2|$ is the distance between $x$ and $2$
Therefore, the given equation says $x$ is equidistant from $1$ and $2$ , so $x=\frac{1+2}2=\frac{3}{2}\Rightarrow\boxed{32}$

Answer: D 32
Problem: What is the value of $x$ if $|x-1|=|x-2|$
$\mathrm{(A) \ } -\frac12 \qquad \mathrm{(B) \ } \frac12 \qquad \mathrm{(C) \ } 1 \qquad \mathrm{(D) \ } \frac32 \qquad \mathrm{(E) \ } 2$

Solution: We know that either $x-1=x-2$ or $x-1=-(x-2)$ . The first equation simplifies to $-1=-2$ , which is false, so $x-1=-(x-2)$ . Solving, we get $x=\frac{3}{2}\Rightarrow\boxed{32}$

Answer: D 32
Problem: Bertha has 6 daughters and no sons.  Some of her daughters have 6 daughters, and the rest have none.  Bertha has a total of 30 daughters and granddaughters, and no great-granddaughters.  How many of Bertha's daughters and grand-daughters have no daughters?
$\mathrm{(A) \ } 22 \qquad \mathrm{(B) \ } 23 \qquad \mathrm{(C) \ } 24 \qquad \mathrm{(D) \ } 25 \qquad \mathrm{(E) \ } 26$

Solution: Since Bertha has $6$ daughters, she has $30-6=24$ granddaughters, of which none have daughters.  Of Bertha's daughters, $\frac{24}6=4$ have daughters, so $6-4=2$ do not have daughters. Therefore, of Bertha's daughters and granddaughters, $24+2=26$ do not have daughters. $\boxed{26}$

Answer: E 26
Problem: Bertha has 6 daughters and no sons.  Some of her daughters have 6 daughters, and the rest have none.  Bertha has a total of 30 daughters and granddaughters, and no great-granddaughters.  How many of Bertha's daughters and grand-daughters have no daughters?
$\mathrm{(A) \ } 22 \qquad \mathrm{(B) \ } 23 \qquad \mathrm{(C) \ } 24 \qquad \mathrm{(D) \ } 25 \qquad \mathrm{(E) \ } 26$

Solution: Bertha has $30 - 6 = 24$ granddaughters, none of whom have any daughters. The granddaughters are the children of $24/6 = 4$ of Bertha's daughters, so the number of women having no daughters is $30 - 4 = \boxed{26}$

Answer:  26
Problem: Bertha has 6 daughters and no sons.  Some of her daughters have 6 daughters, and the rest have none.  Bertha has a total of 30 daughters and granddaughters, and no great-granddaughters.  How many of Bertha's daughters and grand-daughters have no daughters?
$\mathrm{(A) \ } 22 \qquad \mathrm{(B) \ } 23 \qquad \mathrm{(C) \ } 24 \qquad \mathrm{(D) \ } 25 \qquad \mathrm{(E) \ } 26$

Solution: Draw a tree diagram and see that the answer can be found in the sum of $6+6$ granddaughters, $5+5$ daughters, and $4$ more daughters.  Adding them together gives the answer of $\boxed{26}$

Answer: E 26
Problem: A grocer stacks oranges in a pyramid-like stack whose rectangular base is $5$ oranges by $8$ oranges.  Each orange above the first level rests in a pocket formed by four oranges below.  The stack is completed by a single row of oranges.  How many oranges are in the stack?
$\mathrm{(A) \ } 96 \qquad \mathrm{(B) \ } 98 \qquad \mathrm{(C) \ } 100 \qquad \mathrm{(D) \ } 101 \qquad \mathrm{(E) \ } 134$

Solution: There are $5\times8=40$ oranges on the $1^{\text{st}}$ layer of the stack. The $2^{\text{nd}}$ layer that is added on top of the first will be a layer of $4\times7=28$ oranges.  When the third layer is added on top of the $2^{\text{nd}}$ , it will be a layer of $3\times6=18$ oranges, etc.
Therefore, there are $5\times8+4\times7+3\times6+2\times5+1\times4=40+28+18+10+4=100$ oranges in the stack. $\boxed{100}$

Answer: C 100
Problem: A game is played with tokens according to the following rule.  In each round, the player with the most tokens gives one token to each of the other players and also places one token in the discard pile.  The game ends when some player runs out of tokens.  Players $A$ $B$ , and $C$ start with $15$ $14$ , and $13$ tokens, respectively.  How many rounds will there be in the game?
$\mathrm{(A) \ } 36 \qquad \mathrm{(B) \ } 37 \qquad \mathrm{(C) \ } 38 \qquad \mathrm{(D) \ } 39 \qquad \mathrm{(E) \ } 40$

Solution: We look at a set of three rounds, where the players begin with $x+1$ $x$ , and $x-1$ tokens.
After three rounds, there will be a net loss of $1$ token per player (they receive two tokens and lose three). Therefore, after $36$ rounds -- or $12$ three-round sets, $A,B$ and $C$ will have $3$ $2$ , and $1$ tokens, respectively.  After $1$ more round, player $A$ will give away $3$ tokens, leaving them empty-handed, and thus the game will end. We then have there are $36+1=\boxed{37}$ rounds until the game ends.

Answer: B 37
Problem: A game is played with tokens according to the following rule.  In each round, the player with the most tokens gives one token to each of the other players and also places one token in the discard pile.  The game ends when some player runs out of tokens.  Players $A$ $B$ , and $C$ start with $15$ $14$ , and $13$ tokens, respectively.  How many rounds will there be in the game?
$\mathrm{(A) \ } 36 \qquad \mathrm{(B) \ } 37 \qquad \mathrm{(C) \ } 38 \qquad \mathrm{(D) \ } 39 \qquad \mathrm{(E) \ } 40$

Solution: Let's bash a few rounds. The amounts are for players $1,2,$ and $3$ , respectively.
First round: $15,14,13$ (given)
Second round: $12,15,14$ Third round: $13,12,15$ Fourth round: $14,13,12$
We see that after $3$ rounds are played, we have the exact same scenario as the first round but with one token less per player. So, the sequence $1,4,7,10...$ where each of the next members are $3$ greater than the previous one corresponds with the sequence $15,14,13,12...$ where the first sequence represents the round and the second sequence represents the number of tokens player $1$ has. But we note that once player $1$ reaches $3$ coins, the game will end on his next turn as he must give away all his coins. Therefore, we want the $15-3+1=13$ th number in the sequence $1,4,7,10...$ which is $\boxed{37}$

Answer: B 37
Problem: A game is played with tokens according to the following rule.  In each round, the player with the most tokens gives one token to each of the other players and also places one token in the discard pile.  The game ends when some player runs out of tokens.  Players $A$ $B$ , and $C$ start with $15$ $14$ , and $13$ tokens, respectively.  How many rounds will there be in the game?
$\mathrm{(A) \ } 36 \qquad \mathrm{(B) \ } 37 \qquad \mathrm{(C) \ } 38 \qquad \mathrm{(D) \ } 39 \qquad \mathrm{(E) \ } 40$

Solution: Looking at a set of five rounds, you'll see $A$ has $4$ fewer tokens than in the beginning. Looking at four more rounds, you'll notice $A$ has the same amount of tokens, namely $11$ , compared to round five. If you keep doing this process, you'll see a pattern: Every four rounds, the amount of tokens $A$ has either decreased by $4$ or stayed the same compared to the previous four rounds. For example, in round nine, $A$ had $11$ tokens, in round $13$ $A$ had $11$ tokens, and in round $17$ $A$ had $7$ tokens, etc. Using this weird pattern, you can find out that in round $37$ $A$ should have $3$ tokens, but since they would have given them away in that round, the game would end on $\boxed{37}$

Answer: B 37