Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: What are the sign and units digit of the product of all the odd negative integers strictly greater than $-2015$ $\textbf{(A) }$ It is a negative number ending with a 1. $\textbf{(B) }$ It is a positive number ending with a 1. $\textbf{(C) }$ It is a negative number ending with a 5. $\textbf{(D) }$ It is a positive number ending with a 5. $\textbf{(E) }$ It is a negative number ending with a 0.

Solution: Since $-5>-2015$ , the product must end with a $5$
The multiplicands are the odd negative integers from $-1$ to $-2013$ . There are $\frac{|-2013+1|}2+1=1006+1$ of these numbers. Since $(-1)^{1007}=-1$ , the product is negative.
Therefore, the answer must be $\boxed{5.}$

Answer: C 5.
Problem: Let $a$ $b$ , and $c$ be three distinct one-digit numbers. What is the maximum value of the sum of the roots of the equation $(x-a)(x-b)+(x-b)(x-c)=0$
$\textbf{(A) }15\qquad \textbf{(B) }15.5\qquad \textbf{(C) }16\qquad \textbf{(D) }16.5\qquad \textbf{(E) }17$

Solution: Expanding the equation and combining like terms results in $2x^2-(a+2b+c)x+(ab+bc)=0$ . By Vieta's formula the sum of the roots is $\dfrac{-[-(a+2b+c)]}{2}=\dfrac{a+2b+c}{2}$ . To maximize this expression we want $b$ to be the largest, and from there we can assign the next highest values to $a$ and $c$ . So let $b=9$ $a=8$ , and $c=7$ . Then the answer is $\dfrac{8+18+7}{2}=\boxed{16.5}$

Answer: D 16.5
Problem: Let $a$ $b$ , and $c$ be three distinct one-digit numbers. What is the maximum value of the sum of the roots of the equation $(x-a)(x-b)+(x-b)(x-c)=0$
$\textbf{(A) }15\qquad \textbf{(B) }15.5\qquad \textbf{(C) }16\qquad \textbf{(D) }16.5\qquad \textbf{(E) }17$

Solution: Factoring out $(x-b)$ from the equation yields $(x-b)(2x-(a+c))=0 \Rightarrow (x-b)\left(x-\frac{a+c}{2}\right)=0$ . Therefore the roots are $b$ and $\frac{a+c}{2}$ . Because $b$ must be the larger root to maximize the sum of the roots, letting $a,b,$ and $c$ be $8,9,$ and $7$ respectively yields the sum $9+\frac{8+7}{2} = 9+7.5 = \boxed{16.5}$

Answer: D 16.5
Problem: The town of Hamlet has $3$ people for each horse, $4$ sheep for each cow, and $3$ ducks for each person. Which of the following could not possibly be the total number of people, horses, sheep, cows, and ducks in Hamlet?
$\textbf{(A) }41\qquad\textbf{(B) }47\qquad\textbf{(C) }59\qquad\textbf{(D) }61\qquad\textbf{(E) }66$

Solution: Let the amount of people be $p$ , horses be $h$ , sheep be $s$ , cows be $c$ , and ducks be $d$ . 
We know \[3h=p\] \[4c=s\] \[3p=d\] Then the total amount of people, horses, sheep, cows, and ducks may be written as $p+h+s+c+d = 3h+h+4c+c+(3\times3h)$ . This is equivalent to $13h+5c$ . Looking through the options, we see $47$ is impossible to make for integer values of $h$ and $c$ . So the answer is $\boxed{47}$

Answer: B 47
Problem: The town of Hamlet has $3$ people for each horse, $4$ sheep for each cow, and $3$ ducks for each person. Which of the following could not possibly be the total number of people, horses, sheep, cows, and ducks in Hamlet?
$\textbf{(A) }41\qquad\textbf{(B) }47\qquad\textbf{(C) }59\qquad\textbf{(D) }61\qquad\textbf{(E) }66$

Solution: As the solution above says, the total amount of people, horses, sheep, cows, and ducks may be written as $13d+5s$ . However, instead of going through each of the solutions and testing the options, you can use the Chicken McNugget Theorem to find the greatest number of people, horses, sheep, cows, and ducks that cannot be written in the form $13d+5s$ \[13\cdot 5-13-5=47,\] so our answer is $\boxed{47}$

Answer: B 47
Problem: The centers of the faces of the right rectangular prism shown below are joined to create an octahedron. What is the volume of this octahedron?
[asy] import three; size(2inch); currentprojection=orthographic(4,2,2); draw((0,0,0)--(0,0,3),dashed); draw((0,0,0)--(0,4,0),dashed); draw((0,0,0)--(5,0,0),dashed); draw((5,4,3)--(5,0,3)--(5,0,0)--(5,4,0)--(0,4,0)--(0,4,3)--(0,0,3)--(5,0,3)); draw((0,4,3)--(5,4,3)--(5,4,0)); label("3",(5,0,3)--(5,0,0),W); label("4",(5,0,0)--(5,4,0),S); label("5",(5,4,0)--(0,4,0),SE); [/asy]
$\textbf{(A) } \dfrac{75}{12} \qquad\textbf{(B) } 10 \qquad\textbf{(C) } 12 \qquad\textbf{(D) } 10\sqrt2 \qquad\textbf{(E) } 15$

Solution: The octahedron is just two congruent pyramids glued together by their base. The base of one pyramid is a rhombus with diagonals $4$ and $5$ , for an area $A=10$ . The height $h$ , of one pyramid, is $\frac{3}{2}$ , so the volume of one pyramid is $\frac{Ah}{3}=5$ . Thus, the octahedron has volume $2\cdot5=\boxed{10}$

Answer: B 10
Problem: The centers of the faces of the right rectangular prism shown below are joined to create an octahedron. What is the volume of this octahedron?
[asy] import three; size(2inch); currentprojection=orthographic(4,2,2); draw((0,0,0)--(0,0,3),dashed); draw((0,0,0)--(0,4,0),dashed); draw((0,0,0)--(5,0,0),dashed); draw((5,4,3)--(5,0,3)--(5,0,0)--(5,4,0)--(0,4,0)--(0,4,3)--(0,0,3)--(5,0,3)); draw((0,4,3)--(5,4,3)--(5,4,0)); label("3",(5,0,3)--(5,0,0),W); label("4",(5,0,0)--(5,4,0),S); label("5",(5,4,0)--(0,4,0),SE); [/asy]
$\textbf{(A) } \dfrac{75}{12} \qquad\textbf{(B) } 10 \qquad\textbf{(C) } 12 \qquad\textbf{(D) } 10\sqrt2 \qquad\textbf{(E) } 15$

Solution: The "base" of the octahedron is half the base of the rectangular prism because it is connected by the midpoints. Additionally, the volume of an octahedron is $\dfrac{1}{3}$ of its respective prism. Thus, the octahedron's volume is $\dfrac{1}{2} \cdot \dfrac{1}{3} = \dfrac{1}{6}$ of the rectangular prism's volume, meaning that the answer is $3 \cdot 4 \cdot 5 \cdot \dfrac{1}{6} = \boxed{10}$

Answer:  10
Problem: Johann has $64$ fair coins. He flips all the coins. Any coin that lands on tails is tossed again. Coins that land on tails on the second toss are tossed a third time. What is the expected number of coins that are now heads?
$\textbf{(A) } 32 \qquad\textbf{(B) } 40 \qquad\textbf{(C) } 48 \qquad\textbf{(D) } 56 \qquad\textbf{(E) } 64$

Solution: We can simplify the problem first, then apply reasoning to the original problem. Let's say that there are $8$ coins. Shaded coins flip heads, and blank coins flip tails. So, after the first flip;
[asy] filldraw(circle((-5,0),0.35),white); filldraw(circle((-4,0),0.35),white); filldraw(circle((-3,0),0.35),white); filldraw(circle((-2,0),0.35),white); filldraw(circle((-1,0),0.35),black); filldraw(circle((-0,0),0.35),black); filldraw(circle((1,0),0.35),black); filldraw(circle((2,0),0.35),black); [/asy]
Then, after the second (new heads in blue);
[asy] filldraw(circle((-5,0),0.35),white); filldraw(circle((-4,0),0.35),white); filldraw(circle((-3,0),0.35),blue); filldraw(circle((-2,0),0.35),blue); filldraw(circle((-1,0),0.35),black); filldraw(circle((-0,0),0.35),black); filldraw(circle((1,0),0.35),black); filldraw(circle((2,0),0.35),black); [/asy]
And after the third (new head in green);
[asy] filldraw(circle((-5,0),0.35),white); filldraw(circle((-4,0),0.35),green); filldraw(circle((-3,0),0.35),blue); filldraw(circle((-2,0),0.35),blue); filldraw(circle((-1,0),0.35),black); filldraw(circle((-0,0),0.35),black); filldraw(circle((1,0),0.35),black); filldraw(circle((2,0),0.35),black); [/asy]
So in total, $7$ of the $8$ coins resulted in heads. Now we have the ratio of $\frac{7}{8}$ of the total coins will end up heads. Therefore, we have $\frac{7}{8}\cdot64=\boxed{56}$

Answer: D 56
Problem: Johann has $64$ fair coins. He flips all the coins. Any coin that lands on tails is tossed again. Coins that land on tails on the second toss are tossed a third time. What is the expected number of coins that are now heads?
$\textbf{(A) } 32 \qquad\textbf{(B) } 40 \qquad\textbf{(C) } 48 \qquad\textbf{(D) } 56 \qquad\textbf{(E) } 64$

Solution: Every time the coins are flipped, half of them are expected to turn up heads. The expected number of heads on the first flip is $32$ , on the second flip is $16$ , and on the third flip, it is $8$ . Adding these gives $\boxed{56}$

Answer: D 56
Problem: Johann has $64$ fair coins. He flips all the coins. Any coin that lands on tails is tossed again. Coins that land on tails on the second toss are tossed a third time. What is the expected number of coins that are now heads?
$\textbf{(A) } 32 \qquad\textbf{(B) } 40 \qquad\textbf{(C) } 48 \qquad\textbf{(D) } 56 \qquad\textbf{(E) } 64$

Solution: Every time the coins are flipped, each of them has a $1/2$ probability of being tails. Doing this $3$ times, $1/8$ of them will be tails. $64-64*1/8=$ $\boxed{56}$

Answer: D 56
Problem: Johann has $64$ fair coins. He flips all the coins. Any coin that lands on tails is tossed again. Coins that land on tails on the second toss are tossed a third time. What is the expected number of coins that are now heads?
$\textbf{(A) } 32 \qquad\textbf{(B) } 40 \qquad\textbf{(C) } 48 \qquad\textbf{(D) } 56 \qquad\textbf{(E) } 64$

Solution: (Similar to solution 2)
Notice how:
$E(\text{Heads on 1st flip, 2nd flip, 3rd flip}) = E(\text{Heads on 1st flip}) + E(\text{Heads on 2nd flip}) + E(\text{Heads on 3rd flip})$
The expected number of heads for the first flip is simply $64 \cdot \frac{1}{2}$ , since each coin has a 1 in 2 chance of being heads. Then, we are left with $64 - 32 = 32$ coins. Then, half of these coins will be heads again, which leaves us with $32 - 16 = 16$ coins. Then, half of these coins will be heads again, which leaves us with $16 - 8 = 8$ coins.
Hence, the expected number of heads is simply:
$32 + 16 + 8 = \boxed{56} \implies D$

Answer:  56
Problem: problem_id
18bf638a0cb797d222d008b988c16b6d    Erin the ant starts at a given corner of a cub...
18bf638a0cb797d222d008b988c16b6d    From the 2006 AMC 10A Problem 25, they look fo...
Name: Text, dtype: object

Solution: [asy]import three; draw((1,1,1)--(1,0,1)--(1,0,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,1,0)--(0,1,0)--(0,1,1)); draw((0,0,1)--(1,0,1)); draw((1,0,0)--(1,1,0)); draw((0,0,0)--(0,1,0));  label("2",(0,0,0),S); label("A",(1,0,0),W); label("B",(0,0,1),N); label("1",(1,0,1),NW); label("3",(1,1,0),S); label("C",(0,1,0),E); label("D",(1,1,1),SE); label("4",(0,1,1),NE); [/asy]
We label the vertices of the cube as different letters and numbers shown above. We label these so that Erin can only crawl from a number to a letter or a letter to a number (this can be seen as a coloring argument). The starting point is labeled $A$
If we define a "move" as each time Erin crawls along a single edge from one vertex to another, we see that after 7 moves, Erin must be on a numbered vertex. Since this numbered vertex cannot be one unit away from $A$ (since Erin cannot crawl back to $A$ ), this vertex must be $4$
Therefore, we now just need to count the number of paths from $A$ to $4$ . To count this, we can work backwards. There are 3 choices for which vertex Erin was at before she moved to $4$ , and 2 choices for which vertex Erin was at 2 moves before $4$ . All of Erin's previous moves were forced, so the total number of legal paths from $A$ to $4$ is $3 \cdot 2 = \boxed{6}$

Answer: A 6
Problem: problem_id
18bf638a0cb797d222d008b988c16b6d    Erin the ant starts at a given corner of a cub...
18bf638a0cb797d222d008b988c16b6d    From the 2006 AMC 10A Problem 25, they look fo...
Name: Text, dtype: object

Solution: Lets say that this cube is an unit cube and the given corner is $(0,0,0)$ . Because Erin cannot return back to her starting point, she cannot be on $(0,0,1)$ $(0,1,0)$ , or $(1,0,0)$ . She cannot be on $(1,1,0)$ $(1,0,1)$ , or $(0,1,1)$ , because after $7$ moves, the sum of all the coordinates has to be odd. Thus, Erin has to be at $(1,1,1)$ . Now, we draw a net and see that there are $3$ choices for the first move, $2$ for the second, and the rest are forced. Thus the answer is $3\cdot2 = \boxed{6}$

Answer: A 6
Problem: problem_id
18bf638a0cb797d222d008b988c16b6d    Erin the ant starts at a given corner of a cub...
18bf638a0cb797d222d008b988c16b6d    From the 2006 AMC 10A Problem 25, they look fo...
Name: Text, dtype: object

Solution: Let's suppose the given corner on the cube is $(0,0,0)$ . Erin has 3 identical ways to proceed. Suppose she goes to $(0,1,0)$ . She now has two more identical ways to go. Let's say she goes to $(0,1,1)$ . She has to go to $(0,0,1)$ , otherwise, she will end up on that point after 7 moves. This is because once if she chooses the other path to $(1,1,1)$ , the endpoint is certain to be $(0,0,1)$ , which is directly connected to $(0,0,0)$ . After she goes to $(0,0,1)$ , the only option she has is to go to $(1,0,1)$ , then $(1,0,0)$ , after that $(1,1,0)$ , and finally $(1,1,1)$ . She is forced to go this way because she cannot end up on $(1,0,0)$ . At the start, she had 3 ways to choose, and after that 2 ways to choose, so $3\cdot2=\boxed{6}$

Answer: A 6
Problem: Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps.  However, instead of walking up the steps one at a time, both Cozy and Dash jump.  Cozy goes two steps up with each jump (though if necessary, he will just jump the last step).  Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than $5$ steps left).  Suppose Dash takes $19$ fewer jumps than Cozy to reach the top of the staircase.  Let $s$ denote the sum of all possible numbers of steps this staircase can have.  What is the sum of the digits of $s$
$\textbf{(A) }9\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }15$

Solution: Let $n$ be the number of steps. We have
\[\left\lceil \frac{n}{2} \right\rceil - 19 = \left\lceil \frac{n}{5} \right\rceil\]
We will proceed to solve this equation via casework.
Case $1$ $\left\lceil \frac{n}{2} \right\rceil = \frac{n}{2}$
Our equation becomes $\frac{n}{2} - 19 = \frac{n}{5} + \frac{j}{5}$ , where $j \in \{0,1,2,3,4\}$ Using the fact that $n$ is an integer, we quickly find that $j=1$ and $j=4$ yield $n=64$ and $n=66$ , respectively.
Case $2$ $\left\lceil \frac{n}{2} \right\rceil = \frac{n}{2}+\frac{1}{2}$
Our equation becomes $\frac{n}{2} +\frac{1}{2} - 19 = \frac{n}{5} + \frac{j}{5}$ , where $j \in \{0,1,2,3,4\}$ Using the fact that $n$ is an integer, we quickly find that $j=2$ yields $n=63$ . Summing up we get $63+64+66=193$ . The sum of the digits is $\boxed{13}$

Answer: D 13
Problem: Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps.  However, instead of walking up the steps one at a time, both Cozy and Dash jump.  Cozy goes two steps up with each jump (though if necessary, he will just jump the last step).  Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than $5$ steps left).  Suppose Dash takes $19$ fewer jumps than Cozy to reach the top of the staircase.  Let $s$ denote the sum of all possible numbers of steps this staircase can have.  What is the sum of the digits of $s$
$\textbf{(A) }9\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }15$

Solution: We know from the problem that Dash goes $3$ steps further than Cozy per jump (assuming they aren't within $4$ steps from the top). That means that if Dash takes $19$ fewer jumps than Cozy to get to the top of the staircase, the staircase must be at least $3 \cdot 19=57$ steps high. We then start using guess-and-check:
$57$ steps: $\left \lceil {57/2} \right \rceil = 29$ jumps for Cozy, and $\left \lceil {57/5} \right \rceil = 12$ jumps for Dash, giving a difference of $17$ jumps.
$58$ steps: $\left \lceil {58/2} \right \rceil = 29$ jumps for Cozy, and $\left \lceil {58/5} \right \rceil = 12$ jumps for Dash, giving a difference of $17$ jumps.
$59$ steps: $\left \lceil {59/2} \right \rceil = 30$ jumps for Cozy, and $\left \lceil {59/5} \right \rceil = 12$ jumps for Dash, giving a difference of $18$ jumps.
$60$ steps: $\left \lceil {60/2} \right \rceil = 30$ jumps for Cozy, and $\left \lceil {60/5} \right \rceil = 12$ jumps for Dash, giving a difference of $18$ jumps.
$\vdots$
By the time we test $61$ steps, we notice that when the number of steps exceeds a multiple of $2$ , the difference in jumps increases. So, we have to find the next number that will increase the difference. $62$ doesn't because both both Cozy's and Dash's number of jumps increases, but $63$ does, and $64$ $65$ actually gives a difference of $20$ jumps, but $66$ goes back down to $19$ (because Dash had to take another jump when Cozy didn't). We don't need to go any further because the difference will stay above $19$ onward.
Therefore, the possible numbers of steps in the staircase are $63$ $64$ , and $66$ , giving a sum of $193$ . The sum of those digits is $13$ , so the answer is $\boxed{13}$

Answer: D 13
Problem: Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps.  However, instead of walking up the steps one at a time, both Cozy and Dash jump.  Cozy goes two steps up with each jump (though if necessary, he will just jump the last step).  Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than $5$ steps left).  Suppose Dash takes $19$ fewer jumps than Cozy to reach the top of the staircase.  Let $s$ denote the sum of all possible numbers of steps this staircase can have.  What is the sum of the digits of $s$
$\textbf{(A) }9\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }15$

Solution: We're looking for natural numbers $x$ such that $\left \lceil{\frac{x}{5}}\right \rceil + 19 = \left \lceil{\frac{x}{2}}\right \rceil$
Let's call $x = 10a + b$ . We now have $2a  + \left \lceil{\frac{b}{5}}\right \rceil + 19 = 5a + \left  \lceil{\frac{b}{2}}\right \rceil$ , or
$19 - 3a = \left \lceil{\frac{b}{2}}\right \rceil - \left \lceil{\frac{b}{5}}\right \rceil$
Obviously, since $b \le 10$ , this will not work for any value of $a$ under $6$ . In addition, since obviously $\frac{b}{2} \ge \frac{b}{5}$ , this will not work for any value over six, so we have $a = 6$ and $\left \lceil{\frac{b}{2}}\right \rceil - \left \lceil{\frac{b}{5}}\right \rceil = 1.$
This can be achieved when $\left \lceil{\frac{b}{5}}\right \rceil = 1$ and $\left \lceil{\frac{b}{2}}\right \rceil = 2$ , or when $\left \lceil{\frac{b}{5}}\right \rceil = 2$ and $\left \lceil{\frac{b}{2}}\right \rceil = 3$
Case One:
We have $b \le 5$ and $3 \le b \le 4$ , so $b = 3, 4$
Case Two:
We have $6 \le b \le 9$ and $5 \le b \le 6$ , so $b = 6$
We then have $63 + 64 + 66 = 193$ , which has a digit sum of $\boxed{13}$

Answer:  13
Problem: Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps.  However, instead of walking up the steps one at a time, both Cozy and Dash jump.  Cozy goes two steps up with each jump (though if necessary, he will just jump the last step).  Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than $5$ steps left).  Suppose Dash takes $19$ fewer jumps than Cozy to reach the top of the staircase.  Let $s$ denote the sum of all possible numbers of steps this staircase can have.  What is the sum of the digits of $s$
$\textbf{(A) }9\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }15$

Solution: Translate the problem into following equation:
$n = 5D - \{0 \sim 4\} = 2C - \{0 \sim 1\}$
Since $C = D + 19$ , we have
$5D - \{0 \sim 4\} = 2D + 38 - \{0 \sim 1\}$
i.e.,
$3D = 38 + \{0 \sim 4\} - \{0 \sim 1\}$
We then have $D = 13$ when $\{1\} - \{0\}$ or $\{2\} - \{1\}$ (the dog's last jump has $2$ steps and the cat's last jump has $1$ step), which yields $n = 64$ and $n = 63$ respectively.
Another solution is $D = 14$ when $\{4\} - \{0\}$ , which yields $n = 66$
Therefore, with $63 + 64 + 66 = 193$ , the digit sum is $\boxed{13}$

Answer:  13
Problem: Let $n$ be a positive integer greater than 4 such that the decimal representation of $n!$ ends in $k$ zeros and the decimal representation of $(2n)!$ ends in $3k$ zeros. Let $s$ denote the sum of the four least possible values of $n$ . What is the sum of the digits of $s$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$

Solution: A trailing zero requires a factor of two and a factor of five. Since factors of two occur more often than factors of five, we can focus on the factors of five. We make a chart of how many trailing zeros factorials have:
\[\begin{array}{c|c|c|c|c|c|c} \mathrm{Factorial}&0!-4!&5!-9!&10!-14!&15!-19!&20!-24!&25!-29!\\\hline \mathrm{Zeros}&0&1&2&3&4&6 \end{array}\]
We first look at the case when $n!$ has $1$ zero and $(2n)!$ has $3$ zeros. If $n=5,6,7$ $(2n)!$ has only $2$ zeros. But for $n=8,9$ $(2n)!$ has $3$ zeros. Thus, $n=8$ and $n=9$ work.
Secondly, we look at the case when $n!$ has $2$ zeros and $(2n)!$ has $6$ zeros. If $n=10,11,12$ $(2n)!$ has only $4$ zeros. But for $n=13,14$ $(2n)!$ has $6$ zeros. Thus, the smallest four values of $n$ that work are $n=8,9,13,14$ , which sum to $44$ . The sum of the digits of $44$ is $\boxed{8}$

Answer: B 8
Problem: Let $n$ be a positive integer greater than 4 such that the decimal representation of $n!$ ends in $k$ zeros and the decimal representation of $(2n)!$ ends in $3k$ zeros. Let $s$ denote the sum of the four least possible values of $n$ . What is the sum of the digits of $s$
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11$

Solution: By Legendre's Formula and the information given, we have that $3\left(\left\lfloor{\frac{n}{5}}\right\rfloor+\left\lfloor{\frac{n}{25}}\right\rfloor\right)=\left\lfloor{\frac{2n}{5}}\right\rfloor+\left\lfloor{\frac{2n}{25}}\right\rfloor$
We have $n<100$ as there is no way that if $n>100$ $(2n)!$ would have $3$ times as many zeroes as $n!$
First, let's plug in the number $5$ .
We get that $3(1)=1$ , which is obviously not true. Hence, $n>5$
After several attempts, we realize that the RHS needs $1$ to $2$ more "extra" zeroes than the LHS. Hence, $n$ is greater than a multiple of $5$
We find that the least four possible $n$ are $8,9,13,14$
$8+9+13+14=17+27=44\implies 4+4=8\implies\boxed{8}$

Answer: B 8