Difference between revisions of "2015 AMC 10B Problems/Problem 23"

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===Solution 2===
 
===Solution 2===
By Legendre's Formula and the information given, we have that <math>3(\lfloor{\frac{n}{5}}\rfloor+\lfloor{\frac{n}{25}}\rfloor)=\lfloor{\frac{2n}{5}}\rfloor+\lfloor{\frac{2n}{25}}\rfloor</math>.
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By Legendre's Formula and the information given, we have that <math>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</math>.
  
 
Trivially, it is obvious that <math>n<100</math> as there is no way that if <math>n>100</math>, <math>n!</math> would have <math>3</math> times as many zeroes as <math>(2n)!</math>.
 
Trivially, it is obvious that <math>n<100</math> as there is no way that if <math>n>100</math>, <math>n!</math> would have <math>3</math> times as many zeroes as <math>(2n)!</math>.

Revision as of 11:07, 20 May 2018

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

Solution 1

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 $\boxed{\mathbf{(B)\ }8}$

Solution 2

By Legendre's Formula and the information given, we have that $3\left(\left\lfloor{\frac{n}{5}}\right\rfloor+\left\lfloor{\frac{n}{25}}\right\rfloor\right)=\left\lfloor{\frac{2n}{5}}\right\rfloor+\left\lfloor{\frac{2n}{25}}\right\rfloor$.

Trivially, it is obvious that $n<100$ as there is no way that if $n>100$, $n!$ would have $3$ times as many zeroes as $(2n)!$.

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 $5$.

Very quickly, we find that the least $4 n's$ are $8,9,13,14$.

$8+9+13+14=17+27=44\implies 4+4=8\implies\boxed{B}$.

See Also

2015 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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