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

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==Solution==
 
==Solution==
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===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:
 
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:
  
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Secondly, we look at the case when <math>n!</math> has <math>2</math> zeros and <math>(2n)!</math> has <math>6</math> zeros. If <math>n=10,11,12</math>, <math>(2n)!</math> has only <math>4</math> zeros. But for <math>n=13,14</math>, <math>(2n)!</math> has <math>6</math> zeros. Thus, the smallest four values of <math>n</math> that work are <math>n=8,9,13,14</math>, which sum to <math>44</math>. The sum of the digits of <math>44</math> is <math>\boxed{\mathbf{(B)\ }8}</math>
 
Secondly, we look at the case when <math>n!</math> has <math>2</math> zeros and <math>(2n)!</math> has <math>6</math> zeros. If <math>n=10,11,12</math>, <math>(2n)!</math> has only <math>4</math> zeros. But for <math>n=13,14</math>, <math>(2n)!</math> has <math>6</math> zeros. Thus, the smallest four values of <math>n</math> that work are <math>n=8,9,13,14</math>, which sum to <math>44</math>. The sum of the digits of <math>44</math> is <math>\boxed{\mathbf{(B)\ }8}</math>
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 +
===Solution 2===
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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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 +
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>.
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 +
First, let's plug in the number <math>5</math>
 +
We get that <math>3(1)=1</math>, which is obviously not true. Hence, <math>n>5</math>
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 +
After several attempts, we realize that the RHS needs <math>1</math> to <math>2</math> more "extra" zeroes than the LHS. Hence, <math>n</math> is greater than a multiple of <math>5</math>.
 +
 +
Very quickly, we find that the least <math>4 n's</math> are <math>8,9,13,14</math>.
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 +
<math>8+9+13+14=17+27=44\implies 4+4=8\implies\boxed{B}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2015|ab=B|num-b=22|num-a=24}}
 
{{AMC10 box|year=2015|ab=B|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 10:06, 26 July 2017

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 $3(\lfloor{\frac{n}{5}}\rfloor+\lfloor{\frac{n}{25}}\rfloor)=\lfloor{\frac{2n}{5}}\rfloor+\lfloor{\frac{2n}{25}}\rfloor$.

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 (Problems • Answer Key • Resources)
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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