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

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<math> \textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11 </math>
 
<math> \textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad\textbf{(E) }11 </math>
  
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==Solution==
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==Solution 1==
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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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==Solution 2==
 
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>.
 
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>.
  
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<math>8+9+13+14=17+27=44\implies 4+4=8\implies\boxed{B}</math>.
 
<math>8+9+13+14=17+27=44\implies 4+4=8\implies\boxed{B}</math>.
  
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===Solution 3 (Bashing)===
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==Solution 3 (Bashing)==
 
We notice that for a <math>0</math> to be at the end of a factorial, one multiple of five must be there. Therefore, it is intuitive to start at <math>5</math> and work up. If you bash enough you get <math>8</math>, <math>9</math>, <math>13</math>, and <math>14</math>. Going any higher will give too many zeros, and then we can stop going higher. <math>8+9+13+14=17+27=44\implies 4+4=8\implies\boxed{\textbf{(B) }8}</math>.
 
We notice that for a <math>0</math> to be at the end of a factorial, one multiple of five must be there. Therefore, it is intuitive to start at <math>5</math> and work up. If you bash enough you get <math>8</math>, <math>9</math>, <math>13</math>, and <math>14</math>. Going any higher will give too many zeros, and then we can stop going higher. <math>8+9+13+14=17+27=44\implies 4+4=8\implies\boxed{\textbf{(B) }8}</math>.
  
  
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=== Solution 4 ===
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== Solution 4 ==
  
 
Let <math>n=5m+k</math> for some natural numbers <math>m</math>, <math>k</math> such that <math>k\in\{0,1,2,3,4\}</math>. Notice that <math>n<5^3=125</math>. Thus  
 
Let <math>n=5m+k</math> for some natural numbers <math>m</math>, <math>k</math> such that <math>k\in\{0,1,2,3,4\}</math>. Notice that <math>n<5^3=125</math>. Thus  

Revision as of 18:14, 13 July 2020

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

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 $4 n's$ are $8,9,13,14$.

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

Solution 3 (Bashing)

We notice that for a $0$ to be at the end of a factorial, one multiple of five must be there. Therefore, it is intuitive to start at $5$ and work up. If you bash enough you get $8$, $9$, $13$, and $14$. Going any higher will give too many zeros, and then we can stop going higher. $8+9+13+14=17+27=44\implies 4+4=8\implies\boxed{\textbf{(B) }8}$.


Solution 4

Let $n=5m+k$ for some natural numbers $m$, $k$ such that $k\in\{0,1,2,3,4\}$. Notice that $n<5^3=125$. Thus \[3(\lfloor\frac{n}{5}\rfloor+\lfloor\frac{n}{25}\rfloor)=\lfloor\frac{2n}{5}\rfloor+\lfloor\frac{2n}{25}\rfloor+\lfloor\frac{2n}{125}\rfloor\] For smaller $n$, we temporarily let $\lfloor\frac{2n}{125}\rfloor=0$ \[3(\lfloor\frac{n}{5}\rfloor+\lfloor\frac{n}{25}\rfloor)=\lfloor\frac{2n}{5}\rfloor+\lfloor\frac{2n}{25}\rfloor\] \[3(\lfloor\frac{5m+k}{5}\rfloor+\lfloor\frac{5m+k}{25}\rfloor)=\lfloor\frac{2(5m+k)}{5}\rfloor+\lfloor\frac{2(5m+k)}{25}\rfloor\] \[3(\lfloor\frac{5m+k}{5}\rfloor+\lfloor\frac{5m+k}{25}\rfloor)=\lfloor\frac{10m+2k}{5}\rfloor+\lfloor\frac{10m+2k}{25}\rfloor\] \[3m+3\lfloor\frac{5m+k}{25}\rfloor=2m+\lfloor\frac{2k}{5}\rfloor+\lfloor\frac{10m+2k}{25}\rfloor\] \[m+3\lfloor\frac{5m+k}{25}\rfloor=\lfloor\frac{2k}{5}\rfloor+\lfloor\frac{10m+2k}{25}\rfloor\] To minimize $n$, we let $\lfloor\frac{5m+k}{25}\rfloor=\lfloor\frac{10m+2k}{25}\rfloor=0$, then \[m=\lfloor\frac{2k}{5}\rfloor\] Since $k<5$, $m>0$, the only integral value of $m$ is $1$, from which we have $k=3,4\Longrightarrow n=8,9$.

Now we let $\lfloor\frac{5m+k}{25}\rfloor=0$ and $\lfloor\frac{10m+2k}{25}\rfloor=1$, then \[m=\lfloor\frac{2k}{5}\rfloor+\lfloor\frac{10m+2k}{25}\rfloor\] Since $k<5$, $10m>15\Longrightarrow m\ge2$.

If $m>2$, then \[m>\lfloor\frac{2k}{5}\rfloor+\lfloor\frac{10m+2k}{25}\rfloor\] which is a contradiction.

Thus $m=2\Longrightarrow\lfloor\frac{2k}{5}\rfloor=1\Longrightarrow n=13,14$

Finally, the sum of the four smallest possible $n=8+9+13+14=44$ and $4+4=8$. $\boxed{\mathrm{(B)}}$

~ Nafer

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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