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

(→‎Solution 3: use \left and \right to make brackets bigger)
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~ Nafer
 
~ Nafer
  
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== Solution 4==
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We first note that the number of 0's in <math>n!</math> is determined by how many 5's are in the prime factorization. We use Legendre's Formula and split up into two cases:
 +
 +
CASE ONE: 5\geq 2n < 25.
 +
The only way we can fulfill the requirements is if <math>\lfloor \dfrac{n}{5} \rfloor = 1</math> and <math>\lfloor \dfrac{2n}{5} \rfloor=3</math> which means that <math>5\geq n <10</math> and <math>15\geq 2n 20</math>. The only way this works is if <math>n = 8 \text{ or } 9.</math>
 +
 +
CASE TWO: <math>25 \geq 2n</math>.
 +
Since we want the smallest values of <math>n</math>, we first try it when <math>2n<30.</math> Thus <math>(2n)!</math> has 6 zeros, which implies that <math>n!</math> must have 2. The only way to do this while maintaining our restrictions for <math>2n</math> is if <math>n = 13 \text{ or } 14.</math>
 +
 +
So the sum of the four values is <math>8+9+13+14=44</math> so the digit is sum is <math>\boxed{\mathbf{(B)\ }8}.</math>
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-ConfidentKoala4
 
==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 18:15, 6 October 2022

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 four possible $n$ are $8,9,13,14$.

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

Solution 3

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 \[3(\left\lfloor\frac{n}{5}\right\rfloor+\left\lfloor\frac{n}{25}\right\rfloor)=\left\lfloor\frac{2n}{5}\right\rfloor+\left\lfloor\frac{2n}{25}\right\rfloor+\left\lfloor\frac{2n}{125}\right\rfloor\] For smaller $n$, we temporarily let $\left\lfloor\frac{2n}{125}\right\rfloor=0$ \[3(\left\lfloor\frac{n}{5}\right\rfloor+\left\lfloor\frac{n}{25}\right\rfloor)=\left\lfloor\frac{2n}{5}\right\rfloor+\left\lfloor\frac{2n}{25}\right\rfloor\] \[3(\left\lfloor\frac{5m+k}{5}\right\rfloor+\left\lfloor\frac{5m+k}{25}\right\rfloor)=\left\lfloor\frac{2(5m+k)}{5}\right\rfloor+\left\lfloor\frac{2(5m+k)}{25}\right\rfloor\] \[3(\left\lfloor\frac{5m+k}{5}\right\rfloor+\left\lfloor\frac{5m+k}{25}\right\rfloor)=\left\lfloor\frac{10m+2k}{5}\right\rfloor+\left\lfloor\frac{10m+2k}{25}\right\rfloor\] \[3m+3\left\lfloor\frac{5m+k}{25}\right\rfloor=2m+\left\lfloor\frac{2k}{5}\right\rfloor+\left\lfloor\frac{10m+2k}{25}\right\rfloor\] \[m+3\left\lfloor\frac{5m+k}{25}\right\rfloor=\left\lfloor\frac{2k}{5}\right\rfloor+\left\lfloor\frac{10m+2k}{25}\right\rfloor\] To minimize $n$, we let $\left\lfloor\frac{5m+k}{25}\right\rfloor=\left\lfloor\frac{10m+2k}{25}\right\rfloor=0$, then \[m=\left\lfloor\frac{2k}{5}\right\rfloor\] 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 $\left\lfloor\frac{5m+k}{25}\right\rfloor=0$ and $\left\lfloor\frac{10m+2k}{25}\right\rfloor=1$, then \[m=\left\lfloor\frac{2k}{5}\right\rfloor+\left\lfloor\frac{10m+2k}{25}\right\rfloor\] Since $k<5$, $10m>15\Longrightarrow m\ge2$.

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

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

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

~ Nafer

Solution 4

We first note that the number of 0's in $n!$ is determined by how many 5's are in the prime factorization. We use Legendre's Formula and split up into two cases:

CASE ONE: 5\geq 2n < 25. The only way we can fulfill the requirements is if $\lfloor \dfrac{n}{5} \rfloor = 1$ and $\lfloor \dfrac{2n}{5} \rfloor=3$ which means that $5\geq n <10$ and $15\geq 2n 20$. The only way this works is if $n = 8 \text{ or } 9.$

CASE TWO: $25 \geq 2n$. Since we want the smallest values of $n$, we first try it when $2n<30.$ Thus $(2n)!$ has 6 zeros, which implies that $n!$ must have 2. The only way to do this while maintaining our restrictions for $2n$ is if $n = 13 \text{ or } 14.$

So the sum of the four values is $8+9+13+14=44$ so the digit is sum is $\boxed{\mathbf{(B)\ }8}.$

-ConfidentKoala4

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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