Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page with "==Problem== Let <math>n</math> be a positive integer greater than 4 such that the decimal representation of <math>n!</math> ends in <math>k</math> zeros and the decimal repres...")
 
m (→‎Problem)
 
(74 intermediate revisions by 18 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
Let <math>n</math> be a positive integer greater than 4 such that the decimal representation of <math>n!</math> ends in <math>k</math> zeros and the decimal representation of <math>(2n)!</math> ends in <math>3k</math> zeros. Let <math>s</math> denote the sum of the four least possible values of <math>n</math>. What is the sum of the digits of <math>s</math>?
+
Let <math>n</math> be a positive integer greater than <math>4</math> such that the decimal representation of <math>n!</math> ends in <math>k</math> zeros and the decimal representation of <math>(2n)!</math> ends in <math>3k</math> zeros. Let <math>s</math> denote the sum of the four least possible values of <math>n</math>. What is the sum of the digits of <math>s</math>?
  
 
<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>
  
−
==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:
 
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:
  
Line 12: Line 12:
 
\end{array}</cmath>
 
\end{array}</cmath>
  
−
We first look at the case when <math>k!</math> has <math>1</math> zero and <math>(2k)!</math> has <math>3</math> zeros. If <math>k=5,6,7</math>, <math>(2k)!</math> has only <math>2</math> zeros. But for <math>k=8,9</math>, <math>(2k)!</math> has <math>3</math> zeros. Thus, <math>k=8</math> and <math>k=9</math> work.
+
We first look at the case when <math>n!</math> has <math>1</math> zero and <math>(2n)!</math> has <math>3</math> zeros. If <math>n=5,6,7</math>, <math>(2n)!</math> has only <math>2</math> zeros. But for <math>n=8,9</math>, <math>(2n)!</math> has <math>3</math> zeros. Thus, <math>n=8</math> and <math>n=9</math> work.
  
−
Secondly, we look at the case when <math>k!</math> has <math>2</math> zeros and <math>(2k)!</math> has <math>6</math> zeros. If <math>k=10,11,12</math>, <math>(2k)!</math> has only <math>4</math> zeros. But for <math>k=13,14</math>, <math>(2k)!</math> has <math>6</math> zeros. Thus, the smallest four values of <math>k</math> that work are <math>k=8,9,13,14</math>, which sum to <math>44</math>. The sum of the digits of <math>44</math> is <math>\boxed{\textbf{(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>
 +
 
 +
==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>.
 +
 
 +
We have <math>n<100</math> as there is no way that if <math>n>100</math>, <math>(2n)!</math> would have <math>3</math> times as many zeroes as <math>n!</math>.
 +
 
 +
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>
 +
 
 +
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>.
 +
 
 +
We find that the least four possible <math>n</math> are <math>8,9,13,14</math>.
 +
 
 +
<math>8+9+13+14=17+27=44\implies 4+4=8\implies\boxed{B}</math>.
 +
 
 +
== Solution 3 ==
 +
 
 +
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
 +
<cmath>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</cmath>
 +
For smaller <math>n</math>, we temporarily let <math>\left\lfloor\frac{2n}{125}\right\rfloor=0</math>
 +
<cmath>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</cmath>
 +
<cmath>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</cmath>
 +
<cmath>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</cmath>
 +
<cmath>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</cmath>
 +
<cmath>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</cmath>
 +
To minimize <math>n</math>, we let <math>\left\lfloor\frac{5m+k}{25}\right\rfloor=\left\lfloor\frac{10m+2k}{25}\right\rfloor=0</math>, then
 +
<cmath>m=\left\lfloor\frac{2k}{5}\right\rfloor</cmath>
 +
Since <math>k<5</math>, <math>m>0</math>, the only integral value of <math>m</math> is <math>1</math>, from which we have <math>k=3,4\Longrightarrow n=8,9</math>.
 +
 
 +
Now we let <math>\left\lfloor\frac{5m+k}{25}\right\rfloor=0</math> and <math>\left\lfloor\frac{10m+2k}{25}\right\rfloor=1</math>, then
 +
<cmath>m=\left\lfloor\frac{2k}{5}\right\rfloor+\left\lfloor\frac{10m+2k}{25}\right\rfloor</cmath>
 +
Since <math>k<5</math>, <math>10m>15\Longrightarrow m\ge2</math>.
 +
 
 +
If <math>m>2</math>, then
 +
<cmath>m>\left\lfloor\frac{2k}{5}\right\rfloor+\left\lfloor\frac{10m+2k}{25}\right\rfloor</cmath>
 +
which is a contradiction.
 +
 
 +
Thus <math>m=2\Longrightarrow\left\lfloor\frac{2k}{5}\right\rfloor=1\Longrightarrow n=13,14</math>
 +
 
 +
Finally, the sum of the four smallest possible <math>n=8+9+13+14=44</math> and <math>4+4=8</math>. <math>\boxed{\mathrm{(B)}}</math>
 +
 
 +
~ Nafer
 +
 
 +
== Solution 4==
 +
 
 +
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:
 +
 
 +
<math>\textbf{CASE ONE: } 5\leq 2n < 25.</math>
 +
 
 +
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\leq n <10</math> and <math>15\leq 2n < 20</math>. The only way this works is if <math>n = 8 \text{ or } 9.</math>
 +
 
 +
<math>\textbf{CASE TWO: } 25 \leq 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>
 +
 
 +
-ConfidentKoala4
 +
 
 +
 
 +
== Solution 5 ==
 +
 
 +
We will use trial and error to determine the answer to this problem. If <math>n = 5</math>, then <math>n!</math> has  <math>1</math> zero, and <math>2n!</math> will have <math>2</math> zeros. But <math>3 \cdot 1 \neq 2</math> so <math>n = 5</math> does not work. Similarly <math>n = 6, 7</math> do not work either. But <math>n = 8</math> works because <math>8!</math> has <math>1</math> zero and <math>16!</math> has <math>3</math> zeros. Note that <math>n = 9</math> also works because <math>9!</math> has <math>1</math> zero and <math>18!</math> has <math>3</math> zeros. After performing trial and error several times, we find that <math>n = 10, 11, 12</math> do not work but <math>n = 13, 14</math> do work. Therefore, the four smallest values of <math>n</math> are <math>8, 9, 13, 14</math>. Therefore adding them together gives <math>44</math> and our answer is <math>\boxed{\mathbf{(B)\ }8}.</math>
 +
 
 +
-[[User: Yiyj1|Yiyj1]]
 +
 
 +
==See Also==
 +
{{AMC10 box|year=2015|ab=B|num-b=22|num-a=24}}
 +
{{MAA Notice}}

Latest revision as of 00:16, 19 July 2024

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:

$\textbf{CASE ONE: } 5\leq 2n < 25.$

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 $5\leq n <10$ and $15\leq 2n < 20$. The only way this works is if $n = 8 \text{ or } 9.$

$\textbf{CASE TWO: } 25 \leq 2n$.

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


Solution 5

We will use trial and error to determine the answer to this problem. If $n = 5$, then $n!$ has $1$ zero, and $2n!$ will have $2$ zeros. But $3 \cdot 1 \neq 2$ so $n = 5$ does not work. Similarly $n = 6, 7$ do not work either. But $n = 8$ works because $8!$ has $1$ zero and $16!$ has $3$ zeros. Note that $n = 9$ also works because $9!$ has $1$ zero and $18!$ has $3$ zeros. After performing trial and error several times, we find that $n = 10, 11, 12$ do not work but $n = 13, 14$ do work. Therefore, the four smallest values of $n$ are $8, 9, 13, 14$. Therefore adding them together gives $44$ and our answer is $\boxed{\mathbf{(B)\ }8}.$

-Yiyj1

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_23&oldid=223097"