Difference between revisions of "2017 AMC 8 Problems/Problem 19"

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==Problem 19==
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==Problem==
 
For any positive integer <math>M</math>, the notation <math>M!</math> denotes the product of the integers <math>1</math> through <math>M</math>. What is the largest integer <math>n</math> for which <math>5^n</math> is a factor of the sum <math>98!+99!+100!</math> ?
 
For any positive integer <math>M</math>, the notation <math>M!</math> denotes the product of the integers <math>1</math> through <math>M</math>. What is the largest integer <math>n</math> for which <math>5^n</math> is a factor of the sum <math>98!+99!+100!</math> ?
  
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==Solution 1==
 
==Solution 1==
Factoring out <math>98!</math>, we have <math>98!(10,000)</math>. Next, <math>98!</math> has <math>\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{98}{25}\right\rfloor = 19 + 3 = 22</math> factors of <math>5</math>. Now <math>10,000</math> has <math>4</math> factors of <math>5</math>, so there are a total of <math>22 + 4 = \boxed{\textbf{(D)}\ 26}</math> factors of <math>5</math>.
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Factoring out <math>98!+99!+100!</math>, we have <math>98!(1+99+99*100)</math> which is <math>98!(10000)</math> Next, <math>98!</math> has <math>\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{98}{25}\right\rfloor = 19 + 3 = 22</math> factors of <math>5</math>. The <math>19</math> is because of all the multiples of <math>5</math>. Now <math>10,000</math> has <math>4</math> factors of <math>5</math>, so there are a total of <math>22 + 4 = \boxed{\textbf{(D)}\ 26}</math> factors of <math>5</math>.
  
 
==Solution 2==
 
==Solution 2==
The number of 5's in the factorization of <math>98! + 99! + 100!</math> is the same as the number fo trailing zeroes. The number of zeroes is taken by the floor value of each number divided by 5, until you can't divide by 5 anymore. Factorizing <math>98! + 99! + 100!</math>, you get <math>98!(1+99+9900)=98!(1000)</math>. To find the number of trailing seroes in 98!, we do <math>\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{19}{5}\right\rfloor= 19 + 3=22</math>. Now since <math>10000</math> has 4 zeroes, we add <math>22 + 4</math> to get <math>\boxed{\textbf{(D)}\ 26}</math> factors of <math>5</math>.
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Also keep in mind that number of <math>5</math>’s in <math>98!(10,000)</math> is the same as the number of trailing zeros. Number of zeros is <math>98!</math> means we need pairs of <math>5</math>’s and <math>2</math>’s; we know there will be many more <math>2</math>’s, so we seek to find number of <math>5</math>’s in <math>98!</math> which solution tells us and that is <math>22</math> factors of <math>5</math>. <math>10,000</math> has <math>4</math> trailing zeros, so it has <math>4</math> factors of <math>5</math> and <math>22 + 4 = 26</math>.
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== Video Solution ==
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https://youtu.be/alj9Y8jGNz8
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https://youtu.be/HISL2-N5NVg?t=817
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~ pi_is_3.14
  
 
==See Also==
 
==See Also==

Latest revision as of 14:37, 28 June 2021

Problem

For any positive integer $M$, the notation $M!$ denotes the product of the integers $1$ through $M$. What is the largest integer $n$ for which $5^n$ is a factor of the sum $98!+99!+100!$ ?

$\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$

Solution 1

Factoring out $98!+99!+100!$, we have $98!(1+99+99*100)$ which is $98!(10000)$ Next, $98!$ has $\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{98}{25}\right\rfloor = 19 + 3 = 22$ factors of $5$. The $19$ is because of all the multiples of $5$. Now $10,000$ has $4$ factors of $5$, so there are a total of $22 + 4 = \boxed{\textbf{(D)}\ 26}$ factors of $5$.

Solution 2

Also keep in mind that number of $5$’s in $98!(10,000)$ is the same as the number of trailing zeros. Number of zeros is $98!$ means we need pairs of $5$’s and $2$’s; we know there will be many more $2$’s, so we seek to find number of $5$’s in $98!$ which solution tells us and that is $22$ factors of $5$. $10,000$ has $4$ trailing zeros, so it has $4$ factors of $5$ and $22 + 4 = 26$.

Video Solution

https://youtu.be/alj9Y8jGNz8

https://youtu.be/HISL2-N5NVg?t=817

~ pi_is_3.14

See Also

2017 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AJHSME/AMC 8 Problems and Solutions

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