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

(Video Solution by OmegaLearn)
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==Problem==
 
==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> ?
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For any positive integer <math>M</math>, the notation <math>M!</math> denotes the product of the integers <math>1</math> through
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<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> ?
  
<math>\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27</math>  
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<math>\textbf{(A) }23 \qquad \textbf{(B) }24 \qquad \textbf{(C) }25 \qquad \textbf{(D) }26 \qquad \textbf{(E) }27</math>
  
 
==Solution 1==
 
==Solution 1==
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~CHECKMATE2021
 
~CHECKMATE2021
  
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
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Note: Can you say what formula this uses? most AMC 8 test takers won't know it.  
https://youtu.be/WKux87BEO1U
 
  
~Education, the Study of Everything
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==Video Solution (Omega Learn)==
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https://www.youtube.com/watch?v=HISL2-N5NVg&t=817s
  
== Video Solution ==
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~ GeometryMystery
https://youtu.be/alj9Y8jGNz8
 
 
 
https://youtu.be/meEuDzrM5Ac
 
 
 
~savannahsolver
 
  
 
==See Also==
 
==See Also==

Latest revision as of 03:20, 13 October 2024

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 $5$.The $3$ is because of all the multiples of $25$. Now, $10,000$ has $4$ factors of $5$, so there are a total of $22 + 4 = \boxed{\textbf{(D)}\ 26}$ factors of $5$.

~CHECKMATE2021

Note: Can you say what formula this uses? most AMC 8 test takers won't know it.

Video Solution (Omega Learn)

https://www.youtube.com/watch?v=HISL2-N5NVg&t=817s

~ GeometryMystery

See Also

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

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