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

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~CHECKMATE2021
 
~CHECKMATE2021
  
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Note: Do you know what formula this uses? Most AMC 8 test takers won't know it.
+
Note: Do you know what formula this uses? Most AMC 8 test takers won't know it. It's Legendre's Formula.
 +
https://artofproblemsolving.com/wiki/index.php/Legendre%27s_Formula?srsltid=AfmBOoovTeAI0nS9kUtxSrX27kDSKO89Fw2xxDngnBwnXcKW5eql12H5
 +
 
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- Rayansh Mankad
  
 
==Video Solution (Omega Learn)==
 
==Video Solution (Omega Learn)==

Revision as of 18:42, 2 January 2025

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: Do you know what formula this uses? Most AMC 8 test takers won't know it. It's Legendre's Formula. https://artofproblemsolving.com/wiki/index.php/Legendre%27s_Formula?srsltid=AfmBOoovTeAI0nS9kUtxSrX27kDSKO89Fw2xxDngnBwnXcKW5eql12H5

- Rayansh Mankad

Video Solution (Omega Learn)

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

~ GeometryMystery

See Also

2017 AMC 8 (Problems • Answer Key • Resources)
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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