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

(Video Solution (Omega Learn))
(Solution 1)
Line 13: Line 13:
 
~CHECKMATE2021
 
~CHECKMATE2021
  
Note: Can you say what formula this uses? most AMC 8 test takers won't know it. Also, can someone unvandalize this page?
+
Note: Can you say what formula this uses? most AMC 8 test takers won't know it. Also, can someone unvandalize this page at the top?
  
 
==Video Solution (Omega Learn)==
 
==Video Solution (Omega Learn)==

Revision as of 13:47, 16 August 2024

wertesryrtutyrudtu

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 $98!+99!+100!$ ?

$\textbf{(A) }23 \qquad \textbf{(B) }24 \qquad \textbf{(C) }25 \qquad \textbf{(D) }26 \qquad \textbf{(E) }327$


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. Also, can someone unvandalize this page at the top?

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

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