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

m (Solution 1)
(Solution 1)
Line 1: Line 1:
 
wertesryrtutyrudtu
 
wertesryrtutyrudtu
 +
 +
==Problem==
 +
For any positive integer <math>M</math>, the notation M! denotes the product of the integers 1 through
 +
M. 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) }327</math>
 +
  
 
==Solution 1==
 
==Solution 1==

Revision as of 20:49, 4 July 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?

Video Solution (CREATIVE THINKING + ANALYSIS!!!)

https:/90ijn bidxrfgv