2017 AMC 8 Problems/Problem 19

Revision as of 18:45, 20 April 2021 by Putu2003 (talk | contribs) (Solution 2)

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/HISL2-N5NVg?t=817

~ pi_is_3.14

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

Invalid username
Login to AoPS