2017 AMC 8 Problems/Problem 19

Problem 19

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 $98!$ , we have $98!(10,000)$ . Next, $98!$ has $\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{98}{25}\right\rfloor = 19 + 3 = 22$ factors 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

The number of 5's in the factorization of $98! + 99! + 100!$ is the same as the number fo trailing zeroes. The number of zeroes is taken by the floor value of each number divided by 5, until you can't divide by 5 anymore. Factorizing $98! + 99! + 100!$ , you get $98!(1+99+9900)=98!(1000)$ . To find the number of trailing seroes in 98!, we do $\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{19}{5}\right\rfloor= 19 + 3=22$ . Now since $10000$ has 4 zeroes, we add 22 +4 to get

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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2017 AMC 8 Problems/Problem 19

Contents

Problem 19

Solution 1

Solution 2

See Also