Question

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: $98!+99!+100!$

$23$

$24$

$25$

$26$

$27$

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Written solution:

The expression can be factored as follows:

$\begin{align*} &98! + 99! + 100! \\ &= 98! + 99 \cdot 98! + 100 \cdot 99 \cdot 98! \\ &= 98!(1 + 99 + 100 \cdot 99) \\ &= 98!(100 + 100 \cdot 99) \\ &= 98! \cdot 100 \cdot (1 + 99) \\ &= 98! \cdot 100^2 \end{align*}$

Each $100$ possesses two factors of $5.$ The number of factors of $5$ in $98!$ is $\lfloor 98 / 5 \rfloor + \lfloor 98 / 25 \rfloor= 19 + 3 = 22.$ This counts the number of numbers divisible by $5$ and the number of numbers divisible by $25$ to get both factors of $5.$ The total number of factors of $5$ is $2 + 2 + 22 = 26.$

Thus, D is the correct answer.

2017 AMC 8 Problem 19

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Written solution:

Problem 19 in Other Years