2017 AMC 8 Problem 19

Attempt Problem 19 of the 2017 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

19.

For any positive integer M,M, the notation M!M! denotes the product of the integers 11 through M.M. What is the largest integer nn for which 5n5^n is a factor of the sum: 98!+99!+100!98!+99!+100!
23 23
24 24
25 25
26 26
27 27
Answer: D
Concepts:factorialLegendre’s Formula
Small hint:
Factor 98!98! out of all three terms.
Big hint:
After factoring, count the powers of 55 in 98!98! and in 1002100^2.
Video solution:
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Written solution:
Factor out 98!98!: 98!+99!+100!=98!(1+99+100⋅99)=98!⋅1002.\begin{aligned} 98!+99!+100! &=98!\bigl(1+99\\ &\qquad+100\cdot99\bigr)\\ &=98!\cdot100^2. \end{aligned}
The exponent of 55 in 98!98! is ⌊985⌋+⌊9825⌋\left\lfloor\frac{98}{5}\right\rfloor+\left\lfloor\frac{98}{25}\right\rfloor, which equals 19+3=22.19+3=22. The factor 1002100^2 contributes four more factors of 55, so the total exponent is 22+4=26.22+4=26.
Thus, D is the correct answer.
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