AMC 8 · 2017 · #19

Grade 6 number-theory

Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation

factorialprime-factorizationexponentsfactors identify-subproblemspattern-recognition ↑ Prerequisites: factorialprime-factorizationexponents

📏 Medium solution 💡 4 insights

Problem

Recall that M! = 1 · 2 · 3 … M. We are told to find the largest integer n such that 5ⁿ divides the sum 98! + 99! + 100!. Equivalently: how many copies of the prime 5 appear in the prime factorization of that sum?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Tool #7 (Identify Subproblems) is the workhorse here: the sum 98! + 99! + 100! shares a common factor 98!, so we factor it out and split the problem into two independent counting jobs — "how many 5s are inside 98!?" and "how many 5s are inside what's left in the parentheses?". Tool #5 (Look for a Pattern) handles the first subproblem: every 5th number contributes one 5, every 25th number contributes an extra 5, and so on — a clean, repeating pattern we can list. Tool #3 (Eliminate Possibilities) gives a final sanity check against the five answer choices.

1STEP 1

Factor the smallest factorial 98! out of every term, turning the sum into 98! times a small whole number.

98! + 99! + 100! = 98! (1 + 99 + 100 · 99)

2STEP 2

Simplify the bracket: 100 · 99 = 9900, then 1 + 99 + 9900 = 10000, so the sum is 98! × 10000.

1 + 99 + 100 · 99 = 1 + 99 + 9900 = 10000

3STEP 3

Write 10000 = 10⁴ = (2 · 5)⁴ = 2⁴ · 5⁴, so it holds exactly four factors of 5.

10000 = 10⁴ = (2 · 5)⁴ = 2⁴ · 5⁴

4STEP 4

Among 1…98, every 5th number gives a 5 and every 25th gives an extra: ⌊ $\frac{98}{5}$ ⌋ + ⌊ $\frac{98}{25}$ ⌋ = 19 + 3 = 22 fives in 98!.

multiples of 5 up to 98: ⌊

\frac{98}{5}

⌋ = 19; multiples of 25: ⌊

\frac{98}{25}

⌋ = 3; total = 19 + 3 = 22

5STEP 5

A product adds exponents, so the counts add: 22 + 4 = 26 fives total, giving n = 26 → choice (D).

n = 22 + 4 = 26 → (D)

Answer

The answer 26 has to sit between the natural lower and upper bounds. Lower bound: 98! alone already has 22 factors of 5, so the sum (which is a multiple of 98!) has at least 22 — eliminating any answer below 22. Upper bound: the bracketed factor 10000 = 2⁴ · 5⁴ adds exactly four more 5s and nothing else introduces new 5s, so the count tops out at 22 + 4 = 26. The answer 26 sits right at that bound, matching choice (D).

💡Key takeaway

This AMC 8 problem only needs the Grade 6 exponent and factoring ideas you already know — pull out the common factor, then count the 5s in each piece!

More like this