Problem 19 · 2017 AMC 8
Hard
Number Theory
factorizationdivisibility
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 5n is a factor of the sum
98! + 99! + 100! ?
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Answer: D — 26.
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Hint 1 of 2
Don't fear the factorials — the smallest, 98!, is a factor of all three (99! = 99·98!, 100! = 100·99·98!). Pull it out and the leftover bracket collapses to something startlingly round.
Still stuck? Show hint 2 →
Hint 2 of 2
Factors of 5 add up from two independent sources: the round leftover, plus the 5's buried inside 98!. For 98!, use Legendre's count: ⌊98/5⌋ + ⌊98/25⌋ (multiples of 5, then the extra 5 in multiples of 25).
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Approach: factor out 98!, then count 5's from each piece
- Factor out the common 98!: 98! + 99! + 100! = 98!(1 + 99 + 100·99) = 98!(100 + 9900) = 98! · 10,000. That clean 10,000 is the payoff for factoring — it's loaded with 5's.
- Source 1, the leftover: 10,000 = 104 = 24·54, giving 4 factors of 5.
- Source 2, inside 98!: count multiples of 5 (⌊98/5⌋ = 19) plus the second 5 each multiple of 25 contributes (⌊98/25⌋ = 3): that's 19 + 3 = 22 factors of 5.
- Add the two sources: 4 + 22 = 26. So the largest such n is 26.
- Why this transfers: sums of factorials always factor out the smallest one; and the prime-5 count in any k! is ⌊k/5⌋ + ⌊k/25⌋ + ⌊k/125⌋ + … (Legendre's formula) — the engine behind every "trailing zeros" / "power of a prime in a factorial" question.
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