AMC 10 · 2019 · #2
Grade 5 arithmeticPick an answer.
AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Tool #9 (Easier Problem): we cannot compute 20! or 15! directly, so we ask a smaller question — what are the last three digits of each? A number's hundreds digit is decided by its remainder when divided by 1000. Tool #5 (Pattern): count factors of 10 inside each factorial. Each pair of a 2 and a 5 inside the product gives one trailing zero. Tool #16 (Change Focus): instead of asking 'what is the hundreds digit of 20! - 15!', we ask 'is 20! - 15! divisible by 1000?' — if yes, the hundreds digit must be 0.
Inside 15!, the numbers 5, 10, 15 give three factors of 5, paired with plenty of 2s, so 1000 ∣ 15!.
Each factor of 5 paired with a factor of 2 creates a trailing zero — 15! has at least three.
4.OA.B.4Solve An Easier Related ProblemThe same count in 20! gives four factors of 5 (from 5, 10, 15, 20), so 1000 ∣ 20! too.
Same pattern as 15!, just more factors of 5 — 20! has even more trailing zeros.
4.OA.B.4Look For A PatternBoth are multiples of 1000, so 20! - 15! also ends in 000 — the hundreds digit is 0.
If a number ends in … 000, the hundreds place is 0.
5.NBT.A.1Count The ComplementSanity check: 4! - 2! = 22 shares no trailing zeros, but 20! and 15! share three — plenty to lock the last three digits.
Working with tiny factorials first builds confidence in the trailing-zero rule.
5.NBT.A.1Solve An Easier Related ProblemThis AMC 10 problem only needs Grade 5 "place value" you already know — both 20! and 15! end in at least three zeros, so their difference ends in three zeros and the hundreds digit is 0.