AMC 8 · 2020 · #17
Grade 6 number-theorycountingPick an answer.
AMC 8 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Counting divisors with "more than 3" divisors directly would force us to check 12 separate divisors of 2020 one by one. Tool #16 (Complement) flips the question: count the divisors with 3 or fewer divisors and subtract from the total 12. That smaller class is tiny because numbers with ≤ 3 divisors have a clean structure — exactly 1, the primes p, and the prime-squares p². Tool #2 (Systematic List) then enumerates the three categories cleanly so nothing is missed and nothing is double-counted.
Break 2020 down to its prime building blocks: 2020 = 2² × 5 × 101.
Breaking a number into its prime building blocks is the Grade 4 "prime or composite / factor pairs" idea.
4.OA.B.4Make A Systematic ListThe divisor-count rule (a+1)(b+1)(c+1) on exponents 2, 1, 1 gives 2020 a total of 12 divisors.
Working with whole-number exponents like 2² inside an expression is exactly the Grade 6 exponents standard.
6.EE.A.1Make A Systematic ListThe only numbers with 3 or fewer divisors are 1, a prime, or a prime square — the "bad" cases to subtract.
Classifying small divisor counts uses the Grade 4 idea that primes have exactly two factors.
4.OA.B.4Count The ComplementAmong divisors of 2020 the bad cases are 1, the primes 2, 5, 101, and the prime-square 4 — 5 in all.
Listing in a fixed order — by category, then by prime — guarantees no duplicates and no misses.
4.OA.B.4Make A Systematic ListSubtract the complement from the total: 12 - 5 = 7 divisors of 2020 have more than 3 divisors.
The final "total minus bad cases" subtraction is a Grade 4 multi-step word-problem move.
4.OA.A.3Count The ComplementThis AMC 8 problem only needs Grade 6 exponent thinking — once you write 2020 = 2² × 5 × 101, everything else is counting you already know!