Competition · AMC preparation · step 4 of 4
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.
Factor 2020 into primes
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 ListCount all the divisors
The 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.
The number 2020 has exactly 12 positive integer divisors.
▸ Why?
Because 2020 = 2² × 5 × 101, every divisor is made by picking a power of 2 (none, one, or two), a power of 5 (none or one), and a power of 101 (none or one); each different pick gives a different divisor and every divisor comes from exactly one pick, so counting the divisors is the same as counting these picks.
▸ Why?
There are 3 choices for the power of 2, 2 choices for the power of 5, and 2 choices for the power of 101, and the three choices are made independently, so the number of combinations is 3 × 2 × 2 = 12.
▸ Why?
The power of 2, the power of 5, and the power of 101 are chosen independently — no pick restricts the others — so each of the 3 powers of 2 can be paired with each of the 2 powers of 5 and each of the 2 powers of 101, and every combination arises exactly once; when independent choices are combined this way their option counts multiply, giving 3 × 2 × 2.
Name the small-divisor cases
The 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.4Change Focus Count The ComplementList the bad divisors
Among 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 bad cases
Subtract 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.3Change Focus Count 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!
- Factor 2020 into primes
- Count all the divisors
- Name the small-divisor cases
- List the bad divisors
- Subtract the bad cases
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