AMC 10 · 2019 · #11

Grade 6 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

prime-factorizationdivisor-countperfect-squaresexponentsprinciple-of-inclusion-exclusion caseworkcomplementary-counting ↑ Prerequisites: prime-factorizationdivisor-count

📏 Medium solution 💡 3 insights

Problem

201 = 3 × 67, so 201⁹ = 3⁹ · 67⁹. Count the positive divisors of 201⁹ that are perfect squares, perfect cubes, or both.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

Tool #2 (Systematic List): list the legal exponents a, b for each shape — even values for squares, multiples of 3 for cubes. Tool #12 (Venn): squares and cubes overlap on 6th powers, so use inclusion-exclusion: |sq ∪ cu| = |sq| + |cu| - |both|. Tool #16 (Complement) frames "both" as the overlap to subtract. Tool #3 matches the final count to the five choices.

1STEP 1

Since 201 = 3 × 67 with 67 prime, 201⁹ = 3⁹ · 67⁹, so every divisor is 3^a · 67^b with 0 ≤ a, b ≤ 9.

201⁹ = 3⁹ · 67⁹ → divisor = 3^a · 67^b, 0 ≤ a, b ≤ 9

2STEP 2

A perfect square needs every exponent even: a, b ∈ {0,2,4,6,8}, so 5 × 5 = 25 square divisors.

squares: a ∈ {0,2,4,6,8}, b ∈ {0,2,4,6,8} → 5 × 5 = 25

3STEP 3

A perfect cube needs every exponent a multiple of 3: a, b ∈ {0,3,6,9}, so 4 × 4 = 16 cube divisors.

cubes: a ∈ {0,3,6,9}, b ∈ {0,3,6,9} → 4 × 4 = 16

4STEP 4

Both at once means every exponent a multiple of 6: a, b ∈ {0, 6}, so 2 × 2 = 4 sixth-power divisors.

both: a ∈ {0, 6}, b ∈ {0, 6} → 2 × 2 = 4

5STEP 5

Inclusion-exclusion: union = squares + cubes - overlap = 25 + 16 - 4 = 37.

|sq ∪ cu| = 25 + 16 - 4 = 37

6STEP 6

37 matches answer choice (C).

37 → (C)

Answer

Squares (25) and cubes (16) are both small subsets of the 10 × 10 = 100 total divisors, and they overlap on a tiny set (the 4 sixth powers). The union should sit between 25 and 25 + 16 = 41, and closer to 41 since the overlap is small. 37 lands right in that band.

💡Key takeaway

This AMC 10 problem only needs Grade 6 exponents and LCM you already know! Write 201⁹ = 3⁹ · 67⁹, so each divisor is 3^a · 67^b. Squares need even a, b (5 × 5 = 25); cubes need a, b multiples of 3 (4 × 4 = 16); both at once needs multiples of 6 (2 × 2 = 4). Add and subtract the overlap: 25 + 16 - 4 = 37, answer (C).

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