AMC 8 · 2018 · #18

Grade 6 number-theory

Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation

prime-factorizationdivisor-countexponents identify-subproblems ↑ Prerequisites: prime-factorizationexponents

📏 Medium solution 💡 3 insights

Problem

Count every positive whole number that divides 23,232 evenly (including 1 and 23,232 itself), then pick the matching answer choice.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Listing all divisors of 23,232 by hand is hopeless. Tool #7 (Identify Subproblems) splits the job into two clean pieces: (a) find the prime factorization of 23,232, then (b) turn that factorization into a divisor count. To justify the counting step we lean on Tool #9 (Easier Problem) and Tool #5 (Pattern): try the same procedure on a small number like 12 = 2² · 3, list its 6 divisors, notice that 6 = (2+1)(1+1), and generalize. That's much friendlier than memorizing a formula and keeps the reasoning at an elementary level.

1STEP 1

23,232 is even, so keep dividing by 2 until it turns odd — that happens six times, leaving 23232 = 2⁶ · 363.

23232 ÷ 2 = 11616, 11616 ÷ 2 = 5808, 5808 ÷ 2 = 2904, 2904 ÷ 2 = 1452, 1452 ÷ 2 = 726, 726 ÷ 2 = 363. Six divisions, so 23232 = 2⁶ · 363.

2STEP 2

The odd part 363 = 3 × 121 = 3 × 11² (digit sum 12 shows 3 divides it), so 23232 = 2⁶ · 3¹ · 11².

363 = 3 × 121 = 3 × 11², so 23232 = 2⁶ · 3¹ · 11².

3STEP 3

Check on 12 = 2² · 3: it has six divisors, and (2+1)(1+1) = 6 — each prime's power is chosen independently, so the counts multiply.

12 = 2² · 3¹ → divisors = (2+1)(1+1) = 6 ✓

4STEP 4

For 23232 = 2⁶ · 3¹ · 11², the exponents give 7, 2, and 3 independent choices, so the divisor count is (6+1)(1+1)(2+1) = 42.

number of divisors = (6+1)(1+1)(2+1) = 7 · 2 · 3 = 42

5STEP 5

The count 42 matches answer choice (E).

42 → (E)

Answer

Quick sanity check on the prime factorization: 2⁶ = 64, 64 × 3 = 192, and 192 × 121 = 192 × 121 = 23,232 (since 192 × 100 = 19,200 and 192 × 21 = 4,032, summing to 23,232). Also, 42 is the largest answer choice — that fits a number with three different prime factors and the relatively large exponent 6 on the prime 2. A number this size with so much 2-content should have many divisors, so 42 is far more reasonable than the smaller choices like 9 or 12.

💡Key takeaway

This AMC 8 problem only needs Grade 6 whole-number exponents and a sprinkle of Grade 4 prime-factor know-how that you already have!

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