AMC 10 · 2020 · #24

Grade 7 counting

Archetype Branching State Enumeration └ Sub-archetype Numeric Operation Tree

prime-factorizationrecursive-sequencecombinations-basicdivisor-count easier-related-problemcaseworkidentify-subproblems ↑ Prerequisites: prime-factorizationcombinations-basic

📏 Long solution 💡 3 insights

Problem

Write 96 as a product of factors each greater than one, counting order as significant so that two times three differs from three times two. Find how many such factorizations there are.

Pick an answer.

(A)

112

(B)

128

(C)

144

(D)

172

(E)

184

How to solve

Strategy Solve an Easier Related Problem

Tool #9 (Easier Problem): the verifier D(6) = 3 is given — anchor on this small case. Tool #2 (Systematic List): casework on k = number of factors (k = 1, 2, 3, 4, 5, 6) — at most 6 factors since the smallest factor is 2 and 2⁶ = 64 ＜ 96 ＜ 128 = 2⁷. Tool #7 (Subproblems): for each k, separately count (a) where to place the single 3 and (b) how to distribute the remaining powers of 2 among the factors. Tool #5 (Pattern): the per-k count factors into k · C(5, k-1) via stars-and-bars. Tool #13 (Algebra): sum the closed form using Σ k C(n, k-1) = (n+2) 2ⁿ⁻¹ — actually here a clean direct sum. Tool #6 (Guess and Check) cross-verifies via D(6) = 3.

1STEP 1

Describe by prime exponents

Each factor is two exponents.

f_i = 2^a_i 3^b_i, a_i + b_i ≥ 1, Σ a_i = 5, Σ b_i = 1

2STEP 2

Place the three

The three goes in one slot.

Place the 3: k choices

3STEP 3

Distribute the twos

No other slot may be left empty.

Distribute the 2's: C(5, k - 1)

4STEP 4

Count by the number of factors

Count each factor-count case.

N_k = k C(5, k-1): 1, 10, 30, 40, 25, 6

5STEP 5

Add them up

Adding gives 112.

D(96) = 1 + 10 + 30 + 40 + 25 + 6 = 112 → (A)

6STEP 6

Sanity-check on a small number

Check the method on a small number.

D(6) = 1 + 2 = 3 ✓

Answer

112

Two layers of sanity. (1) The D(6) = 3 check matches the problem's worked example, confirming the formula N_k = k C(5, k-1) generalizes correctly. (2) The total 112 is sandwiched between the choices 100 (too low, would miss the k = 4 bulk) and 144 (too high). Distribution of N_k peaks at k = 4 with 40, consistent with 96 having 5 twos and 1 three (typical factorization length ∼ log₂ 96 ≈ 6.6 at minimum factor size 2). Choice (A) confirmed.

💡Key takeaway

This AMC 12 problem only needs Grade 7 counting: 96 = 2⁵ · 3, so for each number of factors k pick where the 3 lives (k ways) and distribute the 5 twos among the k slots (C(5, k-1) ways via stars and bars) — sum k C(5, k-1) from k = 1 to 6: 1 + 10 + 30 + 40 + 25 + 6 = 112.