AMC 8 · 2022 · #3

Grade 4 number-theorycounting

Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration

prime-factorizationfactorssystematic-enumeration caseworksystematic-enumerationbound-inequality-then-enumerate ↑ Prerequisites: factorsmulti-digit-arithmetic

📏 Medium solution 💡 3 insights

📘 View easy version →

Problem

Count the number of triples of positive integers (a, b, c) with a ＜ b ＜ c whose product a · b · c equals 100.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

"How many ways" with a small finite product (100) is a textbook trigger for Tool #2 (Systematic List). To keep the listing organized and guarantee we miss nothing, we use Tool #7 (Identify Subproblems) to split the big question into cases by the smallest value a. Because a ＜ b ＜ c forces a³ ＜ a · b · c = 100, only a ∈ {1, 2, 3, 4} are even possible, and within those only divisors of 100 work — so the casework is tiny (a = 1, 2, 4), and inside each case the remaining list of factor pairs of $\frac{100}{a}$ is short enough to write out completely.

1STEP 1

Since a ＜ b and a ＜ c, multiplying gives a³ ＜ 100, so a is one of 1, 2, 3, 4.

a³ ＜ 100 → a ∈ {1, 2, 3, 4}

2STEP 2

Only divisors of 100 survive, and 3 is not one — so the possible smallest values narrow to 1, 2, 4.

a ∈ {1, 2, 3, 4} ∩ div(100) = {1, 2, 4}

3STEP 3

With a = 1, b · c = 100 and 1 ＜ b ＜ c gives 3 increasing triples: (1, 2, 50), (1, 4, 25), (1, 5, 20).

a = 1: (1, 2, 50), (1, 4, 25), (1, 5, 20) — 3 triples

4STEP 4

With a = 2, b · c = 50 and 2 ＜ b ＜ c leaves only (5, 10), so 1 triple: (2, 5, 10).

a = 2: (2, 5, 10) — 1 triple

5STEP 5

With a = 4, b · c = 25 needs 4 ＜ b ＜ c, but the only pair (5, 5) has b = c — so 0 triples.

a = 4: no triples

6STEP 6

Adding the cases: 3 + 1 + 0 = 4, matching choice (E).

3 + 1 + 0 = 4 → (E)

Answer

Sanity check each triple by multiplying: 1 · 2 · 50 = 100, 1 · 4 · 25 = 100, 1 · 5 · 20 = 100, 2 · 5 · 10 = 100 — all four check out, and each has strictly increasing values. The bound a³ ＜ 100 also guarantees we couldn't have missed an a ≥ 5 triple, so 4 is the complete count and answer (E) fits.

💡Key takeaway

This AMC 8 problem only needs Grade 4 factor pairs you already know — just split 100 into three growing pieces!

More like this