AMC 8 · 2025 · #22

Grade 4 number-theoryalgebra

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

factorsdivisibility-ruleslinear-diophantineequal-spacing systematic-enumerationconvert-to-algebrapattern-recognition ↑ Prerequisites: factorsdivisibility-rules

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Problem

A row of 35 coat hooks must be filled so the empty hooks are arranged in equal groups: one group before the first coat, one group after the last coat, and one equal group between every pair of adjacent coats. Both the number of coats and the size of each empty-hook group must be at least 1. How many different values for the number of coats are possible?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

The question asks "how many different numbers of coats" — a perfect trigger for Tool #2 (Systematic List). For each possible coat count c, the empty hooks left over are 35 - c, and they must split evenly into c+1 equal groups. So for each c we just check whether c+1 divides 35 - c evenly — pure Guess & Check (Tool #6) on a small, finite list. As we walk the list we notice a clean divisibility pattern (Tool #5): the leftover 35 - c equals 36 - (c+1), so (c+1) divides 35 - c exactly when (c+1) divides 36. That insight turns the search into "which divisors of 36 are ≥ 2?" — no algebra needed.

1STEP 1

For c coats, split the 35 - c empty hooks into c+1 equal groups: each holds k = (35 - c)/(c + 1), which must be a whole number ≥ 1.

k = (35 - c)/(c + 1), with k ≥ 1 whole

2STEP 2

List candidates c from small to large, computing (35 - c) ÷ (c + 1) each time; c can be at most 17, since c + (c+1) ≤ 35.

Try c = 1, 2, 3, …, 17

3STEP 3

Check each c: it works when the division gives a whole number ≥ 1. The winners are c = 1, 2, 3, 5, 8, 11, 17.

c=1:

\frac{34}{2}

=17 ✓; c=2:

\frac{33}{3}

=11 ✓; c=3:

\frac{32}{4}

=8 ✓; c=4:

\frac{31}{5}

no; c=5:

\frac{30}{6}

=5 ✓; c=6:

\frac{29}{7}

no; c=7:

\frac{28}{8}

no; c=8:

\frac{27}{9}

=3 ✓; c=9,10: no; c=11:

\frac{24}{12}

=2 ✓; c=12…16: no; c=17:

\frac{18}{18}

=1 ✓

4STEP 4

Rewrite 35 - c = 36 - (c + 1), so the quotient is 36/(c+1) - 1 — a whole number exactly when c + 1 divides 36.

(35 - c)/(c + 1) = 36/(c+1) - 1

5STEP 5

From c + 1 ∈ {2, 3, 4, 6, 9, 12, 18} we get c ∈ {1, 2, 3, 5, 8, 11, 17} — 7 different values, choice (D).

|{1, 2, 3, 5, 8, 11, 17}| = 7 → (D)

Answer

Sanity-check the two extremes. With c = 1 coat, the gap size is k = 17: pattern is 17 empty + 1 coat + 17 empty = 35 hooks. With c = 17 coats, k = 1: pattern is 1-empty-1-coat repeated, total 1 + 17 · 2 = 35. Both check out exactly, so the answer set {1, 2, 3, 5, 8, 11, 17} is real and complete. Also 7 is choice (D) and isn't the trap value 9 (= number of divisors of 36 before we throw out the 1-and-36 edge cases) or 5 (the symmetric solution count if we mistakenly only counted up to c = 5). The two edge-case eliminations exactly account for the difference 9 - 2 = 7.

💡Key takeaway

This AMC 8 problem only needs the Grade 4 "factor pairs of 36" skill you already know — no algebra required!

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