AMC 8 · 2000 · #11

Grade 4 number-theory
Archetype Casework by Position / Vertex Type └ Sub-archetype Digit Place Position
divisibility-rulesdigit-constraintssystematic-enumeration caseworksystematic-enumerationdigit-constraints ↑ Prerequisites: divisibility-rulesmulti-digit-arithmetic
📏 Medium solution 💡 3 insights
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Problem
The number 64 is divisible by its units digit because 64 ÷ 4 = 16. How many whole numbers strictly between 10 and 50 share this property — that is, the number is divisible by whichever digit happens to sit in its units place?

Pick an answer.

(A)
15
(B)
16
(C)
17
(D)
18
(E)
20

AMC 8 2000 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

There are only 39 numbers in the range, so a brute-force check is plausible — but smarter is Tool #2 (Make a Systematic List) grouped by the units digit u. Each u from 1 to 9 gives at most 4 candidates (1u, 2u, 3u, 4u), and the question "is N divisible by u?" is identical for the whole group. Tool #3 (Eliminate Possibilities) then knocks out non-multiples within each group with a single divisibility rule. Splitting by units digit turns one 39-item check into nine tiny checks of at most 4 each — much easier to count without errors.

1STEP 1

Fix the units digit u (skip 0). For each u the candidates are always 1u, 2u, 3u, 4u — four numbers — and we ask which are multiples of u.

Candidates for units digit u: {10+u, 20+u, 30+u, 40+u}
2STEP 2

Easy digits pass all four: u=1 (every number is divisible by 1), u=2 (ends in 2, even), u=5 (ends in 5) — that is 12 winners.

u=1: {11,21,31,41} → 4. u=2: {12,22,32,42} → 4. u=5: {15,25,35,45} → 4.
3STEP 3

Medium cases: u=4 keeps 24 and 44; u=3 only 33; u=6 only 36; u=8 only 48 — that adds 5 more.

u=4: {24,44} → 2. u=3: {33} → 1. u=6: {36} → 1. u=8: {48} → 1.
4STEP 4

Hard but empty: multiples of 7 and of 9 never end in 7 or 9 in range — the next would be 77 and 99 — so both give 0.

u=7: { } → 0. u=9: { } → 0.
5STEP 5

Add every case: 4+4+1+2+4+1+0+1+0 = 17, choice (C).

4_u=1 + 4_u=2 + 1_u=3 + 2_u=4 + 4_u=5 + 1_u=6 + 0_u=7 + 1_u=8 + 0_u=9 = 17 → (C)
Answer
17
Write the 17 winners out by tens digit to confirm none was missed or counted twice. Tens 1: 11, 12, 15 (3 numbers — 14 fails because 14 / 4 is not whole, 13, 16, 17, 18, 19 also fail). Tens 2: 21, 22, 24, 25 (4). Tens 3: 31, 32, 33, 35, 36 (5). Tens 4: 41, 42, 44, 45, 48 (5). Total 3 + 4 + 5 + 5 = 17 — matches the count by units digit. Order-of-magnitude check: each units digit averages roughly 17 / 9 ≈ 1.9 winners out of 4 candidates, about 47%, which is exactly what you would expect when half the digits divide "easy" and half divide almost nothing.
💡Key takeaway

Sort the 39 numbers by their last digit and check each group with one divisibility rule. Three easy digits (1, 2, 5) give 12 winners on their own, and the rest contribute 5 more — total 17, answer (C).

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