AMC 8 · 2020 · #19

Grade 4 number-theorycounting
divisibility-rulesdigit-sumdigit-constraintsmodular-arithmetic digit-constraintscaseworksystematic-enumeration ↑ Prerequisites: divisibility-rulesdigit-sum
📏 Medium solution 💡 4 insights
📘 View easy version →
Problem
A "flippy" number is one whose digits strictly alternate between two distinct digits — like 2020 or 37373, but not 3883 (two 3s in a row) or 123123 (three different digits). How many five-digit flippy numbers are multiples of 15?

Pick an answer.

(A)
3
(B)
4
(C)
5
(D)
6
(E)
8

AMC 8 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Identify Subproblems

The condition "divisible by 15" splits cleanly into two smaller, very different sub-conditions: divisible by 5 (a last-digit rule) and divisible by 3 (a digit-sum rule). Tool #7 (Identify Subproblems) lets us pin down one digit at a time — the 5-rule forces the value of a, then the 3-rule narrows down b. Once both digits are constrained, Tool #2 (Make a Systematic List) finishes the job: write the few remaining candidate numbers in order and count them. We deliberately avoid Tool #13 (Algebra) — divisibility rules and listing handle this with no equations needed.

1STEP 1

A five-digit flippy number must read ababa — positions 1, 3, 5 are a and positions 2, 4 are b — with a ≠ 0 and a ≠ b.

a b a b a, with a ≠ b and a ≠ 0
2STEP 2

Divisibility by 5 needs the last digit 0 or 5; that last digit is a and a ≠ 0, so a = 5 and the number is 5b5b5 (b ≠ 5).

last digit = a ∈ {0, 5}, and a ≠ 0, so a = 5
3STEP 3

The digits of 5b5b5 sum to 15 + 2b; since 15 is a multiple of 3, we only need 2b to be a multiple of 3 too.

5 + b + 5 + b + 5 = 15 + 2b, need 3 ∣ (15 + 2b), so 3 ∣ 2b
4STEP 4

2b is a multiple of 3 exactly when b is, so b ∈ {0, 3, 6, 9}, and every one of these already satisfies b ≠ 5.

b ∈ {0, 3, 6, 9}, and each satisfies b ≠ 5
5STEP 5

List them in order — 50505, 53535, 56565, 59595 — giving 4 flippy multiples of 15, which is choice (B).

b=0 → 50505; b=3 → 53535; b=6 → 56565; b=9 → 59595. Count = 4, so the answer is (B).
Answer
4
Quick sanity check: 50505 ÷ 15 = 3367, 53535 ÷ 15 = 3569, 56565 ÷ 15 = 3771, 59595 ÷ 15 = 3973 — all whole numbers, so each really is a multiple of 15. There is no fifth candidate to miss: any other b ∈ {0,…,9} either fails the digit-sum test (b ∈ {1,2,4,7,8}) or equals 5 (which would collapse the number to 55555, violating a ≠ b). The count 4 matches answer (B), well within the answer-choice range 3 to 8.
💡Key takeaway

This AMC 8 problem only needs Grade 4 divisibility rules — "ends in 0 or 5" for 5, and "digits add up to a multiple of 3" for 3 — that you already know!