AMC 10 · 2022 · #17

Grade 7 arithmetic

fraction-decimal-conversiondigit-decompositiondivisibility-rules convert-to-algebrasystematic-enumeration ↑ Prerequisites: fraction-decimal-conversion

📏 Medium solution 💡 3 insights

Problem

Count three-digit positive integers abc whose digits a, b, c are all nonzero and satisfy 0.abc =

\frac{1}{3}

(0.a + 0.b + 0.c), where the bars mean infinite repetition.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Convert to Algebra

First reduce the decimal equation to an integer relation. Repeating-decimal-to-fraction conversion (Tool #13 Convert to Algebra) turns the unwieldy equation into 7a = 3b + 4c. Once we have a clean equation in three nonzero digits, Tool #2 (Systematic List) sweeps every value of a from 1 to 9 and counts (b, c) pairs satisfying 3b + 4c = 7a with b, c ∈ {1, …, 9}. Tool #9 (Easier Problem) shows up as the observation: trying a=b=c first reveals a whole family of solutions and orients the search.

1STEP 1

Rewrite each bar as a fraction — 0.d = $\frac{d}{9}$ and 0.abc = $\frac{100a+10b+c}{999}$ — then substitute into the equation.

\frac{100a+10b+c}{999}

\frac{1}{3}

\frac{a+b+c}{9}

\frac{a+b+c}{27}

2STEP 2

Multiply by 999 = 37 · 27, expand 37(a+b+c), and divide by 9 to reach 7a = 3b + 4c.

100a + 10b + c = 37(a+b+c) ⟹ 63a = 27b + 36c ⟹ 7a = 3b + 4c

3STEP 3

Try a = b = c: 7a = 3a + 4a holds for every digit, so (1,1,1)…(9,9,9) give 9 solutions.

(a,a,a) for a ∈ {1, …, 9} — 9 solutions.

4STEP 4

Fix a and b, and the last digit is forced: c = $\frac{7a - 3b}{4}$ must be an integer in {1, …, 9}; scan for valid off-diagonal triples.

c =

\frac{7a - 3b}{4}

, scan a = 1, …, 9 and b = 1, …, 9.

5STEP 5

The scan yields (4,8,1), (5,1,8), (5,9,2), (6,2,9), each satisfying 7a = 3b + 4c — 4 solutions.

Off-diagonal: {(4,8,1), (5,1,8), (5,9,2), (6,2,9)} — 4 solutions.

6STEP 6

Add both counts: 9 + 4 = 13 ordered triples, each a distinct three-digit integer abc — choice (D).

9 + 4 = 13 → (D)

Answer

Did the off-diagonal search miss anything? Bounds: b, c ∈ {1, …, 9} gives 3b + 4c ∈ {7, …, 63}, so 7a ∈ {7, …, 63} allows a ∈ {1, …, 9} — all in range. The shift identity (a, b, c) → (a, b+4, c-3) preserves 3b + 4c, so from each diagonal (a,a,a) we can step at most a few times before leaving {1, …, 9}. Stepping up from (4,4,4) reaches (4,8,1) (valid) then (4,12,-2) (invalid). Stepping up from (5,5,5) reaches (5,9,2) (valid); stepping down reaches (5,1,8) (valid). Stepping down from (6,6,6) reaches (6,2,9) (valid). Stepping from (1,1,1), (2,2,2), (3,3,3), (7,7,7), (8,8,8), (9,9,9) leaves the digit range immediately. Total off-diagonal: exactly 4, confirming the count.

💡Key takeaway

Every repeating decimal is just a fraction over 9, 99, 999, …. Convert and simplify, and the messy decimal equation collapses to 7a = 3b + 4c. The nine diagonal triples (a, a, a) always work; a quick scan over a turns up four more — (4,8,1), (5,1,8), (5,9,2), (6,2,9) — for a total of (D) 13.