AMC 8 · 2014 · #21

Grade 6 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

divisibility-rulesdigit-summodular-arithmetic digit-constraintsmodular-arithmeticsystematic-enumeration ↑ Prerequisites: divisibility-rulesmodular-arithmetic

📏 Medium solution 💡 3 insights

Problem

Two 7-digit numbers, 74A52B1 and 326AB4C, are each multiples of 3. The unknown digits A, B, C are each between 0 and 9. Which value among the choices could C be?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Two unknowns (A, B) appear in BOTH numbers, but only C shows up in the second one. Tool #7 (Identify Subproblems) handles this by splitting the question into two clean jobs: (1) extract what the first number's divisibility rule says about A + B, then (2) plug that fact into the second number's rule to pin down C alone. Tool #3 (Eliminate Possibilities) is what we use at the end — scan the five answer choices and keep only the one that fits the rule we derived for C. Tool #6 (Guess and Check) is the natural fallback: try each answer choice with the divisibility rule and see which one passes.

1STEP 1

First number 74A52B1: its known digits sum to 19, so 19 + A + B is a multiple of 3, which means 1 + A + B is a multiple of 3.

7 + 4 + 5 + 2 + 1 = 19, 19 + A + B = 18 + (1 + A + B) → 1 + A + B is a multiple of 3

2STEP 2

Second number 326AB4C: its known digits sum to 15 (already a multiple of 3), so the leftover A + B + C is a multiple of 3.

3 + 2 + 6 + 4 = 15, 15 + A + B + C is a multiple of 3 → A + B + C is a multiple of 3

3STEP 3

Combine: Step 1 gives A + B ≡ 2 (mod 3) and Step 2 gives A + B + C ≡ 0 (mod 3); subtracting isolates C ≡ 1 (mod 3).

A + B ≡ 2 (mod 3), A + B + C ≡ 0 (mod 3) → C ≡ 1 (mod 3)

4STEP 4

Test each choice mod 3 for remainder 1: 1→1, 2→2, 3→0, 5→2, 8→2 — so only 1 leaves remainder 1.

1 ÷ 3 → r = 1, 2 ÷ 3 → r = 2, 3 ÷ 3 → r = 0, 5 ÷ 3 → r = 2, 8 ÷ 3 → r = 2

5STEP 5

Only C = 1 leaves remainder 1 on division by 3, so the answer is C = 1, choice (A).

C = 1 → (A)

Answer

Test the answer with a concrete A and B. Pick A = 1, B = 1, so A + B = 2 (leaves remainder 2 mod 3, matching Step 1). First number: 7411521. Digit sum = 7+4+1+1+5+2+1 = 21 = 3 × 7, divisible by 3. Second number with C = 1: 3261141. Digit sum = 3+2+6+1+1+4+1 = 18 = 3 × 6, divisible by 3. Both checks pass, so C = 1 really works and choice (A) is correct.

💡Key takeaway

This AMC 8 problem only needs the Grade 4 divisibility-by-3 trick (add the digits) plus a Grade 6 "which value fits both rules?" check that you already know!

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