Competition · AMC preparation · step 4 of 4

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

Use the first divisibility rule

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

Use the second divisibility rule

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 the two facts

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)

Subtracting one constraint from another to isolate a single variable is a Grade 6 "use equations as questions about which values work" move — exactly the substitution idea.

The digit C must leave a remainder of 1 when divided by 3.

▸ Why?

The first number 74A52B1 is a multiple of 3, and that pins the pair A+B down to leaving a remainder of 2 when divided by 3.

▸ Why?

A whole number is a multiple of 3 exactly when its digit sum is, because each place value 1, 10, 100, … leaves remainder 1 under division by 3, so the number and its digit sum sit at the same spot in the repeating length-3 remainder cycle; thus 7+4+A+5+2+B+1 = 19+A+B must be a multiple of 3.

A cycle returns after a full periodA fixed repeating cycle of length p comes back to the same position every p steps.

▸ Why?

Since 19 breaks into 18 plus 1 with 18 already a multiple of 3, only the leftover 1+A+B can make up the rest, so 1+A+B is a multiple of 3 and A+B lands one short of a multiple of 3 — remainder 2.

Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.

▸ Why?

The second number 326AB4C is a multiple of 3, and that forces A+B+C to be a multiple of 3.

▸ Why?

By the same digit-sum test, the second number's digit sum 3+2+6+A+B+4+C = 15+A+B+C must be a multiple of 3.

A cycle returns after a full periodA fixed repeating cycle of length p comes back to the same position every p steps.

▸ Why?

Since 15 is already a multiple of 3, removing it leaves A+B+C as the part that must itself be a multiple of 3.

Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.

▸ Why?

Writing C as (A+B+C) - (A+B) subtracts a remainder-2 amount from a multiple of 3, and because remainders under division by 3 run in the repeating cycle 0, 1, 2, stepping back by 2 from 0 lands on 1.

A cycle returns after a full periodA fixed repeating cycle of length p comes back to the same position every p steps.

6.EE.B.5Identify Subproblems

4STEP 4

Test the five choices

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

Pick the one that works

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!

Use the first divisibility rule
Use the second divisibility rule
Combine the two facts
Test the five choices
Pick the one that works

A parent dashboard for the family lives at sensimlab.com.

Use the first divisibility rule

Use the second divisibility rule

Combine the two facts

Test the five choices

Pick the one that works

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