AMC 10 · 2021 · #11

Grade 6 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration

modular-arithmeticplace-valuedivisibility-rulespolynomial-factoring identify-subproblemssystematic-enumeration ↑ Prerequisites: modular-arithmetic

📏 Medium solution 💡 2 insights

Problem

Treat 2021_b and 221_b as base-b numerals. For which choice of b from {3, 4, 6, 7, 8} is the subtraction 2021_b - 221_b NOT a multiple of 3?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Tool #7 (Subproblems) breaks the question into two clean pieces: (a) write each base-b numeral as a polynomial in b, (b) subtract and factor. Tool #9 (Easier Related Problem) lets us check one base by hand first (b=3) to confirm the formula. Then Tool #3 (Eliminate) plugs every choice into the simplified expression and Tool #6 (Guess and Check) just reads off divisibility by 3. No algebra heavier than expanding place values is needed.

1STEP 1

Expand by place value: 2021_b = 2b³ + 2b + 1 and 221_b = 2b² + 2b + 1.

2021_b = 2b³ + 2b + 1, 221_b = 2b² + 2b + 1

2STEP 2

Subtract: the +2b and +1 terms cancel, leaving 2021_b - 221_b = 2b³ - 2b².

2021_b - 221_b = (2b³ + 2b + 1) - (2b² + 2b + 1) = 2b³ - 2b²

3STEP 3

Factor out the common 2b²: 2b³ - 2b² = 2b²(b - 1) — now the divisibility structure is visible.

2b³ - 2b² = 2b²(b - 1)

4STEP 4

2 shares no factor with 3, so 2b²(b-1) is a multiple of 3 exactly when b ≡ 0 or 1 (mod 3).

3 ∣ 2b²(b-1) ⇔ b ≡ 0 or 1 (mod 3)

5STEP 5

Sort the choices mod 3: 3, 6 ≡ 0; 4, 7 ≡ 1; only 8 ≡ 2, so 8 is the base whose difference is not a multiple of 3.

3, 6 ≡ 0; 4, 7 ≡ 1; 8 ≡ 2 (mod 3)

6STEP 6

Check b = 8 directly: 2 · 8² · 7 = 896, and 896 ÷ 3 leaves remainder 2 — not an integer, confirming (E).

b = 8: 2 · 8² · 7 = 896, 896 mod 3 = 2 → (E)

Answer

Sanity-check a 'good' base too. Try b = 4: 2 · 16 · 3 = 96 = 3 · 32, divisible by 3. Try b = 6: 2 · 36 · 5 = 360 = 3 · 120, divisible by 3. The only failure among the five choices really is b = 8, so (E) holds up. The structure also matches intuition: a factor of (b-1) means whenever b leaves remainder 1 when divided by 3, the difference is automatically a multiple of 3; and a factor of b² catches the b ≡ 0 case.

💡Key takeaway

This AMC 10 problem only needs Grade 6 place value and factoring you already know! Once you write 2021_b - 221_b = 2b²(b-1), the question becomes 'which b gives a non-multiple of 3?' — that's just sorting 3, 4, 6, 7, 8 by remainder mod 3. Only 8 leaves remainder 2, so the answer is (E).

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