AMC 10 · 2021 · #13

Grade 6 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

base-conversionsystems-of-equationsplace-valuepolynomial-roots convert-to-algebraidentify-subproblems ↑ Prerequisites: place-valuesystems-of-equations

📏 Medium solution 💡 3 insights

Problem

Find a positive integer base n and a digit d so that (i) the three-digit numeral 32d read in base n equals 263 in base 10, and (ii) the three-digit numeral 324 read in base n equals the four-digit numeral 11d1 read in base 6. Then report n + d.

Pick an answer.

(A)

~10

(B)

~11

(C)

~13

(D)

~15

(E)

~16

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

How to solve

Strategy Convert to Algebra

Tool #13 (Algebra): translate each numeral by its place-value formula. Tool #7 (Subproblems) sets two equations 3n² + 2n + d = 263 and 3n² + 2n + 4 = 253 + 6d. Tool #7 again: subtract one from the other to kill 3n² + 2n, leaving a linear equation in d that gives d = 2. Tool #6 (Guess and Check): plug small bases into 3n² + 2n = 261; n = 9 works on the first try. Tool #3 (Eliminate) matches n + d = 11 to the choices.

1STEP 1

Expand by place value: 32d_n = 3n² + 2n + d, 324_n = 3n² + 2n + 4, and 11d1 in base 6 = 253 + 6d.

32d_n = 3n² + 2n + d, 324_n = 3n² + 2n + 4, 11d1₆ = 253 + 6d

2STEP 2

Two equations: (E₁) 3n² + 2n + d = 263 from (i), and (E₂) 3n² + 2n + 4 = 253 + 6d from (ii).

(E₁): 3n² + 2n + d = 263 (E₂): 3n² + 2n + 4 = 253 + 6d

3STEP 3

Subtract (E₂) from (E₁): 3n² + 2n cancels, leaving d - 4 = 10 - 6d, so 7d = 14 and d = 2.

d - 4 = 10 - 6d → 7d = 14 → d = 2

4STEP 4

Put d = 2 into (E₁): 3n² + 2n = 261. Test small bases — n = 8 gives 208, n = 10 gives 320, and n = 9 gives exactly 261.

n = 9: 3 · 9² + 2 · 9 = 243 + 18 = 261 ✓

5STEP 5

Check: 32d₉ = 322₉ = 263 ✓, and 324₉ = 265 = 1121₆ ✓; also d = 2 ≤ 5 and digits 3, 2, 4 are all under 9.

322₉ = 263, 324₉ = 265 = 1121₆

6STEP 6

Add: n + d = 9 + 2 = 11, which is choice (B).

n + d = 9 + 2 = 11 → (B)

Answer

~11

All three numerals come out integer and consistent: 322₉ = 263 matches the given 263. 324₉ = 265 = 1121₆ — both four- and three-digit numerals agree. Digit constraints hold: d = 2 is a legal digit in base 6 and base 9; the digits 3, 2, 4 in 324₉ are all less than 9. The base n = 9 is reasonable: it sits comfortably between 5 (the smallest base where 32d makes sense) and 10 (where 32d_n = 263 would force d = 263 - 320 ＜ 0).

💡Key takeaway

This AMC 10 problem only needs Grade 6 one-variable equations you already know! Write each numeral by place value: 32d_n = 3n² + 2n + d = 263 and 324_n = 253 + 6d. Subtract to kill the n terms — you get 7d = 14, so d = 2. Then 3n² + 2n = 261 checks at n = 9. So n + d = 11, answer (B).

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