AMC 10 · 2012 · #11

Grade 8 algebra

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

base-conversionplace-valuequadratic-equationsdigit-constraints convert-to-algebrabound-inequality-then-enumeratesystematic-enumeration ↑ Prerequisites: base-conversionplace-value

📏 Long solution 💡 3 insights

Problem

Three numerals in different bases satisfy one addition, and two bases are consecutive. Find their sum.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Convert to Algebra

Tool #13 (Algebra): every numeral has a place-value formula, so rewriting all three in base ten turns the puzzle into one equation in A and B. Tool #15 (Reorganize): that equation is only quadratic in A but linear in B, so solving it for B instead of for A gives a clean formula B = (A-4)(A+1)/2. Tool #14 (Extreme Principle): that formula grows much faster than A+1, which caps A from above and leaves only a handful of cases. Tool #3 (Eliminate) and #6 (Check): the digit rules and a direct plug-back kill the survivors that do not work, and confirm the one that does.

1STEP 1

Read the digits before the algebra

The digits alone put a floor under each base.

A ≥ 4, B ≥ 5, A+B ≥ 10

2STEP 2

Rewrite all three numerals in base ten

Rewriting in base ten gives one equation.

A² + 3A + 2 + 4B + 3 = 6A + 6B + 9

3STEP 3

Solve for B, not for A

Solving for the linear base is easier.

2B = A² - 3A - 4 ⟹ B = (A-4)(A+1)/2

4STEP 4

Compare the forced B against A+1

Comparing it with the neighbour is one difference.

B - (A+1) = (A+1)(A-6)/2

5STEP 5

The gap caps A at 6

That difference caps the first base.

A ≥ 7 → B ＞ A+1 → not consecutive. So A ∈ {4, 5, 6}

6STEP 6

Test the three survivors

Only three candidates survive.

A=4 → B=0 ×, A=5 → B=3 ×, A=6 → B=7 ✓

7STEP 7

Plug back in to confirm

Substituting back confirms 13, choice (C).

132₆ + 43₇ = 56 + 31 = 87 = 78 + 9 = 69₁₃ → A+B = (C) 13

8STEP 8

Why no other choice can work

The pattern shows no other choice can work.

A=5 → 3, A=6 → 7, A=7 → 12, A=8 → 18, A=9 → 25

Answer

The arithmetic closes exactly: 56 + 31 = 87 and 69₁₃ = 87, so the found pair satisfies the original equation, not merely a rearranged one. Every legality condition holds too — 6 ≥ 4 for the digit 3, 7 ≥ 5 for the digit 4, and 13 ≥ 10 for the digit 9 — which matters because the digit 9 is exactly what makes choice (A) impossible no matter what the algebra says. Size-wise the answer is believable: 132_A is roughly A², while the right side is only about 6(A+B) ≈ 12A, so A has to be near 12-ish divided by A's own growth, i.e. small — and A=6 is right in that window. Uniqueness does not lean on the answer list: the gap identity B - (A+1) = (A+1)(A-6)/2 forces A ≤ 6 on its own, and only A=6 then produces a consecutive partner.

💡Key takeaway

Write each numeral out by place value, then solve for the variable that appears only to the first power — B = (A-4)(A+1)/2 — and "consecutive" squeezes A down to 6, so A+B = 13, choice (C).

Read the digits before the algebra
Rewrite all three numerals in base ten
Solve for B, not for A
Compare the forced B against A+1
The gap caps A at 6
Test the three survivors
Plug back in to confirm
Why no other choice can work