AMC 8 · 2024 · #15

Grade 4 algebra

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

place-valuemulti-digit-arithmeticdivisibility-rules digit-constraintssystematic-enumeration ↑ Prerequisites: multi-digit-arithmeticplace-value

📏 Medium solution 💡 4 insights

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Problem

Six different letters F, L, Y, B, U, G each stand for a different digit (0-9). The six-digit number FLYFLY (the three-digit block FLY written twice) is the greatest number for which 8 · FLYFLY = BUGBUG. Find FLY + BUG (treating FLY and BUG as three-digit numbers).

Pick an answer.

(A)

1089

(B)

1098

(C)

1107

(D)

1116

(E)

1125

AMC 8 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Guess and Check

The six-digit number FLYFLY looks scary, but Tool #9 (Easier Related Problem) cracks it: try a small three-digit number first, say 123, and compute 123,123. Notice 123,123 = 1000 × 123 + 123 = 1001 × 123 — the same trick works for every three-digit block. So the messy six-digit equation simplifies to 8 · FLY = BUG. From there the task becomes: find the greatest three-digit FLY such that 8 · FLY is also a three-digit number and all six digits are different. That is a perfect job for Tool #6 (Guess and Check) — start with the biggest legal FLY and step down one at a time. Tool #2 keeps the descending search organized, and Tool #3 confirms the final sum sits in the answer list.

1STEP 1

A block written twice is 1001 × the block, so FLYFLY = 1001 × FLY and BUGBUG = 1001 × BUG; dividing by 1001 gives 8 · FLY = BUG.

254,254 = 254 × 1001 → 8 · FLY = BUG

2STEP 2

Since BUG = 8 · FLY must stay a three-digit number, 8 · FLY ≤ 999, so FLY ≤ 999 ÷ 8 = 124.875 — meaning FLY ≤ 124.

FLY ≤

\frac{999}{8}

= 124.875 → FLY ≤ 124

3STEP 3

Start at the top, FLY = 124: BUG = 8 × 124 = 992, but its digits repeat 9 and clash with L = 2, so 124 is rejected.

FLY = 124 → BUG = 8 × 124 = 992, digits = {1,2,4,9,9,2} (repeats)

4STEP 4

Step down to FLY = 123: BUG = 8 × 123 = 984, and the six digits {1,2,3,9,8,4} are all different — the greatest valid FLY.

FLY = 123 → BUG = 8 × 123 = 984, digits = {1,2,3,9,8,4} (all distinct) ✓

5STEP 5

Add the two: FLY + BUG = 123 + 984 = 1107, which matches choice (C).

123 + 984 = 1107 → (C)

Answer

1107

Double-check by plugging FLY = 123 back into the original six-digit equation: the left side is 8 · 123,123 = 984,984, and the right side BUGBUG = 984,984 — exact match. The six digits {F, L, Y, B, U, G} = {1, 2, 3, 9, 8, 4} are all different. And FLY = 123 being one short of the arithmetic ceiling 124 is reasonable, since exactly one tiny case (124) had to be skipped for the distinctness rule. The sum 1107 sits squarely in the middle of the answer choices, which is also where guessing-and-checking with a 'first failure, then success' pattern tends to land.

💡Key takeaway

This AMC 8 problem only needs Grade 4 multi-digit place value, multiplication, division, and addition you already know!

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