Problem 15 · 2024 AMC 8
Hard
Number Theory
factorizationcaseworkwork-backward
Let the letters F, L, Y, B, U, G represent distinct digits. Suppose FLYFLY is the greatest number that satisfies the equation
8 · FLYFLY = BUGBUG.
What is the value of FLY + BUG?
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Answer: C — 1107.
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Hint 1 of 2
A number that repeats a 3-digit block, like ABCABC, isn't random — it's the block times something. What number, when multiplied by 123, gives 123123?
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Hint 2 of 2
Key fact: ABCABC = 1001 × ABC. So FLYFLY = 1001·FLY and BUGBUG = 1001·BUG, and the whole equation collapses to 8 · FLY = BUG.
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Approach: strip the repeat, then maximize digit-by-digit
- The repeated block is the doorway: ABCABC = ABC×1000 + ABC = 1001·ABC. So FLYFLY = 1001·FLY, BUGBUG = 1001·BUG, and dividing both sides by 1001 leaves the tiny equation 8 · FLY = BUG.
- BUG is still only 3 digits, so 8 · FLY < 1000 → FLY ≤ 124. That forces F = 1, and the tens digit L ≤ 2 (if L = 3, 8·13Y already exceeds 1040).
- We want the GREATEST FLY, so push digits up. L = 2 (1 is taken by F). Now test the units Y from high down, needing all six digits F,L,Y,B,U,G distinct: Y = 4 gives 8·124 = 992 (repeated 9, fails); Y = 3 gives 8·123 = 984, digits {1,2,3,9,8,4} all different ✓.
- FLY + BUG = 123 + 984 = 1107. You'll see it again: 1001 = 7×11×13 is the workhorse behind every ABCABC pattern — spotting the repeated block lets you divide a 6-digit monster down to 3 digits.
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