2024 AMC 8 Problem 15

Attempt Problem 15 of the 2024 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

15.

Let the letters F,F, L,L, Y,Y, B,B, U,U, GG represent distinct digits. Suppose FLYFLY\text{FLYFLY} is the greatest number that satisfies the equation 8⋅FLYFLY=BUGBUG. 8 \cdot \text{FLYFLY} = \text{BUGBUG}. What is the value of FLY+BUG?\text{FLY} + \text{BUG}?
10891089
10981098
11071107
11161116
11251125
Answer: C
Concepts:cryptarithmplace valueoptimization
Small hint:
FLYFLY=1001⋅FLY\text{FLYFLY}=1001\cdot\text{FLY}
Big hint:
The equation reduces to 8⋅FLY=BUG8\cdot\text{FLY}=\text{BUG}
Video solution:
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Written solution:
Firstly, note that FLYFLY=1001(FLY)\text{FLYFLY} = 1001(\text{FLY}) and, similarly, BUGBUG=1001(BUG)\text{BUGBUG}=1001(\text{BUG}) so the equation can be simplified to 8⋅FLY=BUG.8 \cdot \text{FLY} = \text{BUG}.
Because BUG\text{BUG} has three digits, FLY≤124.\text{FLY}\le124. The largest possible three-digit number at most 124124 with distinct digits is 124,124, but 8⋅124=9928\cdot124=992 repeats a digit and also reuses the digit 2.2.
The next candidate is 123,123, and 8⋅123=984.8\cdot123=984. The six digits 1,1, 2,2, 3,3, 9,9, 8,8, and 44 are all distinct, so 123123 is the greatest possible value of FLY.\text{FLY}.
Hence, FLY+BUG=123+984=1107. \begin{gathered} \text{FLY} + \text{BUG} = 123 + 984 \\ = 1107. \end{gathered}
Thus, C is the correct answer.
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