2020 AIME I Problems/Problem 3

Revision as of 16:03, 12 March 2020 by Jhawk0224 (talk | contribs) (Solution)

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Problem

Solution

Since $a$, $b$, and $c$ are digits in base eight, they are all 7 or less. Now, from the given equation, $121a+11b+c=512+64b+8c+a \implies 120a-53b-7c=512$. Since $a$, $b$, and $c$ have to be positive, $a \geq 5$. If $a = 5$, then by casework we see that $b = 1$ and $c = 5$ is the only solution. Since the question asks for the lowest working value, we can stop here. Finally, $515_{11} = 621_{10}$, so our answer is $\boxed{621}$.

See Also

2020 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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