2020 AIME I Problems/Problem 3

Note: Please do not post problems here until after the AIME.

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 $\boxed{621}$ .

2020 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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2020 AIME I Problems/Problem 3

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