<s>[INST] Base $11$ representation of a number must equal to another representation in base $8$. Credit: 2020 AIME I (Problem #$3$ on the test)
Question: A positive integer $N$ has base-eleven representation $\underline{a}\kern 0.1em\underline{b}\kern 0.1em\underline{c}$ and base-eight representation $\underline1\kern 0.1em\underline{b}\kern 0.1em\underline{c}\kern 0.1em\underline{a},$ where $a,b,$ and $c$ represent (not necessarily distinct) digits. Find the least such $N$ expressed in base ten.
My solution: The base-$11$ representation of $abc$ has a value of $11^2(a) + 11b + c = 121a + 11b + c$. (In base $10$)
The base-$8$ representation of the number of $1bca$ has a value of $8^3 + 8^2(b) + 8c + a = 512 + 64b + 8c + a$. (In base $10$)
Therefore, $121a + 11b + c = 64b + 8c + a + 512 \Longrightarrow 120a - 53b - 7c = 512$. So, $120a > 512$, because $a, b$, and $c$ are all positive. $a, b,$ and $c$ are all less than or equal to $7$ because they are in base $8$, and $120a > 512$, the minimum possible value of $a$ is $5$, and $a = 5$ to create the minimum possible $N$. ($6$ and $7$ would be too high).
So, $53b + 7c = 88$. The only pair of positive integers $b$ and $c$ that works is $b = 1, c = 5$.
Therefore, Substituting into either of the $2$ equations; $N = 121a + 11b + c = 512 + 64b + 8c + a = \boxed{621}.$
I was wondering, is this solution valid and/or is there a better way to solve it, if it is.
 [/INST] Yes your solution is correct. Another solution also suggested by AoPS is
The conditions of the problem imply that $121a + 11b + c = 512 + 64b + 8 c + a$, so $120 a = 512+ 53b+7c$. The maximum digit in base eight is $7,$ and because $120a \ge 512$, it must be that $a$ is $5, 6,$ or $7.$ When $a = 5$, it follows that $600=512 + 53b+7c$, which implies that $88 = 53b+7c$. Then $b$ must be $0$ or $1.$ If $b = 0$, then $c$ is not an integer, and if $b = 1$, then $7c = 35$, so $c = 5$. Thus $N = 515_{11}$, and $N=5\cdot 121 + 1\cdot 11 + 5 = 621$. The number $637_{11} =1376_{8} = 766$ also satisfies the conditions of the problem, but $621$ is the least such number.
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