2022 AIME I Problems/Problem 2

Problem

Find the three-digit positive integer $\underline{a}\,\underline{b}\,\underline{c}$ whose representation in base nine is $\underline{b}\,\underline{c}\,\underline{a}_{\,\text{nine}},$ where $a,$ $b,$ and $c$ are (not necessarily distinct) digits.

Solution 1

We are given that $100a + 10b + c = 81b + 9c + a,$ which rearranges to $99a = 71b + 8c.$ Taking both sides modulo $71,$ we have $\begin{align*} 28a &\equiv 8c \pmod{71} \\ 7a &\equiv 2c \pmod{71}. \end{align*}$ The only solution occurs at $(a,c)=(2,7),$ from which $b=2.$

Therefore, the requested three-digit positive integer is $\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}.$

~MRENTHUSIASM

Solution 2

As shown in Solution 1, we get $99a = 71b+8c$ . Note that $99$ and $71$ are big numbers comparatively to $8$ , so we hypothesize that $a$ and $b$ are equal and $8c$ fills the gap between them. The difference between $99$ and $71$ is $28$ , which is a multiple of $4$ . So, if we multiply this by $2$ , it will be a multiple of $8$ and thus the gap can be filled. Therefore, a viable solution is $(a,b,c)=(2,2,7)$ , and the answer is $\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}$ .

~KingRavi

Video Solution (Mathematical Dexterity)

https://www.youtube.com/watch?v=z5Y4bT5rL-s

Video Solution

https://www.youtube.com/watch?v=CwSkAHR3AcM

~Steven Chen (www.professorchenedu.com)

2022 AIME I (Problems • Answer Key • Resources)
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