2022 AIME I Problems/Problem 2


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 $\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}.$


Solution 2

As shown in Solution 1, we get $99a = 71b+8c$.

Note that $99$ and $71$ are large 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, the only solution is $(a,b,c)=(2,2,7)$, and the answer is $\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}$.


Solution 3

As shown in Solution 1, we get $99a = 71b+8c.$

We list a few multiples of $99$ out: \[99,198,297,396.\] Of course, $99$ can't be made of just $8$'s. If we use one $71$, we get a remainder of $28$, which can't be made of $8$'s either. So $99$ doesn't work. $198$ can't be made up of just $8$'s. If we use one $71$, we get a remainder of $127$, which can't be made of $8$'s. If we use two $71$'s, we get a remainder of $56$, which can be made of $8$'s. Therefore we get $99\cdot2=71\cdot2+8\cdot7$ so $a=2,b=2,$ and $c=7$. Plugging this back into the original problem shows that this answer is indeed correct. Therefore, $\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}.$


Solution 4

As shown in Solution 1, we get $99a = 71b+8c$.

We can see that $99$ is $28$ larger than $71$, and we have an $8c$. We can clearly see that $56$ is a multiple of $8$, and any larger than $56$ would result in $c$ being larger than $9$. Therefore, our only solution is $a = 2, b = 2, c = 7$. Our answer is $\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}$.


Solution 5

As shown in Solution 1, we get $99a = 71b+8c,$ which rearranges to \[99(a – b) = 8c – 28 b = 4(2c – 7b) \le 4(2\cdot 9 - 0 ) = 72.\] So $a=b, 2c = 7b \implies c=7, b=2,a=2.$

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See Also

2022 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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