2022 AIME I Problem 2

Below is the professionally curated solution for Problem 2 of the 2022 AIME I, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AIME I solutions, or check the answer key.

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Concepts:number basedigitsmodular arithmetic

Difficulty rating: 1950

2.

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

Solution:

The condition says 100a+10b+c=81b+9c+a,100a + 10b + c = 81b + 9c + a, which simplifies to 99a=71b+8c.99a = 71b + 8c. Since the digits also appear in a base-nine numeral, each is at most 8.8. Reducing modulo 88 gives 3ab(mod8),3a \equiv -b \pmod 8, so b5a(mod8).b \equiv 5a \pmod 8.

For a=1,a = 1, b=5b = 5 makes 71b71b exceed 99;99; for a=2,a = 2, b=2b = 2 gives 8c=198142=56,8c = 198 - 142 = 56, so c=7.c = 7. For each a3,a \ge 3, the required bb forces 99a71b99a - 71b outside the range [0,64],[0, 64], so there is no other solution.

The number is 227,227, and indeed 227=281+79+2=272nine.227 = 2 \cdot 81 + 7 \cdot 9 + 2 = 272_{\text{nine}}.

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