Difference between revisions of "2022 AIME I Problems/Problem 2"

(Problem 2)
 
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/*Problem 2*/
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== Problem ==
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Find the three-digit positive integer <math>\underline{a}\,\underline{b}\,\underline{c}</math> whose representation in base nine is <math>\underline{b}\,\underline{c}\,\underline{a}_{\,\text{nine}},</math> where <math>a,</math> <math>b,</math> and <math>c</math> are (not necessarily distinct) digits.
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== Solution ==
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We are given that <cmath>100a + 10b + c = 81b + 9c + a,</cmath> which rearranges to <cmath>99a = 71b + 8c.</cmath>
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Taking both sides modulo <math>71,</math> we have
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<cmath>\begin{align*}
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28a &\equiv 8c \pmod{71} \\
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7a &\equiv 2c \pmod971}.
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\end{align*}</cmath>
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The only solution occurs at <math>(a,c)=(2,7),</math> from which <math>b=2.</math>
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Therefore, the requested three-digit positive integer is <math>\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}.</math>
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~MRENTHUSIASM
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==See Also==
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{{AIME box|year=2022|n=I|num-b=2|num-a=4}}
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{{MAA Notice}}

Revision as of 16:18, 17 February 2022

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

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 \pmod971}.
\end{align*} (Error compiling LaTeX. Unknown error_msg)

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

See Also

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

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