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

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==Video Solution (Mathematical Dexterity)==
 
==Video Solution (Mathematical Dexterity)==
 
https://www.youtube.com/watch?v=z5Y4bT5rL-s
 
https://www.youtube.com/watch?v=z5Y4bT5rL-s
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==Video Solution==
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https://www.youtube.com/watch?v=CwSkAHR3AcM
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~Steven Chen (www.professorchenedu.com)
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2022|n=I|num-b=1|num-a=3}}
 
{{AIME box|year=2022|n=I|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:31, 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 \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 (Simple)

As in the previous solution, we get $99a = 71b+8c$. 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 $a = 2, b = 2, c = 7$ and the answer is $\boxed{227}$

~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)

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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