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

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

Revision as of 00:13, 18 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 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)

See Also

2022 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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All AIME Problems and Solutions

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