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

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Of course, <math>99</math> can't be made of just <math>8</math>'s. If we use one <math>71</math>, we get a remainder of <math>28</math>, which can't be made of <math>8</math>'s either. So <math>99</math> doesn't work.  
 
Of course, <math>99</math> can't be made of just <math>8</math>'s. If we use one <math>71</math>, we get a remainder of <math>28</math>, which can't be made of <math>8</math>'s either. So <math>99</math> doesn't work.  
 
<math>198</math> can't be made up of just <math>8</math>'s. If we use one <math>71</math>, we get a remainder of <math>127</math>, which can't be made of <math>8</math>'s. If we use two <math>71</math>'s, we get a remainder of <math>56</math>, which can be made of <math>8</math>'s.  
 
<math>198</math> can't be made up of just <math>8</math>'s. If we use one <math>71</math>, we get a remainder of <math>127</math>, which can't be made of <math>8</math>'s. If we use two <math>71</math>'s, we get a remainder of <math>56</math>, which can be made of <math>8</math>'s.  
Therefore we get <math>99\cdot2=71\cdot2+8\cdot7</math> so <math>a=2,b=2,</math> and <math>c=7</math>. Plugging this back into the original problem shows that this answer is indeed correct. Therefore, <math>\underline{a}\underline{b}\underline{c}=227.</math>
+
Therefore we get <math>99\cdot2=71\cdot2+8\cdot7</math> so <math>a=2,b=2,</math> and <math>c=7</math>. Plugging this back into the original problem shows that this answer is indeed correct. Therefore, <math>\underline{a}\underline{b}\underline{c}=\boxed {227}.</math>
 
~Technodoggo
 
~Technodoggo
  

Revision as of 22:31, 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

Solution 3

As shown in Solution 1, we get $99a = 71b+8c.$ We list a few multiples of $99$ out: \begin{align*} 99\\198\\297\\396\\ \end{align*} 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, $\underline{a}\underline{b}\underline{c}=\boxed {227}.$ ~Technodoggo

Solution 4

As shown in Solution 1, the equation simplifies to $99a = 71b+8c$.

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 $\boxed {227}$.

~Arcticturn

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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