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

Revision as of 12:13, 19 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 large 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: $99,198,297,396.$ 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, we get $99a = 71b+8c$ .

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 $\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}$ .

~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)

@@ Line 27: / Line 27: @@
 == Solution 3 ==
-As shown in Solution 1, we get <math>99a = 71b+8c.</math> We list a few multiples of <math>99</math> out: <cmath>99,198,297,396.</cmath>
+As shown in Solution 1, we get <math>99a = 71b+8c.</math>
+We list a few multiples of <math>99</math> out: <cmath>99,198,297,396.</cmath>
 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.

2022 AIME I (Problems • Answer Key • Resources)
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