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2022 AIME I Problems/Problem 2

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, the only solution is $(a,b,c)=(2,2,7)$, and the answer is $\underline{a}\,\underline{b}\,\underline{c}=\boxed{227}$.

~KingRavi

Solution 2a

A little bit more motivation: taking mod $8$ on both sides, we find that $7a\equiv7b\pmod8\implies a\equiv b\pmod8$, so either $a=b$ or they differ by a multiple of $8$ (okok technically $a=b$ is a subcase of the other one but shh for the sake of clarity). $99$ and $71$ differ by $28$, so a difference of merely $8$ in $a$ and $b$ accounts for a huge difference on the scale of $28\cdot8=224$ (which is obviously unfillable by the mere $8c$), so $a=b$ is very reasonable.

~Technodoggo

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

Solution 5

As shown in Solution 1, we get $99a = 71b+8c,$ which rearranges to \[99(a – b) = 8c – 28 b = 4(2c – 7b) \le 4(2\cdot 9 - 0 ) = 72.\] So $a=b, 2c = 7b \implies c=7, b=2,a=2.$

vladimir.shelomovskii@gmail.com, vvsss

Solution 6

As shown in Solution 1, we have that $99a = 71b + 8c$.

Note that by the divisibility rule for $9$, we have $a+b+c \equiv a \pmod{9}$. Since $b$ and $c$ are base-$9$ digits, we can say that $b+c = 0$ or $b+c=9$. The former possibility can be easily eliminated, and thus $b+c=9$. Next, we write the equation from Solution 1 as $99a = 63b + 8(b+c)$, and dividing this by $9$ gives $11a = 7b+8$. Taking both sides modulo $7$, we have $4a \equiv 1 \pmod{7}$. Multiplying both sides by $2$ gives $a\equiv 2 \pmod{7}$, which implies $a=2$. From here, we can find that $b=2$ and $c=7$, giving an answer of $\boxed{227}$.

~Sedro

Video Solution by OmegaLearn

https://youtu.be/SCGzEOOICr4?t=340

~ pi_is_3.14

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)

Video Solution

https://youtu.be/MJ_M-xvwHLk?t=392

~ThePuzzlr

Video Solution by MRENTHUSIASM (English & Chinese)

https://www.youtube.com/watch?v=v4tHtlcD9ww&t=360s&ab_channel=MRENTHUSIASM

~MRENTHUSIASM

Video Solution

https://youtu.be/YcZzxez-j-c

~AMC & AIME Training

See Also

2022 AIME I (Problems • Answer Key • Resources)
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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