# 2017 AIME I Problems/Problem 5

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## Problem 5

A rational number written in base eight is $\underline{ab} . \underline{cd}$, where all digits are nonzero. The same number in base twelve is $\underline{bb} . \underline{ba}$. Find the base-ten number $\underline{abc}$.

## Solution 1

First, note that the first two digits will always be a positive number. We will start with base twelve because of its repetition. List all the positive numbers in base eight that have equal tens and ones digits in base 12.

$11_{12}=15_8$

$22_{12}=32_8$

$33_{12}=47_8$

$44_{12}=64_8$

$55_{12}=101_8$

We stop because we only can have two-digit numbers in base 8 and 101 is not a 2 digit number. Compare the ones places to check if they are equal. We find that they are equal if $b=2$ or $b=4$. Evaluating the places to the right side of the decimal point gives us $22.23_{12}$ or $44.46_{12}$. When the numbers are converted into base 8, we get $32.14_8$ and $64.30_8$. Since $d\neq0$, the first value is correct. Compiling the necessary digits leaves us a final answer of $\boxed{321}$

## Solution 2

The parts before and after the decimal points must be equal. Therefore $8a + b = 12b + b$ and $c/8 + d/64 = b/12 + a/144$. Simplifying the first equation gives $a = (3/2)b$. Plugging this into the second equation gives $3b/32 = c/8 + d/64$. Multiplying both sides by 64 gives $6b = 8c + d$. $a$ and $b$ are both digits between 1 and 7 (they must be a single non-zero digit in base eight) so using $a = 3/2b$, $(a,b) = (3,2)$ or $(6,4)$. Testing these gives that $(6,4)$ doesn't work, and $(3,2)$ gives $a = 3, b = 2, c = 1$, and $d = 4$. Therefore $abc = \boxed{321}$

## Solution 3

Converting to base $10$ we get

$4604a+72c+9d=6960b$

Since $72c$ and $9d$ are much smaller than the other two terms, dividing by $100$ and approximating we get

$46a=70b$

Writing out the first few values of $a$ and $b$, the first possible tuple is

$a=3, b=2, c=1, d=4$

and the second possible tuple is

$a=6, b=4, c=3, d=0$

Note that $d$ can not be $0$, therefore the answer is $\boxed{321}$

By maxamc

## Solution 4 (modular congruency/nt bash)

In the problem, we are given that $$8a+b+\dfrac c8+\dfrac d{64}=12b+b+\dfrac b{12}+\dfrac a{144}.$$ We multiply by the LCM of the denominators, which is $576$ to get $$4608a+576b+72c+9d=6912b+576b+48b+4a.$$ We then group like terms and factor to get $$4604a+72c+9d=48b(144+1)=145\cdot48b=24\cdot290b.$$

Observe that the coefficients of $a$, $c$, and $b$ are all divisible by $4$. Therefore, we know that $d$ must also be divisible by $4$ to compensate. (To observe this, one could rearrange the terms to see $9d=24\cdot290b-4604a-72c=4(6\cdot290b-1151a-18c)$. At this point, it is obvious that $9d$ must be divisible by $4$, so $d$ must be divisible by $4$.) Thus, we let $d=4y$ to see $4604a+72c+9\cdot4y=24\cdot290b$. We divide by $4$ in the equation to get $$1151a+18c+9y=6\cdot290b.$$

Observe that the coefficients of $c$, $y$, and $b$ are all divisible by $3$. By similar reasoning, $a$ must be divisible by $3$, so we let $a=3x$. Substituting and dividing by $3$, we get $$1151x+6c+3y=2\cdot290b.$$ We observe that this can no longer be reduced by similar means.

We know that $0. We examine $0<4y<8$; this becomes $0. It is apparent that $y=1\implies d=4$. However, the problem does not even ask for $d$, so it may appear that this find is meaningless. However, we substitute our value of $y$ in to get $$1151x+6c+3=580b.$$

We know that $x$ is either $1$ or $2$, since $0<3x<8$. We take our equation modulo $6$ to find $$5x+3\equiv4b\pmod{6}\implies3\equiv4b+x\pmod{6}.$$ If $x=2$, then $1\equiv4b\pmod{6}$ - this is equivalent to saying that $1+6n=4b\implies 1=4b-6n$. However, $4b-6n$ is always even, and $1$ is odd. Thus, this case is not possible, so $x=1\implies a=3$. We know that $x=1$; thus, $3\equiv4b+1\pmod{6}\implies2\equiv4b\pmod{6}\implies1\equiv2b\pmod{3}$. Obviously, $b=2$ and $b=5$ work; however, if $b=5$, then the RHS of our original equation ($2\cdot290b$) is much too large to be equal to the LHS (the maximum possible value of the LHS is $1151\cdot1+6\cdot7+3\cdot1$, which is less than $1200$ while the RHS would become $2\cdot290\cdot5=2900$), so we have $b=2$.

We recall our equation of $1151x+6c+3=580b$. Plugging in what we know, we have $1151\cdot1+6c+3=580\cdot2\implies6c=6\implies c=1$.

Therefore, $\overline{abc}=\boxed{321}$.

~~this insane timesink solution was composed by Technodoggo~~

## Video Solution

https://youtu.be/Mk-MCeVjSGc?t=298 ~Shreyas S