2017 AIME I Problems/Problem 5

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.






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


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


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 $0<a,b,c<8\implies0<3x,b,4y<8$. We examine $0<4y<8$; this becomes $0<y<2$. 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

See Also

2017 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png