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 twelve that have equal twelves and ones digits in base 8.

$11_{12}=15_8$

$22_{12}=32_8$

$33_{12}=47_8$

$44_{12}=64_8$

$55_{12}=101_8$

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 the decimal points must be equal as must the parts after. Therefore 8 $a$ + $b$ = 12 $b$ + $b$ 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 3 $b$ /32 = $c$ /8 + $d$ /64. Multiply by 64: 6 $b$ = 8 $c$ + $d$ . $a$ and $b$ are both digits between 1 and 7 (they must be a single non-zero digit in base eight) so using $a$ = 3/2 $b$ , $a$ , $b$ = (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 $abc$ = $\boxed{321}$

2017 AIME I (Problems • Answer Key • Resources)
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2017 AIME I Problems/Problem 5

Contents

Problem 5

Solution 1

Solution 2

See Also