2017 AIME I Problems/Problem 5
Contents
Problem 5
A rational number written in base eight is , where all digits are nonzero. The same number in base twelve is . Find the base-ten number .
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.
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 or . Evaluating the places to the right side of the decimal point gives us or . When the numbers are converted into base 8, we get and . Since , the first value is correct. Compiling the necessary digits leaves us a final answer of
Solution 2
The parts before the decimal points must be equal as must the parts after. Therefore 8 + = 12 + and /8 + /64 = /12 + /144. Simplifying the first equation gives: = 3/2. Plugging this into the second equation gives 3/32 = /8 + /64. Multiply by 64: 6 = 8 + . and are both digits between 1 and 7 (they must be a single non-zero digit in base eight) so using = 3/2, , = (3,2) or (6,4). Testing these gives that (6,4) doesn't work, and (3,2) gives = 3, = 2, = 1, and = 4. Therefore =
See Also
2017 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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