2017 AIME I Problems/Problem 5
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 tens 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 and after the decimal points must be equal. Therefore and . Simplifying the first equation gives . Plugging this into the second equation gives . Multiplying both sides by 64 gives . and are both digits between 1 and 7 (they must be a single non-zero digit in base eight) so using , or . Testing these gives that doesn't work, and gives , and . Therefore
Solution 3
Converting to base we get
Since and are much smaller than the other two terms, dividing by and approximating we get
Writing out the first few values of and , the first possible tuple is
and the second possible tuple is
Note that can not be , therefore the answer is
By maxamc
Video Solution
https://youtu.be/Mk-MCeVjSGc?t=298 ~Shreyas S
See Also
2017 AIME I (Problems • Answer Key • Resources) | ||
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Followed by Problem 6 | |
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