Difference between revisions of "2017 AIME I Problems/Problem 5"
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A rational number written in base eight is <math>\underline{ab} . \underline{cd}</math>, where all digits are nonzero. The same number in base twelve is <math>\underline{bb} . \underline{ba}</math>. Find the base-ten number <math>\underline{abc}</math>. | A rational number written in base eight is <math>\underline{ab} . \underline{cd}</math>, where all digits are nonzero. The same number in base twelve is <math>\underline{bb} . \underline{ba}</math>. Find the base-ten number <math>\underline{abc}</math>. | ||
| − | ==Solution== | + | ==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. | 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. | ||
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Evaluating the places to the right side of the decimal point gives us <math>22.23_{12}</math> or <math>44.46_{12}</math>. | Evaluating the places to the right side of the decimal point gives us <math>22.23_{12}</math> or <math>44.46_{12}</math>. | ||
When the numbers are converted into base 8, we get <math>32.14_8</math> and <math>64.30_8</math>. Since <math>d\neq0</math>, the first value is correct. Compiling the necessary digits leaves us a final answer of <math>\boxed{321}</math> | When the numbers are converted into base 8, we get <math>32.14_8</math> and <math>64.30_8</math>. Since <math>d\neq0</math>, the first value is correct. Compiling the necessary digits leaves us a final answer of <math>\boxed{321}</math> | ||
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| + | ==Solution 2== | ||
| + | The parts before the decimal points must be equal as must the parts after. Therefore: 8a * b = 12b + b and c/8 + d/64 = b/12 + a/144. Simplifying the first equation gives: a = 3/2b. Plugging this into the second equation gives 3b/32 = c/8 + d/64. Multiply by 64: 6b = 8c + d. A and B are both integers between 1 and 7 (they must be a single 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} | ||
Revision as of 17:41, 8 March 2017
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: 8a * b = 12b + b and c/8 + d/64 = b/12 + a/144. Simplifying the first equation gives: a = 3/2b. Plugging this into the second equation gives 3b/32 = c/8 + d/64. Multiply by 64: 6b = 8c + d. A and B are both integers between 1 and 7 (they must be a single 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}
