Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2017 AIME I Problems/Problem 5"
(→‎Solution 2: fixed typo)
Line 22: Line 22:
  
 
==Solution 2==
 
==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 = <math>\boxed{321}</math>
+
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 = <math>\boxed{321}</math>

Revision as of 17:13, 8 March 2017

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: 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 = $\boxed{321}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_5&oldid=84497"