1981 AHSME Problems/Problem 16
Problem
The base three representation of 
is
![\[12112211122211112222\]](http://latex.artofproblemsolving.com/1/9/e/19ee9a81d71f36177bfbd8f9886022b1ae264e14.png)
The first digit (on the left) of the base nine representation of is
Solution 1
Every 
digits in base 
corresponds to 
digit in base 
.
Since this number is 
digits long, which is an even number of digits, the answer must correspond to the first digits on the left. So the answer is $12_{3} = 1 \cdot 3 + 2 = 5\ \fbox{(E)}.
==Solution 2 (Long Way)== Convert$ (Error compiling LaTeX. Unknown error_msg)x$to base 10 then convert the result to base 9. <cmath>12112211122211112222_{3} = 2150029898</cmath>
<cmath>2150029898 = 5484584488_{9}</cmath>
Therefore, the answer is$ (Error compiling LaTeX. Unknown error_msg) \textbf{(E)}\ 5.$-edited by coolmath34
==Solution (Faster Way)== Every 2 numbers in base 3 represents 1 number in base 9. The first 2 numbers on the left,12 = 1(3) + 2(1) = 5.
So the answer is$ (Error compiling LaTeX. Unknown error_msg) \textbf{(E)}\ 5.$
See also
| 1981 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
