Difference between revisions of "1981 AHSME Problems/Problem 16"
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<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 </math> | <math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 </math> | ||
| + | |||
| + | ==See also== | ||
| + | |||
| + | {{AHSME box|year=1981|num-b=15|num-a=17}} | ||
| + | {{MAA Notice}} | ||
==Solution (Long Way)== | ==Solution (Long Way)== | ||
Revision as of 14:03, 28 June 2025
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
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. 
Solution (Long Way)
Convert to base 10 then convert the result to base 9.
![\[12112211122211112222_{3} = 2150029898\]](http://latex.artofproblemsolving.com/6/4/e/64ebb09efcb51d16c956f98d227a8502de88a8b0.png)
![\[2150029898 = 5484584488_{9}\]](http://latex.artofproblemsolving.com/9/d/e/9ded47e3858d6a4a0d1be36ac3ce6e923e00ac6f.png)
Therefore, the answer is 
-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
