1981 AHSME Problems/Problem 16: Difference between revisions
Coolmath34 (talk | contribs) No edit summary |
J314andrews (talk | contribs) No edit summary |
||
| (6 intermediate revisions by 3 users not shown) | |||
| Line 6: | Line 6: | ||
<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> | ||
==Solution (Long Way)== | ==Solution 1== | ||
Since <math>9=3^2</math>, every <math>2</math> digits in base <math>3</math> corresponds to <math>1</math> digit in base <math>9</math>. | |||
Since this number is <math>20</math> digits long, which is an even number of digits, the answer must correspond to the first <math>2</math> digits on the left. So the answer is <math>12_{3} = 1 \cdot 3 + 2 = 5\ \fbox{(E)}</math>. | |||
-edited by coolmath34, j314andrews | |||
==Solution 2 (Long Way)== | |||
Convert <math>x</math> to base 10 then convert the result to base 9. | Convert <math>x</math> to base 10 then convert the result to base 9. | ||
<cmath>12112211122211112222_{3} = | <cmath>12112211122211112222_{3} = 2150029898</cmath> | ||
<cmath> | <cmath>2150029898 = 5484584488_{9}</cmath> | ||
Therefore, the answer is <math> \textbf{( | Therefore, the answer is <math> \textbf{(E)}\ 5.</math> | ||
-edited by coolmath34 | -edited by coolmath34 | ||
==See also== | |||
{{AHSME box|year=1981|num-b=15|num-a=17}} | |||
{{MAA Notice}} | |||
Latest revision as of 13:09, 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
Solution 1
Since 
, 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 
.
-edited by coolmath34, j314andrews
Solution 2 (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
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. 
