1981 AHSME Problems/Problem 16: Difference between revisions

Revision as of 15:44, 28 January 2021

The base three representation of $x$ is $12112211122211112222$ The first digit (on the left) of the base nine representation of $x$ is

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Convert $x$ to base 10 then convert the result to base 9. $12112211122211112222_{3} = 8847859$

$8847859 = 17574874_{9}$

Therefore, the answer is $\textbf{(A)}\ 1.$

-edited by coolmath34

@@ Line 5: / Line 5: @@
 <math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 </math>
-==Solution==
-Notice that the value of the leftmost <math>1</math> is equal to <math>1 \cdot 3^{20}.</math> Converting this to base 9, you would get <math>1 \cdot 9^{10}.</math> Therefore, the leftmost digit is equal to 1, and the answer is <math> \textbf{(A)}\ 1.</math>
 ==Solution (Long Way)==