Difference between revisions of "1981 AHSME Problems/Problem 16"
Coolmath34 (talk | contribs) (Created page with "==Problem== The base three representation of <math>x</math> is <cmath>12112211122211112222</cmath> The first digit (on the left) of the base nine representation of <math>x</m...") |
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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> | ||
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==Solution (Long Way)== | ==Solution (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 | ||
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+ | ==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. | ||
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+ | So the answer is <math> \textbf{(E)}\ 5.</math> |
Latest revision as of 19:56, 5 May 2021
Problem
The base three representation of is The first digit (on the left) of the base nine representation of is
Solution (Long Way)
Convert to base 10 then convert the result to base 9.
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