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Difference between revisions of "1981 AHSME Problems/Problem 16"

(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==
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==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>.
  
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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>
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-edited by coolmath34, j314andrews
  
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==Solution (Long Way)==
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==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} = 8847859</cmath>
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<cmath>12112211122211112222_{3} = 2150029898</cmath>
  
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<cmath>8847859 = 17574874_{9}</cmath>
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<cmath>2150029898 = 5484584488_{9}</cmath>
  
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Therefore, the answer is <math> \textbf{(A)}\ 1.</math>
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Therefore, the answer is <math> \textbf{(E)}\ 5.</math>
  
 
-edited by coolmath34
 
-edited by coolmath34
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==See also==
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{{AHSME box|year=1981|num-b=15|num-a=17}}
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{{MAA Notice}}

Latest revision as of 14:09, 28 June 2025

Problem

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

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

Solution 1

Since $9=3^2$, every $2$ digits in base $3$ corresponds to $1$ digit in base $9$. Since this number is $20$ digits long, which is an even number of digits, the answer must correspond to the first $2$ digits on the left. So the answer is $12_{3} = 1 \cdot 3 + 2 = 5\ \fbox{(E)}$.

-edited by coolmath34, j314andrews

Solution 2 (Long Way)

Convert $x$ to base 10 then convert the result to base 9. \[12112211122211112222_{3} = 2150029898\]

\[2150029898 = 5484584488_{9}\]

Therefore, the answer is $\textbf{(E)}\ 5.$

-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, as part of the American Mathematics Competitions. AMC Logo.png

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