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Difference between revisions of "1954 AHSME Problems/Problem 4"

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<math>\gcd(6432, 132)</math>
 
<math>\gcd(6432, 132)</math>
 
<math>13\cdot 12=132</math>
 
<math>13\cdot 12=132</math>
−
<math>\frac{6432}{6}=1072\implies\frac{1072}{4}=268\implies\frac{268}{4}=67</math>, so <math>\mathoperator{gcd}(2^5\cdot 3\cdot 67, 2^2\cdot 3\cdog 13)=2^2\cdot 3=12\implies 12-8=4, \fbox{E}</math>
+
<math>\frac{6432}{6}=1072\implies\frac{1072}{4}=268\implies\frac{268}{4}=67</math>, so <math>\mathop{gcd}(2^5\cdot 3\cdot 67, 2^2\cdot 3\cdog 13)=2^2\cdot 3=12\implies 12-8=4, \fbox{E}</math>

Revision as of 12:38, 6 June 2016

Problem 4

If the Highest Common Divisor of $6432$ and $132$ is diminished by $8$, it will equal:

$\textbf{(A)}\ -6 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ -2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution

$\gcd(6432, 132)$ $13\cdot 12=132$ $\frac{6432}{6}=1072\implies\frac{1072}{4}=268\implies\frac{268}{4}=67$, so $\mathop{gcd}(2^5\cdot 3\cdot 67, 2^2\cdot 3\cdog 13)=2^2\cdot 3=12\implies 12-8=4, \fbox{E}$ (Error compiling LaTeX. Unknown error_msg)

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