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>
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<math>\frac{6432}{6}=1072\implies\frac{1072}{4}=268\implies\frac{268}{4}=67</math>, so <math>\gcd(2^5\cdot 3\cdot 67, 2^2\cdot 3\cdot 13)=2^2\cdot 3=12\implies 12-8=4, \fbox{E}</math>
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==See Also==
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{{AHSME 50p box|year=1954|num-b=3|num-a=5}}
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{{MAA Notice}}

Latest revision as of 19:36, 17 February 2020

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 $\gcd(2^5\cdot 3\cdot 67, 2^2\cdot 3\cdot 13)=2^2\cdot 3=12\implies 12-8=4, \fbox{E}$

See Also

1954 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
All AHSME Problems and Solutions


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