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>\ | + | <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> |
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AHSME 50p box|year=1954|num-b=3|num-a=5}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 19:36, 17 February 2020
Problem 4
If the Highest Common Divisor of and is diminished by , it will equal:
Solution
, so
See Also
1954 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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All AHSME Problems and Solutions |
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