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 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
