Difference between revisions of "1993 AHSME Problems/Problem 6"
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\text{(E) } 512.5</math> | \text{(E) } 512.5</math> | ||
| − | == Solution == | + | == Solution 1 == |
<math>\sqrt{\frac{ 8^{10}+4^{10} }{8^4 + 4^{11}}} = \sqrt{\frac{2^{30}+2^{20}}{2^{12}+2^{22}}}= \sqrt{\frac{2^{20}(2^{10}+1)}{2^{12}(1+2^{10})}} = \sqrt{2^8} = 2^4</math> | <math>\sqrt{\frac{ 8^{10}+4^{10} }{8^4 + 4^{11}}} = \sqrt{\frac{2^{30}+2^{20}}{2^{12}+2^{22}}}= \sqrt{\frac{2^{20}(2^{10}+1)}{2^{12}(1+2^{10})}} = \sqrt{2^8} = 2^4</math> | ||
<math>\fbox{B}</math> | <math>\fbox{B}</math> | ||
| + | |||
| + | == Solution 2 == | ||
| + | |||
| + | <math>8^{10}+4^{10} = 1074790400</math> | ||
| + | |||
| + | <math>8^4 + 4^{11} = 4198400</math> | ||
| + | |||
| + | <math>\frac{1074790400}{4198400} = 256</math> | ||
| + | |||
| + | <math>\sqrt{256} = 16 = \boxed{B}</math> | ||
== See also == | == See also == | ||
Latest revision as of 12:18, 11 August 2023
Contents
Problem
Solution 1
Solution 2
See also
| 1993 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
