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Difference between revisions of "1993 AHSME Problems/Problem 6"
(→‎Solution 2)
 
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<math>\frac{1074790400}{4198400} = 256</math>
 
<math>\frac{1074790400}{4198400} = 256</math>
  
−
<math>\sqrt{256} = \boxed{16}</math>
+
<math>\sqrt{256} = 16 = \boxed{B}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 12:18, 11 August 2023

Problem

$\sqrt{\frac{8^{10}+4^{10}}{8^4+4^{11}}}=$

$\text{(A) } \sqrt{2}\quad \text{(B) } 16\quad \text{(C) } 32\quad \text{(D) } (12)^{\tfrac{2}{3}}\quad \text{(E) } 512.5$

Solution 1

$\sqrt{\frac{ 8^{10}+4^{10} }{8^4 + 4^{11}}} = \sqrt{\frac{2^{30}+2^{20}}{2^{12}+2^{22}}}= \sqrt{\frac{2^{20}(2^{10}+1)}{2^{12}(1+2^{10})}} = \sqrt{2^8} = 2^4$

$\fbox{B}$

Solution 2

$8^{10}+4^{10} = 1074790400$

$8^4 + 4^{11} = 4198400$

$\frac{1074790400}{4198400} = 256$

$\sqrt{256} = 16 = \boxed{B}$

See also

1993 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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All AHSME Problems and Solutions

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