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Difference between revisions of "1975 AHSME Problems/Problem 29"

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==Solution(Very Stupid)==
 
==Solution(Very Stupid)==
  
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<math>(\sqrt{3}+\sqrt{2})^6=(5+2\sqrt{6})^3=(5+2\sqrt{6})(31+20\sqrt{6})=(395+162\sqrt{6})</math> Then, find that <math>\sqrt{6}</math> is about <math>2.449</math>. Finally, multiply and add to find that the smallest integer higher is <math>\boxed {\textbf{(A) } 972}</math>
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<math>(\sqrt{3}+\sqrt{2})^6=(5+2\sqrt{6})^3=(5+2\sqrt{6})(31+20\sqrt{6})=(395+162\sqrt{6})</math> Then, find that <math>\sqrt{6}</math> is about <math>2.449</math>. Finally, multiply and add to find that the smallest integer higher is <math>\boxed {\textbf{(C) } 970}</math>

Revision as of 03:31, 25 June 2019

Problem

What is the smallest integer larger than $(\sqrt{3}+\sqrt{2})^6$?

$\textbf{(A)}\ 972 \qquad \textbf{(B)}\ 971 \qquad \textbf{(C)}\ 970 \qquad \textbf{(D)}\ 969 \qquad \textbf{(E)}\ 968$

Solution(Very Stupid)

$(\sqrt{3}+\sqrt{2})^6=(5+2\sqrt{6})^3=(5+2\sqrt{6})(31+20\sqrt{6})=(395+162\sqrt{6})$ Then, find that $\sqrt{6}$ is about $2.449$. Finally, multiply and add to find that the smallest integer higher is $\boxed {\textbf{(C) } 970}$

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