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

## 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)

$(\sqrt{3}+\sqrt{2})^6=(5+2\sqrt{6})^3=(5+2\sqrt{6})(49+20\sqrt{6})=(485+198\sqrt{6})$ Then, find that $\sqrt{6}$ is about $2.449$. Finally, multiply and add to find that the smallest integer higher is $\boxed {\textbf{(C) } 970}$

## Solution(The Real Way - Binomial Thereom)

Let's evaluate $(\sqrt{3}+\sqrt{2})^6 + (\sqrt{3}-\sqrt{2})^6)$. We see that all the irrational terms cancel. Then, using binomial theorem, we evaluate all the rational terms in the first expression to get 485. Then, the sum of the rational parts of the 2nd term will be 485 as well. Then, we get a total of 970 and since $(\sqrt{3}-\sqrt{2})^6) < 1$, the greatest integer greater than our original expression is $\boxed {\textbf{(C) } 970}$