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1987 AHSME Problems/Problem 19

Problem

Which of the following is closest to $\sqrt{65}-\sqrt{63}$? $\textbf{(A)}\ .12 \qquad \textbf{(B)}\ .13 \qquad \textbf{(C)}\ .14 \qquad \textbf{(D)}\ .15 \qquad \textbf{(E)}\ .16$

Solution

We have $\sqrt{65} > 8 > 7.5$. Also $7.5^2 = (7 + 0.5)^2 = 7^2 + 2 \cdot 7 \cdot 0.5 + 0.5^2 = 49 + 7 + 0.25 = 56.25 < 63$, so $\sqrt{63} > 7.5$. Thus $\sqrt{65} + \sqrt{63} > 7.5 + 7.5 = 15$. Now notice that $\sqrt{65} - \sqrt{63} = \frac{(\sqrt{65} - \sqrt{63})(\sqrt{65} + \sqrt{63})}{\sqrt{65} + \sqrt{63}} = \frac{2}{\sqrt{65} + \sqrt{63}}$, so $\sqrt{65} - \sqrt{63} < \frac{2}{15} = 0.1333333...$, so the answer must be $A$ or $B$. To determine which, we write $\sqrt{65} - \sqrt{63} > 0.125 \iff 65 - 2\sqrt{65 \cdot 63} + 63 > 0.015625 \iff 128 - 0.015625 > 2\sqrt{4095} \iff \sqrt{4095} < 64 - 0.0078125 \iff 4095 < 4096 - 32 \cdot 0.0078125 + 0.0078125^2 = 4096 - 0.25 + 0.0078125^2$ which is true. Hence as the expression is greater than $0.125$, and less than or equal to $0.13$ (since we showed it is certainly less than $0.1333333...$), it is closest to $0.13$, which is answer $\boxed{\text{B}}$.

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