1983 AHSME Problems/Problem 21

Problem

Find the smallest positive number from the numbers below.

$\textbf{(A)} \ 10-3\sqrt{11} \qquad  \textbf{(B)} \ 3\sqrt{11}-10 \qquad  \textbf{(C)}\ 18-5\sqrt{13}\qquad \textbf{(D)}\ 51-10\sqrt{26}\qquad \textbf{(E)}\ 10\sqrt{26}-51$

Solution 1

Notice that $3\sqrt{11} - 10 = \sqrt{99} - \sqrt{100} < 0$, $18 - 5\sqrt{13} = \sqrt{324} - \sqrt{325} < 0$, and similarly, $10\sqrt{26} - 51 = \sqrt{2600} - \sqrt{2601} < 0$, so these numbers can be excluded immediately as they are negative and we seek the smallest positive number. The remaining choices are $10 - 3\sqrt{11} = \sqrt{100} - \sqrt{99}$ and $51 - 10\sqrt{26} = \sqrt{2601} - \sqrt{2600}$. Observe that $\sqrt{100} - \sqrt{99} = \frac{1}{\sqrt{100} + \sqrt{99}} \approx \frac{1}{10+10} = \frac{1}{20}$, and $\sqrt{2601} - \sqrt{2600} = \frac{1}{\sqrt{2601} + \sqrt{2600}} \approx \frac{1}{51+51} = \frac{1}{102} < \frac{1}{20}$, so the smallest positive number in the list is $\boxed{\textbf{(E)} \ 10 \sqrt{26} - 51}$.

Solution 2

After reaching the two possibilities $\sqrt{100}-\sqrt{99}$ and $\sqrt{2601}-\sqrt{2600}$ as in the first solution, note that these can be written as $\sqrt{x}-\sqrt{x-1}$ for $x=100$ and $x=2601$ respectively. Now, $\frac{\mathrm{d}}{\mathrm{d}x}\left(\sqrt{x}-\sqrt{x-1}\right) = \frac{1}{2\sqrt{x}} - \frac{1}{2\sqrt{x-1}} = \frac{\sqrt{x-1} - \sqrt{x}}{2\sqrt{x}\sqrt{x-1}} < 0$ (as the numerator is always negative and the denominator is always positive for $x \geq 1$). Therefore, with its derivative being negative, $\sqrt{x}-\sqrt{x-1}$ is a strictly decreasing function over its domain, so the smaller value will be the one with the larger value of $x$, namely $x = 2600$. Therefore the answer is $\boxed{\textbf{(E)} \ 10 \sqrt{26} - 51}$ as before.

See Also

1983 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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