Difference between revisions of "1983 AHSME Problems/Problem 21"
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==Solution 1== | ==Solution 1== | ||
− | Notice that <math>3\sqrt{11} - 10 = \sqrt{99} - \sqrt{100} < 0</math>, <math>18 - 5\sqrt{13} = \sqrt{324} - \sqrt{325} < 0</math>, and similarly, <math>10\sqrt{26} - 51 = \sqrt{2600} - \sqrt{2601} < 0</math>, so these numbers can be excluded immediately as they are negative and we seek the smallest '''positive''' number. The remaining choices are <math>10 - 3\sqrt{11} = \sqrt{100} - \sqrt{99}</math> and <math>51 - 10\sqrt{26} = \sqrt{2601} - \sqrt{2600}</math>. Observe that <math>\sqrt{100} - \sqrt{99} = \frac{1}{\sqrt{100} + \sqrt{99}} \approx \frac{1}{10+10} = \frac{1}{20}</math>, and <math>\sqrt{2601} - \sqrt{2600} = \frac{1}{\sqrt{2601} + \sqrt{2600}} \approx \frac{1}{51+51} = \frac{1}{102} < \frac{1}{20}</math>, so the smallest positive number in the list is <math>\boxed{\textbf{( | + | Notice that <math>3\sqrt{11} - 10 = \sqrt{99} - \sqrt{100} < 0</math>, <math>18 - 5\sqrt{13} = \sqrt{324} - \sqrt{325} < 0</math>, and similarly, <math>10\sqrt{26} - 51 = \sqrt{2600} - \sqrt{2601} < 0</math>, so these numbers can be excluded immediately as they are negative and we seek the smallest '''positive''' number. The remaining choices are <math>10 - 3\sqrt{11} = \sqrt{100} - \sqrt{99}</math> and <math>51 - 10\sqrt{26} = \sqrt{2601} - \sqrt{2600}</math>. Observe that <math>\sqrt{100} - \sqrt{99} = \frac{1}{\sqrt{100} + \sqrt{99}} \approx \frac{1}{10+10} = \frac{1}{20}</math>, and <math>\sqrt{2601} - \sqrt{2600} = \frac{1}{\sqrt{2601} + \sqrt{2600}} \approx \frac{1}{51+51} = \frac{1}{102} < \frac{1}{20}</math>, so the smallest positive number in the list is <math>\boxed{\textbf{(D)} \ 51 - 10 \sqrt{26}}</math>. |
==Solution 2== | ==Solution 2== | ||
− | After reaching the two possibilities <math>\sqrt{100}-\sqrt{99}</math> and <math>\sqrt{2601}-\sqrt{2600}</math> as in the first solution, note that these can be written as <math>\sqrt{x}-\sqrt{x-1}</math> for <math>x=100</math> and <math>x=2601</math> respectively. Now, <math>\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</math> (as the numerator is always negative and the denominator is always positive for <math>x \geq 1</math>). Therefore, with its derivative being negative, <math>\sqrt{x}-\sqrt{x-1}</math> is a strictly decreasing function over its domain, so the smaller value will be the one with the larger value of <math>x</math>, namely <math>x = 2600</math>. Therefore the answer is <math>\boxed{\textbf{( | + | After reaching the two possibilities <math>\sqrt{100}-\sqrt{99}</math> and <math>\sqrt{2601}-\sqrt{2600}</math> as in the first solution, note that these can be written as <math>\sqrt{x}-\sqrt{x-1}</math> for <math>x=100</math> and <math>x=2601</math> respectively. Now, <math>\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</math> (as the numerator is always negative and the denominator is always positive for <math>x \geq 1</math>). Therefore, with its derivative being negative, <math>\sqrt{x}-\sqrt{x-1}</math> is a strictly decreasing function over its domain, so the smaller value will be the one with the larger value of <math>x</math>, namely <math>x = 2600</math>. Therefore the answer is <math>\boxed{\textbf{(D)} \ 51 - 10 \sqrt{26}}</math> as before. |
+ | |||
+ | ==See Also== | ||
+ | {{AHSME box|year=1983|num-b=20|num-a=22}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 14:34, 23 December 2020
Contents
Problem
Find the smallest positive number from the numbers below.
Solution 1
Notice that , , and similarly, , so these numbers can be excluded immediately as they are negative and we seek the smallest positive number. The remaining choices are and . Observe that , and , so the smallest positive number in the list is .
Solution 2
After reaching the two possibilities and as in the first solution, note that these can be written as for and respectively. Now, (as the numerator is always negative and the denominator is always positive for ). Therefore, with its derivative being negative, is a strictly decreasing function over its domain, so the smaller value will be the one with the larger value of , namely . Therefore the answer is as before.
See Also
1983 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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All AHSME Problems and Solutions |
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