Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1983 AHSME Problems/Problem 21"

(Added two solutions)
 
(→‎Solution 2)
 
(3 intermediate revisions by 2 users not shown)
Line 11: Line 11:
 
==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{(E)} \ 10 \sqrt{26} - 51}</math>.
+
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{(E)} \ 10 \sqrt{26} - 51}</math> as before.
+
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 = 2601</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 11:07, 31 July 2024

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 $\boxed{\textbf{(D)} \ 51 - 10 \sqrt{26}}$.

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 $x = 2601$. Therefore the answer is $\boxed{\textbf{(D)} \ 51 - 10 \sqrt{26}}$ as before.

See Also

1983 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions


The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1983_AHSME_Problems/Problem_21&oldid=224058"