Difference between revisions of "1978 AHSME Problems/Problem 18"
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==Problem== | ==Problem== | ||
| + | What is the smallest positive integer <math>n</math> such that <math>\sqrt{n}-\sqrt{n-1}<.01</math>? | ||
| + | |||
| + | <math>\textbf{(A) }2499\qquad \textbf{(B) }2500\qquad \textbf{(C) }2501\qquad \textbf{(D) }10,000\qquad \textbf{(E) }\text{There is no such integer}</math> | ||
| + | |||
| + | ==Solution== | ||
| + | Adding <math>\sqrt{n - 1}</math> to both sides, we get | ||
| + | <cmath>\sqrt{n} < \sqrt{n - 1} + 0.01.</cmath> | ||
| + | Squaring both sides, we get | ||
| + | <cmath>n < n - 1 + 0.02 \sqrt{n - 1} + 0.0001,</cmath> | ||
| + | which simplifies to | ||
| + | <cmath>0.9999 < 0.02 \sqrt{n - 1},</cmath> | ||
| + | or | ||
| + | <cmath>\sqrt{n - 1} > 49.995.</cmath> | ||
| + | Squaring both sides again, we get | ||
| + | <cmath>n - 1 > 2499.500025,</cmath> | ||
| + | so <math>n > 2500.500025</math>. The smallest positive integer <math>n</math> that satisfies this inequality is <math>\boxed{2501}</math>. | ||
| + | ==alternative== | ||
| + | Taking reciprocals and flipping the inequality we get <cmath>\sqrt{n}+\sqrt{n-1}>100</cmath> Which is easy to see the answer is | ||
| + | <math>\boxed{2501}</math>. | ||
| + | -bjump | ||
Latest revision as of 20:09, 13 July 2022
Problem
What is the smallest positive integer 
such that 
?
Solution
Adding 
to both sides, we get
![\[\sqrt{n} < \sqrt{n - 1} + 0.01.\]](http://latex.artofproblemsolving.com/3/1/f/31ff4be73d8bd87ac29318b16ebaab7f5cd0b7b6.png)
Squaring both sides, we get
![\[n < n - 1 + 0.02 \sqrt{n - 1} + 0.0001,\]](http://latex.artofproblemsolving.com/7/2/e/72e4a6a9b48d8391e5e10abb9de81223c3432dab.png)
which simplifies to
![\[0.9999 < 0.02 \sqrt{n - 1},\]](http://latex.artofproblemsolving.com/d/f/d/dfdfc974ea20d9ee176e55eb8a8c81d61e4a69d9.png)
or
![\[\sqrt{n - 1} > 49.995.\]](http://latex.artofproblemsolving.com/7/9/b/79ba6ab7b218badd13dcaf346c2e9c4bca98263e.png)
Squaring both sides again, we get
![\[n - 1 > 2499.500025,\]](http://latex.artofproblemsolving.com/2/f/5/2f56d78d79168eafa7caacdcc34b2b17cc96eb62.png)
so 
. The smallest positive integer that satisfies this inequality is 
.
alternative
Taking reciprocals and flipping the inequality we get ![\[\sqrt{n}+\sqrt{n-1}>100\]](http://latex.artofproblemsolving.com/1/8/a/18ae6a4066029d800b2e9051effd0b000277de80.png)
Which is easy to see the answer is
.
-bjump
