Difference between revisions of "1978 AHSME Problems/Problem 18"
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==Solution== | ==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 Squaring both sides, we get which simplifies to or Squaring both sides again, we get so . The smallest positive integer that satisfies this inequality is .
alternative
Taking reciprocals and flipping the inequality we get Which is easy to see the answer is
.
-bjump