1978 AHSME Problems/Problem 18: Difference between revisions
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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>. | |||
Revision as of 21:17, 30 September 2021
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
.