Difference between revisions of "1978 AHSME Problems/Problem 18"

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==Solution==
 
==Solution==
No solutions yet!
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Adding <math>\sqrt{n - 1}</math> to both sides, we get
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<cmath>\sqrt{n} < \sqrt{n - 1} + 0.01.</cmath>
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Squaring both sides, we get
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<cmath>n < n - 1 + 0.02 \sqrt{n - 1} + 0.0001,</cmath>
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which simplifies to
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<cmath>0.9999 < 0.02 \sqrt{n - 1},</cmath>
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or
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<cmath>\sqrt{n - 1} > 49.995.</cmath>
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Squaring both sides again, we get
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<cmath>n - 1 > 2499.500025,</cmath>
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so <math>n > 2500.500025</math>. The smallest positive integer <math>n</math> that satisfies this inequality is <math>\boxed{2501}</math>.

Latest revision as of 21:17, 30 September 2021

Problem

What is the smallest positive integer $n$ such that $\sqrt{n}-\sqrt{n-1}<.01$?

$\textbf{(A) }2499\qquad \textbf{(B) }2500\qquad \textbf{(C) }2501\qquad \textbf{(D) }10,000\qquad  \textbf{(E) }\text{There is no such integer}$

Solution

Adding $\sqrt{n - 1}$ to both sides, we get \[\sqrt{n} < \sqrt{n - 1} + 0.01.\] Squaring both sides, we get \[n < n - 1 + 0.02 \sqrt{n - 1} + 0.0001,\] which simplifies to \[0.9999 < 0.02 \sqrt{n - 1},\] or \[\sqrt{n - 1} > 49.995.\] Squaring both sides again, we get \[n - 1 > 2499.500025,\] so $n > 2500.500025$. The smallest positive integer $n$ that satisfies this inequality is $\boxed{2501}$.

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