1978 AHSME Problems/Problem 18

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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 $\boxed{2501}$.

alternative

Taking reciprocals and flipping the inequality we get \[\sqrt{n}+\sqrt{n-1}>100\] Which is easy to see the answer is

$\boxed{2501}$.

-bjump

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