1978 AHSME Problems/Problem 18

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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1978 AHSME Problems/Problem 18

Problem

Solution