# 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