1978 AMC 12 第 18 题

Rationalize $\sqrt n-\sqrt{n-1}$

Compare the resulting denominator with $100$ near $n=2500$

Rationalizing gives $$\sqrt n-\sqrt{n-1} =\frac1{\sqrt n+\sqrt{n-1}}.$$ For $n=2500,$ the denominator is $50+\sqrt{2499}\lt100,$ so the difference exceeds $0.01.$ For $n=2501,$ the denominator is $\sqrt{2501}+50\gt100,$ so the difference is less than $0.01.$ The denominator increases with $n,$ making $2501$ the least such integer.

Therefore, the correct answer is C.