AIME 2008 I · 第 7 题

Solution

The difference between consecutive squares is $(x + 1)^2 - x^2 = 2x + 1$, which means that all squares above $50^2 = 2500$ are more than $100$ apart.

Then the first $26$ sets ($S_0,\cdots S_{25}$) each have at least one perfect square because the differences between consecutive squares in them are all less than $100$. Also, since $316$ is the largest $x$ such that $x^2 < 100000$ ($100000$ is the upper bound which all numbers in $S_{999}$ must be less than), there are $316 - 50 = 266$ other sets after $S_{25}$ that have a perfect square.

There are $1000 - 266 - 26 = \boxed{708}$ sets without a perfect square.

Video Solution

https://youtu.be/6eBLXnzK0n4

~IceMatrix