2008 AIME I Problems/Problem 7

Problem

Let $S_i$ be the set of all integers $n$ such that $100i\leq n < 100(i + 1)$. For example, $S_4$ is the set ${400,401,402,\ldots,499}$. How many of the sets $S_0, S_1, S_2, \ldots, S_{999}$ do not contain a perfect square?

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. Also, since $316^2 < 100000$ (which is when $i = 1000$), 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

See also

2008 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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