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2008 AIME I Problems/Problem 7

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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 $316^2 < 100000 < 317^2$ , 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.

See also

2008 AIME I (Problems • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2008_AIME_I_Problems/Problem_7"

Category: Intermediate Number Theory Problems

Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.

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