Difference between revisions of "2008 AIME I Problems/Problem 7"

Revision as of 04:11, 4 May 2020

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.

@@ Line 11: / Line 11: @@
 == See also ==
 {{AIME box|year=2008|n=I|num-b=6|num-a=8}}
+Video Solution:
+https://www.youtube.com/watch?v=6eBLXnzK0n4
 [[Category:Intermediate Number Theory Problems]]
 {{MAA Notice}}

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Difference between revisions of "2008 AIME I Problems/Problem 7"

Revision as of 04:11, 4 May 2020

Problem

Solution

See also