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

Revision as of 03:12, 24 February 2012

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 5: / Line 5: @@
 The difference between consecutive squares is <math>(x + 1)^2 - x^2 = 2x + 1</math>, which means that all squares above <math>50^2 = 2500</math> are more than <math>100</math> apart.
-Then the first <math>26</math> sets (<math>S_0,\cdots S_{25}</math>) each have at least one perfect square. Also, since <math>316^2 < 100000 < 317^2</math>, there are <math>316 - 50 = 266</math> other sets after <math>S_{25}</math> that have a perfect square.
+Then the first <math>26</math> sets (<math>S_0,\cdots S_{25}</math>) each have at least one perfect square. Also, since <math>316^2 < 100000</math> (which is when <math>i = 1000</math>), there are <math>316 - 50 = 266</math> other sets after <math>S_{25}</math> that have a perfect square.
 There are <math>1000 - 266 - 26 = \boxed{708}</math> sets without a perfect square.

2008 AIME I (Problems • Answer Key • Resources)
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Difference between revisions of "2008 AIME I Problems/Problem 7"

Revision as of 03:12, 24 February 2012

Problem

Solution

See also