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

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== Solution ==
 
== Solution ==
The difference between consecutive squares is
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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.
<cmath>(x + 1)^2 - x^2 = 2x + 1</cmath>
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which means that all squares above <math>50^2 = 2500</math> are more than 100 apart.
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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 < 10000 < 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 26 sets (<math>S_0,\cdots S_{25}</math>) each have at least one perfect square. Also, since <math>316^2 < 10000 < 317^2</math>, there are <math>316 - 50 = 266</math> other sets after <math>S_{25}</math> that have a perfect square. Then there are <math>1000 - 266 - 26 = 708</math> without a perfect square.
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There are <math>1000 - 266 - 26 = \boxed{708}</math> sets without a perfect square.
  
 
== See also ==
 
== See also ==

Revision as of 15:25, 19 April 2008

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 $316^2 < 10000 < 317^2$, 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 (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