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

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== Solution ==
 
== Solution ==
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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.
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<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 because the differences between consecutive squares in them are all less than <math>100</math>. Also, since <math>316</math> is the largest <math>x</math> such that <math>x^2 < 100000</math> (<math>100000</math> is the upper bound which all numbers in <math>S_{999}</math> must be less than), there are <math>316 - 50 = 266</math> other sets after <math>S_{25}</math> that have a perfect square.
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Then the first 25 sets (<math>S_1,\cdots S_{25}</math>) each have 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 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.
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==Video Solution==
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https://youtu.be/6eBLXnzK0n4
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~IceMatrix
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2008|n=I|num-b=6|num-a=8}}
 
{{AIME box|year=2008|n=I|num-b=6|num-a=8}}
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[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 21:59, 2 January 2021

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 because the differences between consecutive squares in them are all less than $100$. Also, since $316$ is the largest $x$ such that $x^2 < 100000$ ($100000$ is the upper bound which all numbers in $S_{999}$ must be less than), 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 (Problems • Answer Key • Resources)
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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