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

Revision as of 17:27, 23 March 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 square. Then there are $1000 - 266 - 26 = 708$ without a perfect square.

@@ Line 6: / Line 6: @@
 <cmath>(x + 1)^2 - x^2 = 2x + 1</cmath>
 which means that all squares above <math>50^2 = 2500</math> are more than 100 apart.
-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.
+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 square.  Then there are <math>1000 - 266 - 26 = 708</math> without a perfect square.
 == 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
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Difference between revisions of "2008 AIME I Problems/Problem 7"

Revision as of 17:27, 23 March 2008

Problem

Solution

See also