Difference between revisions of "2008 AIME I Problems/Problem 7"
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<cmath>(x + 1)^2 - x^2 = 2x + 1</cmath> | <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. | which means that all squares above <math>50^2 = 2500</math> are more than 100 apart. | ||
− | 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. | + | 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. |
== See also == | == See also == |
Revision as of 18:27, 23 March 2008
Problem
Let be the set of all integers
such that
. For example,
is the set
. How many of the sets
do not contain a perfect square?
Solution
The difference between consecutive squares is
which means that all squares above
are more than 100 apart.
Then the first 26 sets (
) each have at least one perfect square. Also, since
, there are
other sets after
that have a perfect square. Then there are
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 | ||
All AIME Problems and Solutions |