Difference between revisions of "2008 AIME I Problems/Problem 7"
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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. | 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</math> (which | + | 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. |
There are <math>1000 - 266 - 26 = \boxed{708}</math> sets without a perfect square. | There are <math>1000 - 266 - 26 = \boxed{708}</math> sets without a perfect square. |
Latest revision as of 21:59, 2 January 2021
Contents
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
apart.
Then the first sets (
) each have at least one perfect square because the differences between consecutive squares in them are all less than
. Also, since
is the largest
such that
(
is the upper bound which all numbers in
must be less than), there are
other sets after
that have a perfect square.
There are sets without a perfect square.
Video Solution
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.