Difference between revisions of "1997 AIME Problems/Problem 2"
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For <math>s</math>, there are <math>8^2</math> [[unit square]]s, <math>7^2</math> of the <math>2\times2</math> squares, and so on until <math>1^2</math> of the <math>8\times 8</math> squares. Using the sum of squares formula, that gives us <math>s=1^2+2^2+\cdots+8^2=\dfrac{(8)(8+1)(2\cdot8+1)}{6}=12*17=204</math>. | For <math>s</math>, there are <math>8^2</math> [[unit square]]s, <math>7^2</math> of the <math>2\times2</math> squares, and so on until <math>1^2</math> of the <math>8\times 8</math> squares. Using the sum of squares formula, that gives us <math>s=1^2+2^2+\cdots+8^2=\dfrac{(8)(8+1)(2\cdot8+1)}{6}=12*17=204</math>. | ||
− | Thus <math>\frac | + | Thus <math>\frac sr = \dfrac{204}{1296}=\dfrac{17}{108}</math>, and <math>m+n=\boxed{125}</math>. |
== See also == | == See also == |
Revision as of 18:11, 29 August 2011
Problem
The nine horizontal and nine vertical lines on an checkerboard form
rectangles, of which
are squares. The number
can be written in the form
where
and
are relatively prime positive integers. Find
Solution
To determine the two horizontal sides of a rectangle, we have to pick two of the horizontal lines of the checkerboard, or . Similarily, there are
ways to pick the vertical sides, giving us
rectangles.
For , there are
unit squares,
of the
squares, and so on until
of the
squares. Using the sum of squares formula, that gives us
.
Thus , and
.
See also
1997 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |