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 17:11, 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 | ||
