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October 23, 2014

October 23, 2014

Posted by **Kristen** on Monday, September 10, 2007 at 12:01pm.

- Math -
**bobpursley**, Monday, September 10, 2007 at 12:46pmfor any odd number (b), there are (b-1)/2 odd numbers less than it.

check on that thinking. for b= 11, the odd numbers less are 1,3,5,7,9 or five. Yep, it works.

There are 1000 numbers for the choice of b, so there must be this number of ordered pairs:

Sum b=1 to 1999(0dd) of (b-1)/2

Sum 1/2 (b-1)= 1/2 sum b - 1/2 sum

and see if you can sum those from 1 to 1999

- Math -
**Kristen**, Monday, September 10, 2007 at 5:18pmThanks for responding, it really helped.

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