Posted by Kristen on Monday, September 10, 2007 at 12:01pm.
for 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
Thanks for responding, it really helped.
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