Determine all possible sequences of consecutive triangular numbers whose sum is precisely 2014.
Extra Challenge: A non computer program based method.
(In reply to
re: Possible Solution by Charlie)
Because I answered the wrong question, Charlie!
(nm) (m^2+mn+3m+n^2+3n+2) = 12084 = 2^2×3×19×53 gives:
1. ((nm)=19:) m=3, n=22 (i.e. T(4)...T(22))
2. ((nm)=2^2*3:) m=11, n=23 (i.e. T(12)...T(23))
corresponding with your solution.
Edited on August 13, 2014, 2:07 am

Posted by broll
on 20140813 01:13:26 