Find all possible arithmetic sequences of integers, with a common difference of 2, whose sum is exactly 2016.

      There is a serious error in my presentation- double counting-
I leave the erroneous text unchanged and recommend to see my revised solution

thus 2*1008==>1007+1009    3*672=670+672+674   etc

-668-666-664-662   -0+2+4+6 +... 666+668  put prior to 670+672+674   creates additional solution.

Therefore there are 2*18 for the a<b case  and 18 solution when a>b

easy to show that those cannot be "mirrored".

out of  54 possible solution the 1st two  i.e .   2016    and  1007,1009

are not "politically correct"- less than 3 members.

SO THERE ARE 52 SERIES - no intention to expand now.

Just hope I did not err.

Edited on January 24, 2015, 3:01 am

Posted by Ady TZIDON on 2015-01-23 12:40:44