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

I have written directly into the space allowed for comment and

by mistake lost everything.

Too tired to restart it - I will just present shortly the method:

since 2016=2^4*3^2*7  there are 5*3*2=36  ways to represent

2016 as a product a*b where a is the # of members and b  the average  of those members :  1*2016, 2*1008, 3*672...21*96...

96*21 ...1008*2, 2016*1.

To each of those pairs corresponds a sequence with positive numbers only, from  the 1st 18 one can create another 18 by adding a series consisting of positive and negative numbers cancelling  each other .

Edited on January 24, 2015, 3:00 am

Posted by Ady TZIDON on 2015-01-23 12:32:14