All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Deux Difference Deduction (Posted on 2015-01-23)
Find all possible arithmetic sequences of integers, with a common difference of 2, whose sum is exactly 2016.

 No Solution Yet Submitted by K Sengupta No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 Enough factors? Solution. | Comment 3 of 8 |
The number 2016 has 36 factors.  Each factor, n, will lead to a sequence except for n=1.

Sum=(n/2)(2*a1+(n-1)d
2016=(n/2)(2*a1+(n-1)*2)
2016=n(a1+n-1)
For each factor, solve for the first term a1
a1 = 2016/n - n + 1
and the rest follows

n=2, 1007+1009
n=3, 670+672+674
...
n=42, 7+9+11+...+49
n=48, -5+-3+-1+...+89
...
n=1008, -1005+-1003+-1002+...+1009
n=2016, -2014+-2012+-2010+...+2016

35 sequences in all, 17 of just positive integers.

Edit:
Those sequences with only positive integers can have extra terms added to the front that sum to zero, including the n=1 case.
So that's 18 more sequences for 53 total.

ex. n=42, -5+-3+-1+1+3+5+7+...+49

Edited on January 23, 2015, 12:50 pm
 Posted by Jer on 2015-01-23 12:45:52

 Search: Search body:
Forums (1)