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 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

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 My revised solution | Comment 6 of 8 |

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 an arithmetic  sequence uniquely defined by  a,b,d (d=2) :
e.g. 3*672=>670,672,674;   21*96=>76,78,80, ..., 96, ...114,116.

There are 36 pairs  of a*b, if listed by increasing a we get first
eighteen consisting of positive numbers only, and for every sequence S(i)  defined by a(i),b(i)  there exists a corresponding
sequence S(j) such that j=37-i and  a(i)=b(j) & a(j)=b(i):

S(2) with a*b=2*1008 corresponds to S(35) generated by 1008*2 .

However the first two sequences hardly fit the definition of arithmetic progression ( sequence of numbers such that the difference of any two successive members of the sequence is a constant):

2016 - there are no successive members,

1007,1009 - no constant difference.

The total number of arithmetic  sequences  - S(3) to S(36) - is 34.

Edited on January 24, 2015, 2:56 am
 Posted by Ady TZIDON on 2015-01-24 02:35:56

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