You must mention that 0≤k<r≤n
Otherwise the question becomes too easy.
And please use subscript.
However assuming 0≤k<r≤n.
Let S1 = a1 , S2 = a1+a2
In general Si = a1+a2+a3+......+ai
Let us divide each of these n sums by n.
We can either have the remainder of one of these (say Sj)=0.For such a case k=1 and r=j-1.
If none of the remainders are = 0,we must have 2 sums which have equal remainders.(Since there are n sums and n-1 possible remainders other than 0).
For such a case the difference of the 2 sums will be divisible by n.
- Shaswata Roy Sn = a1+a2+a3+......+anUpvote·0· Reply ·2013-02-22 02:23:17