Let $a_{1},\dots,a_{n}$ be integers. Show that there exist integers $k$ and $r$ such that the sum $$a_{k}+a_{k+1}+\dots+a_{k+r}$$ is divisible by $n$.
I am unable to find the necessary way to solve this.
Thanks in advance.
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Hint: pigeonhole as here using partial sums instead of partial digits – Bill Dubuque May 03 '15 at 17:41
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Consider the sequence $a_1, a_1+a_2, a_1+a_2+a_3, \dotsc$. There are $n$ terms in this sequence. If any one of them is divisible by $n$, then we are done.
If no sum is divisible, then the $n$ sums have $n-1$ possible remainders when divided by $n$. Thus two sums must have the same remainder, and their difference must be divisible by $n$. But their difference is of the form $a_k+a_{k+1}+\dotsb+a_{k+r}$. For example, $$a_1+a_2+a_3+a_4+a_5+a_6 - (a_1+a_2) = a_3+a_4+a_5+a_6$$
shardulc
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" If any one of them is divisible by n, then we are done". please explain. – gaufler May 03 '15 at 16:03
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Suppose the sum from $a_1$ until $a_m$ is divisible. Then $k=1$ and $k+r=m$. – shardulc May 03 '15 at 17:06