# A problem in May Lunchtime 2014, Codechef.com

So i was trying to solve this problem.

 You are given a multiset of N integers. Please find such a nonempty subset of it that the sum of the subset's elements is divisible by N. Otherwise, state that this subset doesn't exist.


What i did was simply generate the combinations and check if each of them was divisible by its cardinality. This solution turned out just fine.

After the competetion i found an elegant solution here.

It so turns out that i cannot generate a sequence of n numbers without having a consecutive subset which is not a multiple of n.

That is to say, whatever combination of numbers i try..there always exists a consecutive subset which is a multiple of 'No. of numbers'(n).

Can anyone explain this theoritically? Or is there any theorem that states this?

PS: I already tried Googling it...all in vain.

Note by Varshith Reddy
4 years, 1 month ago

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This is a classic application of Pigeonhole. Let the terms be $$a_1$$, $$a_2$$, $$\dots$$, $$a_n$$. Consider the $$n$$ sums \begin{align*} s_1 &= a_1, \\ s_2 &= a_1 + a_2, \\ &\dots, \\ s_n &= a_1 + a_2 + \dots + a_n. \end{align*} If any of these sums are divisible by $$n$$, then we are done.

Otherwise, each of these sums are congruent to one of 1, 2, $$\dots$$, $$n - 1$$ modulo $$n$$. There are $$n$$ sums and $$n - 1$$ possible residues modulo $$n$$, so by Pigeonhole, two residues must be the same. In other words, there exist $$i$$ and $$j$$, $$i < j$$, such that $$s_i \equiv s_j \pmod{n}$$. Then $s_j - s_i = a_{i + 1} + a_{i + 2} + \dots + a_j$ is divisible by $$n$$, as desired.

Furthermore, the example $$a_1 = a_2 = \dots = a_{n - 1} = 1$$ shows that you need at least $$n$$ terms to guarantee that you can find a sum of consecutive terms that are divisible by $$n$$.

- 4 years, 1 month ago

nice

- 4 years ago

Oh. I first though to apply Pigeonhole to a different part of the problem. @Sreejato Bhattacharya remember that Pigeonhole trick/thing you showed me? By the way I'm totally showing that to the class. :D

- 4 years, 1 month ago

Cool! :)

- 4 years, 1 month ago

not a question

- 4 years, 1 month ago

Interesting problem. Possibly a Pigeonhole argument is needed?

Staff - 4 years, 1 month ago