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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
3 years, 5 months 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 \(n\).

Jon Haussmann - 3 years, 4 months ago

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Shaan Ragib - 3 years, 3 months ago

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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

Finn Hulse - 3 years, 4 months ago

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Cool! :)

Sreejato Bhattacharya - 3 years, 4 months ago

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not a question

Rohan Lad - 3 years, 4 months ago

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Interesting problem. Possibly a Pigeonhole argument is needed?

Calvin Lin Staff - 3 years, 4 months ago

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