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.

#NumberTheory #ComputerScience

Easy Math Editor

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Comments

This is a classic application of Pigeonhole. Let the terms be $a_1$ , $a_2$ , $\dots$ , $a_n$ . Consider the $n$ sums $\begin{aligned} s_1 &= a_1, \\ s_2 &= a_1 + a_2, \\ &\dots, \\ s_n &= a_1 + a_2 + \dots + a_n. \end{aligned}$ 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 - 7 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

Finn Hulse - 7 years ago

Cool! :)

Sreejato Bhattacharya - 7 years ago

nice

shaan ragib - 6 years, 11 months ago

Interesting problem. Possibly a Pigeonhole argument is needed?

Calvin Lin Staff - 7 years ago

not a question

Rohan Lad - 7 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$