MHB Divisibility Problem: Prove Existence of $i,j$

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The discussion centers on proving the existence of indices \(i\) and \(j\) such that the sum of a sequence of integers \( (a_1, a_2, \ldots, a_n) \) is divisible by \(n\). A suggested approach involves defining \(b_j = \sum_{k=1}^j a_k\) and applying the pigeonhole principle to the remainders of \(b_j\) when divided by \(n\). The concept of \(\mathbb{Z}^n\) is clarified as the set of all ordered \(n\)-tuples of integers. The discussion emphasizes the importance of understanding these mathematical constructs to solve the divisibility problem. Ultimately, the existence of such indices is guaranteed by these principles.
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$$\text{ Let } n∈N \text{ and } (a1,a2,…,a_{n})∈\mathbb{Z}^{n}.

\text{ Prove that always exist } i,j∈ \underline{n} \text{ with } i≤j \text{ so }

\sum\limits_{k=i}^{\\j} a_{k} \text{ divisible by n} .$$
 
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Hi, and welcome to the forum!

Hint: Consider $$b_j=\sum_{k=1}^ja_k$$, $j=1,\ldots,n$ and apply the pigeonhole principle to remainders when $b_j$ are divided by $n$.
 
What is the meaning of $$\mathbb{Z}^{n}$$?
 
$$\mathbb{Z}$$ denotes the set of integers. If $A$ and $B$ are sets, $A\times B$ denotes the set of all ordered pairs, where the first element comes from $A$ and the second one comes from $B$. More generally, $A_1\times \dots\times A_n$ denotes the set of all $n$-tuples, i.e., of ordered sequences of length $n$, where the $i$th element comes from $A_i$ for $i=1,\ldots,n$. Finally, $A^n=A\times\dots\times A$ ($n$ times). Thus, $$(a_1,\ldots,a_n)\in\mathbb{Z}^n$$ means that all $a_i$ are integers.
 
First trick I learned this one a long time ago and have used it to entertain and amuse young kids. Ask your friend to write down a three-digit number without showing it to you. Then ask him or her to rearrange the digits to form a new three-digit number. After that, write whichever is the larger number above the other number, and then subtract the smaller from the larger, making sure that you don't see any of the numbers. Then ask the young "victim" to tell you any two of the digits of the...

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