### The ring of Numbers

Again: copied from Saurabh Joshi's blog

Highlight the part between the * symbols for the answer.

* Set of k consecutive integers can be uniquely specified by the start position and the direction ( clockwise, or counter clockwise ). We will call it a k-box. Because they are arranged in a circular fashion there are n positions. We will assume clockwise direction. For each of these n position compute sum of its corresponding k-boxes. Sum of all these k-boxes would be kn(n+1)/2. The reason is that \sum_{i=1} ^{n} i = n(n+1)/2. and each number will appear in k different k-boxes.

Or, in other words. If none of the k-box can make contribution of at least k(n+1)/2 then the total of all k-boxes would be strictly less than kn(n+1)/2 which is a contradiction. *

**Problem :**Integers from 1 to n are arranged in a circle in some random permutation. Prove that there exist a set of k consecutive integers on the circle sum of which would be at least k(n+1)/2.**Solution :**Highlight the part between the * symbols for the answer.

* Set of k consecutive integers can be uniquely specified by the start position and the direction ( clockwise, or counter clockwise ). We will call it a k-box. Because they are arranged in a circular fashion there are n positions. We will assume clockwise direction. For each of these n position compute sum of its corresponding k-boxes. Sum of all these k-boxes would be kn(n+1)/2. The reason is that \sum_{i=1} ^{n} i = n(n+1)/2. and each number will appear in k different k-boxes.

So on average, each k-box makes a contribution of k(n+1)/2. So there exist at least one k-box which makes contribution of at least k(n+1)/2.

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