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# Prove that if p ones and q zeros are placed around a circle in an arbitrary manner, where p, q, and k are positive integers satisfying p$\geq$kq, the arrangement must contain at least k consecutive ones.

Solution

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Given:

$p$ ones and $q$ zeros are placed arbitrarily about a circle

$p, q, k$ are positive integers with $p\geq kq$

To proof: The arrangement contains at least $k$ consecutive ones.

$\textbf{PROOF BY CONTRADICTION}$

Let us assume, for the sake of contradiction, that there are not at least $k$ consecutive ones.

Since there are $q$ zeros and since there are not at least $k$ consecutive ones, there are at most $k-1$ ones between each pair of consecutive zeros.

Since there are $q$ pairs of consecutive zeros with at most $k-1$ pairs of consecutive ones between them, there are at most $(k-1)q$ ones placed about the circle.

However, we placed $p$ ones about the circle which is more than $(k-1)q$ ones as $(k-1)q and thus we derived a contradiction.

This then implies that out assumption "there are not at least $k$ consecutive ones" is incorrect and thus there are at leat $k$ consecutive ones in the arrangement.

$\square$

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