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Question

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\geqkq, the arrangement must contain at least k consecutive ones.

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Given:

pp ones and qq zeros are placed arbitrarily about a circle

p,q,kp, q, k are positive integers with pkqp\geq kq

To proof: The arrangement contains at least kk consecutive ones.

PROOF BY CONTRADICTION\textbf{PROOF BY CONTRADICTION}

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

Since there are qq zeros and since there are not at least kk consecutive ones, there are at most k1k-1 ones between each pair of consecutive zeros.

Since there are qq pairs of consecutive zeros with at most k1k-1 pairs of consecutive ones between them, there are at most (k1)q(k-1)q ones placed about the circle.

However, we placed pp ones about the circle which is more than (k1)q(k-1)q ones as (k1)q<kqp(k-1)q<kq\leq p and thus we derived a contradiction.

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

\square

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