Surviving elimination

First, we conjecture that $\frac{v}{c}$ is the answer.

We must show that:

1) $\frac{v}{c} - 1$ will not work.

and

2) $\frac{v}{c}$ votes can guarantee that a contestant will pass to the next round.

First, we prove 1).

This can be proved easily by taking $\frac{v}{c} - 1$ votes for contestant 1, $\frac{v}{c}$ for contestants 2 to $c - 1$ and $\frac{v}{c} + 1$ for contestant $c$ .

Then, we prove 2).

Suppose that a contestant loses with a total of $\frac{v}{c}$ votes. Then, the other contestants must all have at least $\frac{v}{c} + 1$ votes. The total sum is $v + ( c - 1 )$ , which is obviously larger than $v$ . So we have a contradiction.

Now, it would seem that the answer is $\frac{v}{c}$ . However, one question arises : What if all the contestants are tied in a $c$ -way tie? As unlikely as it is, it is still a possibility, and the tiebreaker will result in one contestant losing, thus contradicting our assumption that $\frac{v}{c}$ is sufficient to guarantee that a contestant will proceed to the next round.

So, we take the next smallest value, $\frac{v}{c} + 1$ as our answer. A quick check confirms that this value satisfies the conditions, and we are done. $\blacksquare$