Math Circle with Kevin

Consider $F_{n+4} = F_{n+3} + F_{n+2} = 2F_{n+2} + F_{n+1} = 3F_{n+1} + 2F_n$ . Therefore $F_{n+4}$ is divisible by 3 if $F_n$ is divisible by 3. We note that $F_0 = 0$ is divisible by 3, therefore $F_4$ is divisible by 3, and so as $F_8, F_{12}, F_{16}, ....$ . Note that $F_1$ , $F_2$ , and $F_3$ are not divisible by 3. There $F_n$ is a multiple of 3, if and only if $n \bmod 4 = 0$ . Since Kevin's turn is $5k+1$ , where $k$ is a non-negative integer. Then for Kevin to clap, we must have:

$\begin{aligned} 5k + 1 & \equiv 0 \text{ (mod 4)} \\ k + 1 & \equiv 0 \text{ (mod 4)} \\ \implies k & \equiv 3, 7, 11, ... \end{aligned}$

Therefore, Kevin claps when $5k+1 = 16, 36, 56, 76, 96$ for $\boxed 5$ times.

See a trend? ;) $\\$ So the sequence of "Fibonacci Sequence (mod 3)" repeats every eight terms while having a mutiple of $3$ every four terms . $\\$ Since the math circle has five students. $4 \times 5 = 20$ , $100 \div 20 = 5$ , Kevin would clap $\boxed{5}$ times