Tessellate S.T.E.M.S. (2019) - Mathematics - Category B - Set 1 - Objective Problem 3

1- $g(n)=2$ iff $n$ is a prime number.

2- $g(g(p))=g(2)=2$ , when $p$ is a prime number. Therefore, all the prime numbers, from $2$ to $100$ are part of the solution. there are $25$ of them.

3- If $n=ab$ is a composite number, with $b\geq a$ and $b$ being the greatest proper divisor of $n$ , then $g(g(ab))=g(b+1)$ . Then, $g(b+1)$ would be equal to $2$ , if $b+1$ is a prime (according to 1). Therefore $b=p-1$ , where $p$ is a prime number.

$g\big(g\big(a(p-1)\big)\big)$

notice that $p-1$ is an even number. So, $a$ cannot be greater than $2$ , for $p-1$ has a factor of $2$ and, consequently $p-1$ cannot become the greatest proper divisor. So, $a=2$

$g\big(g\big(2(p-1)\big)\big)=g(p)=2$

there would be $14$ prime numbers that satisfy $2 < 2(p-1) \leq 100$ .

the final solution is $25+14=39$