Prime's Prime

Given that $t(n) +1 = s(n) \rightarrow s(n) - t(n) =1$ .

Since both $t(n)$ and $s(n)$ are prime, then the only possible value of $t(n)$ and $s(n)$ is $s(n) = 3 \space ; \space t(n) = 2$ .

We can clearly see that $n$ is in a form of $2^p \times 3^q$ , for some positive integer $p,$ and $q$ .

For $p =1$ , then $q = \big\{ 1,2,3 \big\}$ .

For $p= 2$ , then $q = \big\{ 1,2 \big\}$ .

For $p = 3$ , then $q = \big\{ 1,2\big\}$ .

For $p = 4$ , then $q = 1$ .

For $p =5$ , then $q = 1$ .

Hence, there are $9$ possible value of $n$ .

Alternative 2

We know that $n = 2^p \times 3^q$ ; then $n$ is/are the multiple of $2 \times 3 = 6$ that is/are not divisible by $r$ , where $r$ is/are positive prime number that satisfy $r < \left\lfloor \dfrac{100}{2 \times 3} \right\rfloor$ .

Then, we have $r = ( 5,7,11,13)$ .

Let $S_n$ denotes the number of possible value of $n$ that satisfy $t(n) +1 = s(n)$ . Then,

$\begin{aligned} S_n & = \left\lfloor \dfrac{100}{6} \right\rfloor - \left( \left\lfloor \dfrac{100}{2 \times 3 \times 5} \right\rfloor + \left\lfloor \dfrac{100}{2 \times 3 \times 7} \right\rfloor + \left\lfloor \dfrac{100}{2 \times 3 \times 11} \right\rfloor + \left\lfloor \dfrac{100}{2 \times 3 \times 13} \right\rfloor \right) \\ & = 16 - (3+2+1+1) = 9 \end{aligned}$ .