2021/046) Find all 2 digit numbers having 6 factors

we shall be using the following formula for number of factors of a number.

If the number N is of the form $p_1^{q_1}p_2^{q_2} \cdots p_n^{q_n}$ when 

$p_1,p_2,p_3\cdots p_n$ are prime numbers and

 $q_1,q_2\cdots q_n$ are the powers of the prime numbers then the number of factors 

 are $(q_1+1)(q_2+1)\cdots(q_n+1)$  

now let us factor 6 is different ways 6 = 6 = 2 * 3

if we take 6 then the number is $p^5$ and for this to be between 10 and 100 this is $2^5= 32$ as $3^5 = 243$ is bigger

If we take 2* 3 then the number is $p_1p_2^2$ and for this to be between 10 and 100 

Taking $p_1=2$ we get $2* 3^2= 18$, $2 * 5^2 = 50$, $2 * 7^2 = 98$.

Taking $p_1=3$ we get $3* 2^2= 12$, $3 * 5^2 = 75$

Taking $p_1=5$ we get $5* 2^2= 20$, $5 * 3^2 = 45$

Taking $p_1=7$ we get $7* 2^2= 28$, $7 * 3^2 = 63$

Taking $p_1=11$ we get $11* 2^2= 44$, $11 * 3^2 = 99$

Taking $p_1= 13$ we get $13* 2^2= 52$

Taking $p_1= 17$ we get $17* 2^2= 68$

Taking $p_1= 19$ we get $19* 2^2= 76$

Taking $p_1= 23$ we get $23* 2^2= 92$

So solution set 

$\{12,18,20,28, 32,44,45,50,52,63,68,75,76,92,98,99\}$