A number theory problem by Madhavarapu Revanth

The question translates to: how many 2 - digit numbers are of the form $p_{1}^2p_2$ or $p_1^5$ , where $p_1$ and $p_2$ are distinct primes? Let's simply write them down:

$2^2\cdot3, 2^2\cdot5, 2^2\cdot7, 2^2\cdot11, 2^2\cdot13, 2^2\cdot17, 2^2\cdot19, 2^2\cdot23,$

$3^2\cdot2, 3^2\cdot5, 3^2\cdot7, 3^2\cdot11,$

$5^2\cdot2, 5^2\cdot3,$

$7^2\cdot2,$

$2^5.$

We know no. of factors of n(=p1^(a1)*p2^(a2).....)=(a1+1)(a2+1)....(an+1) where all ai>0 6=2X3 & 6=1X6 so n=p^5 or n=pXq^2 where p & q are primes. 10<=n<=99 we find possible values of n satisfying either of the given forms are: 32,18,50,98,12,75,20,45,28,63,44,99,52,68,76,92 so answer is 16.

Let there be a two digit number then to have six different factors if we choose three different numbers (as factors)a,b,c(being prime) then if we arrange them then the possible factors are ab,bc,ca,abc,a,b,c,1 (abc being the number itself)which is more than six .... Therefore the number can't have more than two different prime factors ,let it be l,m if arranged it gives l,m,lm,1(lm being the number itself) but it is less than six therefore one must repeat (only one time) to give l,m,lm,lm^2,m^2,1(lm^2 being the number itself) Possible values of l,m are If l=2 ,m=3,5,7 If l=3, m=2,5 If l=5, m=2,3 If l=7, m=2,3 If l=11,m=2,3 If l=13,17,19,23 then m can be 2 only (See I have arrived at this by considering lm^2 to be less than 💯 but greater than 10) Ohhhh have I missed a number the total numbers is coming to be 15!!!! Not actually because if we consider a number such that it has only one prime factors then 32 is only such number having 1,2,4,8,16,32 as it's factors!!!!!! Therefore the answer is 16!!!!!!! (Hey abc denotes a times b times c)