If the umber is of the form $\prod_{n=1}^{k}p_n^{q_n}$ then the number of factors = $\prod_{n=1}^{k}(q_n+1)$
Because 5 is prime the number must be of the form $p^4$ where p is prime
We have the number of digits =2 and we should find p such that $ 9 \lt p^4 \lt 100$
The only number that satsfies the condition is n = 2 and n is a prime and $2^4= 16$
So 16 is the only 2 digit number having 5 factors
HCF is 9 so the 2 numbers are 9x, 9y where HCF(x,y) = 1 and x and y >0 and without loss of generality let x > y
product of the 2 numbers are 9xy = 270
or $xy = 30\cdots(1)$
sum of the 2 numbers are 9(x+y) = 99 or $x + y = 11\cdots(2)
we have $(x-y)^2 = (x+y)^2 - 4 * xy = 11^2 - 120 = 1$
so $x - y = 1\cdots(3)$
so we get x = 6 and y = 5 and numbers are $54, 45$
we have $4^x = 2^{2x}$
now working in mod 3 we get $1-5^y \equiv 0 \pmod 3$
or $5^y \equiv 1 \pmod 3$
as we know $5 \equiv -1 \pmod 3$ so y has to be even say 2m
now $4^x - 5^y = 2^{2x} - 5^{2m} = 39$
Farctoring we get $(2^x + 5^m)(2^x-5^m) = 39 = 39 *1 = 13 * 3$ (39 can be factored in 2 ways)
so we have 2 cases
$2^x+5^m= 39$ and $2^x - 5^m=1$ adding we get $2^x *2 = 40$ and this does not have integer
or
$2^x+5^m= 13$ and $2^x - 5^m=3$ adding we get $2^x *2 = 16$ or x = 3 and subtracting $2 * 5^m = 10$ and m = 1
so x = 3 and y = 2
