
Hi Avinash
It is LCM and not product of the 20 numbers.
Regarding selecting the 20 factors with sum 801, you can invert the question. Just select the number of divisors which are extra and whose sum is equal to the Total sum of divisors  801.
For example, in the options we have smallest number as 360 (i.e. 2^{3}3^{2}5) whose number of factors are 24 (i.e. {3 + 1}{2 + 1}{1 + 1}) and whose sum of all the factors is 1170 (i.e. {2^{3} + 2^{2} + 2 + 1}{3^{2} + 3 + 1}{5 + 1}).
So instead of looking for the 20 divisors with sum equal to 801, I'd prefer to look for 4 divisors (i.e. 24  20) whose sum is 369 (i.e. 1170  801).
And this is pretty easy to do.
Two easily available possible set of 4 divisors are: 360, 5, 3, 1 & 360, 6, 2, 1.
Kamal Lohia
