Any integer number can be represented as product of powers of primes:
I = 2^p1 * 3^p2 * 5^p3 * 7^p4 * 11^p5 * ....
Total number of all its factors is
N = (p1+1) * (p2+1) * (p3+1) * ....
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K multipliers
So you you need to represent the value N
as a product of K
factors larger than 1.
Factorize N into primes, group these primes into K groups.
Imagine you have N=420
with 5 prime factors: 2 2 3 5 7
and K=3
. Make groups 2*2, 3*5, 7
(or any other combination) so corresponding powers to make I
are 3,14,6
For example, having N = 12
and K=3
, you can represent 12 = 2 * 2 * 3
and use a product of any two primes with a square of a third prime as the number I
. The smallest such value is 60 (2^2 * 3 * 5)
, the next one is 90 (2 * 3^2 * 5)
and so on (for example, 3 * 7 * 11^2
is also a solution).
For the case N = 12
and K=2
you can represent 12 = 3 * 4
and get results as p^2*q^3
or 12 = 2 * 6
and get results as p*q^5
where p,q
are distinct primes
For the case N = 12
and K=4
you cannot represent 12
as product of four integers larger than 1
, so it is impossible to produce result with these arguments