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It is not simple. The only systematic way is to find the prime factorisation of the number and write it in exponential form.

So suppose n = (p1^r1)*(p2^r2)*...*(pk^rk) where p1, p2, ... pk are prime numbers and rk are the indices (or powers).

Then the factors of n are (p1^s1)*(p2^s2)*...*(pk^sk) where 0 ≤ sk ≤ rk. And remember that anything raised to the power 0 is 1.

Example:

n = 72 = 2*2*2*3*3 = (2^3)*(3^2)

so, the factors of n are (2^a)*(3^b) where a = 0, 1, 2 or 3 and b = 0, 1 or 2.

When

(a, b) = (0, 0) the factor is 1.

(a, b) = (1, 0) the factor is 2.

(a, b) = (2, 0) the factor is 4.

(a, b) = (3, 0) the factor is 8.

(a, b) = (0, 1) the factor is 3.

(a, b) = (1, 1) the factor is 6.

(a, b) = (2, 1) the factor is 12.

(a, b) = (3, 1) the factor is 24.

(a, b) = (0, 2) the factor is 9.

(a, b) = (1, 2) the factor is 18.

(a, b) = (2, 2) the factor is 36.

(a, b) = (3, 2) the factor is 72.

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