Divisors always involve prime numbers so the first step should be to write \(1324\) in its prime factorization form: $$ 1324 = 2^2 \times 331$$

Now notice that any factor of \(1324\) must be a combination of \(2\)'s and \(331\)'s, with at most two \(2\)'s and one \(331\). In other words, any factor \(d\) of \(1324\) is of the form: $$ d = 2^a \times 331^b \text{ where } a \in \{0,1,2\} \text{ and } b \in \{0,1\} $$

How many values can \(a\) and \(b\) take? \(3\) and \(2\) respectively, so \(d\) can take \(3 \times 2 = 6\) different values. Namely: $$ d_1 = 2^0 \times 331^0 $$ $$ d_2 = 2^0 \times 331^1 $$ $$ d_3 = 2^1 \times 331^0 $$ $$ d_4 = 2^1 \times 331^1 $$ $$ d_5 = 2^2 \times 331^0 $$ $$ d_6 = 2^2 \times 331^1 $$ So \(1324\) has \(6\) distinct divisors.

Notice that you can generalize the argument to any number \(n\). Writing \(n\) in its prime factorization form: $$ n = p_1^{a_1}\times p_2^{a_2}\times ... \times p_k^{a_k} $$ The number of divisors \(n\) has is given by the formula: $$ (a_1+1) \times (a_2+1) \times ... \times (a_k+1)$$