The number of divisors of 2700 including 1 and 2700 equals

Question

I don't really know how to approach this kind of problems, is there any trick or formula for this?

Answer 1

Factor the number in question: $$2700=2^2 \times 3^3 \times 5^2.$$

Any divisor must be a combination of these prime factors, with the exponent of $2$ at most $2$, the exponent of $3$ at most $3$, etc. The number of factors is then $3\times 4\times 3 = 36$ because the first exponent can range from $0$ to $2$ (3 possibilities), the second exponent can range from $0$ to $3$ (4 possibilities), and the third can range from $0$ to $2$ again (3 possibilities).

In general, if we have $$N = \prod_{i\ge 1}{p_i^{e_i}}$$ then the number of factors including $N$ and $1$ is $\prod_{i\ge 1}{e_i+1}$.