How many ways can $720$ be decomposed into a product of two positive integers?

Solution 1:

Since \begin{align*} 720 & = 2 \cdot 360\\ & = 2 \cdot 2 \cdot 180\\ & = 2 \cdot 2 \cdot 2 \cdot 90\\ & = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 45\\ & = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 15\\ & = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 5\\ & = 2^4 \cdot 3^2 \cdot 5 \end{align*} each factor must be of the form $2^a3^b5^c$.

Suppose the factors are $2^{a_1}3^{b_1}5^{c_1}$ and $2^{a_2}3^{b_2}5^{c_2}$. Observe that \begin{align*} a_1 + a_2 & = 4 \tag{1}\\ b_1 + b_2 & = 2 \tag{2}\\ c_1 + c_2 & = 1 \tag{3} \end{align*} are equations in the nonnegative integers.

The equation $$x_1 + x_2 = n$$ has $n + 1$ solutions in the nonnegative integers, namely $$(n, 0), (n - 1, 1), (n - 2, 2), \ldots, (2, n - 2), (1, n - 1), (0, n)$$ Hence, equation 1 has five solutions, equation 2 has three solutions, and equation 2 has two solutions. Hence, there are $5 \cdot 3 \cdot 2 = 30$ ordered pairs of factors with product $720$. Since $720$ is not a perfect square, each ordered pair consists of two distinct numbers. Since we do not care about the order of the factors, we must divide this result by two. Thus, the number of ways to decompose $720$ into two positive integer factors is $$\frac{5 \cdot 3 \cdot 2}{2} = 15$$