How many natural-number factors does $N$ have if $N = 2^4 \cdot 3^3 \cdot 5^2 \cdot 7^2$?
A natural-number factor of $N$ is any product of the form $2^a \cdot 3^b \cdot 5^c \cdot 7^d$, where $a, b, c, d$ are non-negative integers and $a \leq 4, b \leq 3, c \leq 2, d \leq 2$. For $a$, I can choose any of the five values $0, 1, 2, 3, 4$. For $b$, I can choose any of the four values $0, 1, 2, 3$. For $c$, I can choose any of the three values $0, 1, 2$. For $d$, I can choose any of the three values $0, 1, 2$. Therefore, the total number of factors is $5 \cdot 4 \cdot 3 \cdot 3 = 180$. 