How many different positive integer factors does have? How many different positive integer factors does $(2^7)(3^4)(7^3)(23^5)$ have?
Do we have to do any combinations between the powers here? 
 A: Is all the numbers of the form $2^n3^m7^k35^\ell$ with $n\in\{0,...,7\}, m\in\{0,...,4\},k\in\{0,...,3\}$ and $\ell\in\{0,...,5\}$ and that gives $8\cdot 5\cdot 4\cdot 6$ possibilities.
A: Generally, if
$$n = p_1^{\alpha_1}·p_2^{\alpha_2}\ldots p_k^{\alpha_k}$$
Then the amount of positive divisors of $n$ is:
$$d(n) = (\alpha_1+1)(\alpha_2+1)\ldots (\alpha_k+1)$$
In your case, it has $(7+1)(4+1)(3+1)(5+1) = 8·5·4·6 = 960$
A: Note the following:


*

*$\gcd(2,3)=1$

*$\gcd(2,7)=1$

*$\gcd(2,23)=1$

*$\gcd(3,7)=1$

*$\gcd(3,23)=1$

*$\gcd(7,23)=1$


Therefore, the number of different positive integer divisors of $(2^7)(3^4)(7^3)(23^5)$ is:
$$(7+1)\cdot(4+1)\cdot(3+1)\cdot(5+1)=960$$
A: What you are looking for is the number of the positive divisors of $2^7\cdot 3^4\cdot 7^3\cdot 23^5$. Then, the answer is
$$(7+1)(4+1)(3+1)(5+1)=8\cdot 5\cdot 4\cdot 6=960.$$
In general, if
$$N={p_1}^{\color{red}{q_1}}\cdot {p_2}^{\color{red}{q_2}}\cdots {p_k}^{\color{red}{q_k}}$$
where $q_i,k\in\mathbb N$ and $p_1\lt p_2\lt\cdots\lt p_k$ are primes, then the number of the positive divisors of $N$ can be represented as
$$(\color{red}{q_1}+1)(\color{red}{q_2}+1)\cdots (\color{red}{q_k}+1).$$ 
