Is all the numbers of the form $2^n3^m7^k35^ell$ with $nin,...,7, min,...,4,kin,...,3$ and $ellin,...,5$ and that gives $8cdot 5cdot 4cdot 6$ possibilities.

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Generally, if

$$n = p_1^alpha_1·p_2^alpha_2ldots p_k^alpha_k$$

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 $(7+1)(4+1)(3+1)(5+1) = 8·5·4·6 = 960$


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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:

$$(7+1)cdot(4+1)cdot(3+1)cdot(5+1)=960$$


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What you are looking for is the number of the positive divisors of $2^7cdot 3^4cdot 7^3cdot 23^5$. Then, the answer is$$(7+1)(4+1)(3+1)(5+1)=8cdot 5cdot 4cdot 6=960.$$

In general, if$$N=p_1^colorredq_1cdot p_2^colorredq_2cdots p_k^colorredq_k$$where $q_i,kinlifwynnfoundation.orgbb N$ and $p_1lt p_2ltcdotslt p_k$ are primes, then the number of the positive divisors of $N$ can be represented as$$(colorredq_1+1)(colorredq_2+1)cdots (colorredq_k+1).$$


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