Cubic numbers that can be divided by $2^{26} \cdot 3^{16} \cdot 5^8 \cdot 7^4$

Question

There's another set of exercises given by my teacher to work on. My subject is permutation and combinatorics.

The question is as follows:

Which cubic numbers are there that can divide into the number $$2^{26} \cdot 3^{16} \cdot 5^8 \cdot 7^4?$$

I just need the idea of what to search for. Can someone help?

Answer 1

The number of choices $(a,b,c,d)$ that satisfy the inequalities in @Alexey's hint is: $$\left(1+\left\lfloor\frac{26}{3}\right\rfloor\right)\left(1+\left\lfloor\frac{16}{3}\right\rfloor\right)\left(1+\left\lfloor\frac{8}{3}\right\rfloor\right)\left(1+\left\lfloor\frac{4}{3}\right\rfloor\right)=(9)(6)(3)(2)=324$$

Answer 2

I think that the OP should re-interpret the query into three separate questions, as follows:

Let $n = 2^{(26)} \;\times\; 3^{(16)} \;\times\; 5^8 \;\times\; 7^4.$

(1)
Using the prime factorization theorem, what are all the possible factors of $n$?

(2)
Of these factors, which ones will be perfect cubes?

(3)
How many such perfect cubes are there?

Answers:

(1)
Numbers of the form
$\left(2^a \;\times\; 3^b \;\times\; 5^c \;\times\;7^d\right),\;$ where
$a,b,c,d$ are all integers and
$0 \leq a \leq 26$ and
$0 \leq b \leq 16$ and
$0 \leq c \leq 8$ and
$0 \leq d \leq 4$.

(2)
$n$ will be a perfect cube if and only if
each of $a,b,c,d$ is a multiple of 3.

(3)
There are 9 choices for $a$, 6 choices for $b$, 3 choices for $c$, and 2 choices for $d$.

These choices are independent of each other.
Therefore, the # of such satisfying perfect cubes is
$9 \times 6 \times 3 \times 2 = 324.$