Counting a Factorials Factors

• $3! \cdot 5! \cdot 7! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7^1$
• A cube will be of the form $2^{3a} \cdot 3^{3b} \cdot 5^{3c} \cdot 7^{3d}$ , where each power is less than the exponents of the expressions $\begin{cases} a \in \{0, 1, 2\} \implies \boxed{3} \\ b \in \{0, 1\} \implies \boxed{2} \\ c \in \{0\} \implies \boxed{1} \\ d \in \{0\} \implies \boxed{1} \end{cases} \quad \quad \implies \quad 3 \cdot 2 \cdot 1 \cdot 1 = \boxed{6}$

The problem asks how many perfect cubes can be evenly divided into 3!⋅5!⋅7!.

First we find the prime factorization of 3!⋅5!⋅7!

3!⋅5!⋅7!=2x2x2x2x2x2x2x2x3x3x3x3x5x5x7

Then we find the amount of combinations that make perfect cubes. A cube has to have a prime factorization with exponents that are multiples of three.

The possible values of the exponent in the exponent with base 2 are 0,3, and 6

The possible values of the exponent in the exponent with base 3 are 0,3

The only possible value of the exponent in the exponent with base 5 is 0

The only possible value of the exponent in the exponent with base 7 is 0

So there $\boxed{6}$ perfect cubes that are factors 3!⋅5!⋅7!

All the perfect cubes are 1, 8, 27, 64, 216, 1728