Here's the problem:
The super factorial number $1! \cdot 2! \cdot 3! \cdot\cdots\cdot 12!$ can be written as a factorial times a perfect square, that is, in the form $m! \cdot n^2$. What is the value of $m$?
(A) $4 \qquad$ (B) $6\qquad$ (C) $8\qquad$ (D) $10\qquad$ (E) $12$
I don't understand a part of the solution, here is the solution:
In the product $1! \cdot 2! \cdot 3!\cdot\cdots\cdot 12!$ change every second factorial as follows: $$\begin{align} 2! &= 1! \cdot 2 \\ 4! &= 3! \cdot 4 \\ 6! &= 5! \cdot 6 \\ \cdots &= \cdots \\ 12! &= 11! \cdot 12 \end{align}$$ This gives $$ \begin{align} 1! \cdot 2! \cdot 3! \cdot \cdots \cdot 12! &= (1)^2 \cdot 2 \cdot (3!)^2 \cdot 4 \cdot (5!)^2 \cdot 6 \cdot \cdots \cdot (11!)^2 \cdot 12 \\ &= (1 \cdot 3! \cdot 5! \cdot \cdots \cdot 11!)^2 \cdot 2^6 \cdot 6! \\ &= 6! × (1! \cdot 3! \cdot 5! \cdot \cdots \cdot 11! \cdot 2^3)^2 \end{align}$$
So $m = 6$, hence (B).
I'm confused on the third step "$(1 \cdot 3! \cdot 5! \cdot \cdots \cdot 11!)^2 \cdot 2^6 \cdot 6!$". I can see how they group the odd factorials together, but where did the $2^6$ and $6!$ come from?
I just need some clarification on this.
From this step:
$= (1!)^2 \cdot 2 \cdot (3!)^2 \cdot 4 \cdot (5!)^2 \cdot 6 \cdot \ldots \cdot (11!)^2 \cdot 12$
Group factorials and separate even numbers:
$= (1! \cdot 2! \cdot 3! \cdot \ldots \cdot 11!)^2 \cdot 2 \cdot 4 \cdot 6 \cdot \ldots \cdot 12$
Notice that each even number can be written in the form $2 \cdot n$, therefore if you have even numbers up to $k$ then $2 \cdot 4 \cdot 6 \cdot \ldots \cdot k = (2 \cdot 1) \cdot (2 \cdot 2) \cdot (2 \cdot 3) \ldots \cdot (2 \cdot \frac{k}{2}) = 2^{\frac{k}{2}} \cdot (1 \cdot 2 \cdot 3 \cdot \ldots \cdot \frac{k}{2}) = 2^{\frac{k}{2}} \cdot (\frac{k}{2})!$ Since the twos multiple together and you're left with numbers starting from $1$ multiplying to $\frac{k}{2}$, which is the definition of factorial.
Therefore,
$= (1! \cdot 2! \cdot 3! \cdot \ldots \cdot 11!)^2 \cdot 2^{6} \cdot 6!$