Let $a$ and $b$ be positive integers greater than one. With that in mind, $$(a \cdot b)!$$ is not necessarily divisible by:
a) $$a!^b$$ b) $$b!^a$$ c) $$a! \cdot b!$$ d) $${2}^{ab}$$
By brute-forcing, I found that letter D is the answer, can someone solve this in a non-brute-force way?
On
Hint $1:$
For the first two, observe that the product of $n$ consecutive numbers is divisible by $n!$
Hint $2:$
$\frac{(a \cdot b)!}{a! \cdot b!}$ is a multiple of $\binom{a+b}{a}=\binom{a+b}{b}$, both of which clearly are natural numbers, as $ab \geq {a+b}$, for all $a,b \geq 2$
Hint $3:$
$a=b=2$ acts as a simple contradiction as pointed in comments by @MPW.
Option D looks suspicious..
The power of 2 in a factorial $((ab)!) $ is given by:
$$\left|\frac {ab}{2^1}\right|+\left|\frac {ab}{2^2}\right|+\left|\frac {ab}{2^3}\right| + \dots $$
This is definitely lesser than:
$$ab \left(\frac {1}{2^1}+\frac {1}{2^2}+\frac {1}{2^3} + \dots \right) $$
which is equal to $ab $. Hence $ab $ can never be the power of 2 in $(ab)! $