As the question goes , I tried to solve for $2^{2a}$ first but all I got was a mess. If anyone could kindly solve this for me I would be grateful .
Edit :
Thanks for the replies. I tried the problem on my own and went with this approach,
$1000! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot \dots \cdot 1000$
I want to solve for $2^{2a}$ first so I consider the even numbers only
Hence,
$$1000!=(2\cdot4\cdot6\cdot8 \dots 1000) \cdot (1 \cdot 3 \cdot 5 \cdot 7 \cdot 9 \cdot \dots 999)$$
Since, odd numbers cannot be factored by $2$, they will be ignored (in the following steps).
$$1000! = 2^{500}(1 \cdot 2 \cdot 3 \cdot \dots \cdot 500) \\ = 2^{500}(2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot \dots \cdot 500) \\ = 2^{500} \cdot 2^{250} \cdot (1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot \dots \cdot 250) \\ = 2^{500}2^{250}(2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot \dots \cdot 250) \\ = 2^{500}2^{250}2^{125}(1 \cdot 2 \cdot 3 \cdot 4 \cdot \dots \cdot 125)$$
I repeat this process and what i get at the end is
$2^{500} \cdot 2^{250} \cdot 2^{125} \cdot 2^{62} \cdot 2^{31} \cdot 2^{15} \cdot 2^{7} \cdot 2^{3} \cdot 2^{1} = \mathbf{2^{994}}$
Therefore, the answer to the original question would be 497
Is this process correct?
For a number $n$, denote by $v_2(n)$ the highest power of 2 which divides it, such that $v_2(1) = 0$, and $v_2(2^k) = k$ and so on.
Then it is clear that $v_2(1000!) = \sum_{n=1}^{1000}v_2(n)$.
To count the latter, there is a simple trick one can use: write $v_2$ as a sum of indicators. Write $I_k(n) = 1$ if $2^k$ divides $n$ and $0$ otherwise.
Clearly, $v_2 = \sum_{k=1}^{\infty}I_k$.
Now plug this identity into the previous sum: $$v_2(1000!) = \sum_{n=1}^{1000}v_2(n) = \sum_{k=1}^{\infty}\sum_{n=1}^{1000}I_k(n).$$
The inner summation is simply $[1000/2^k]$, hence the largest power of $2$ that divides $1000!$ is
$$[1000/2] + [1000/4] + ... + [1000/512]$$
This is something you can do by hand or plug in a calculator. The point is that it gives you a number $b$ which has the property that $2^b | 1000!$ but $2^{b+1}$ does not divide $1000!$.
The maximal $a$ for which $4^a | 1000!$ is then $[b/2]$.