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Let $$N = \binom{2^{2012}}{0}\binom{2^{2012}}{1}\binom{2^{2012}}{2}\binom{2^{2012}}{3} \cdots \binom{2^{2012}}{2^{2012}}.$$ Let $M$ be the number of $0$'s when $N$ is written in binary. How many $0$'s does $M$ have when written in binary?

I didn't see an easy way of simplifying the product of combinations here. Algebraically, we have $$N = \dfrac{2^{2012} \cdot (2^{2012} \cdot (2^{2012}-1)) \cdots 2^{2012}!}{1! \cdot 2! \cdot 3! \cdots 2^{2012}!} = \dfrac{2^{2012 \cdot 2012} \cdot (2^{2012}-1)^{2011} \cdot (2^{2012}-2)^{2010} \cdots 1}{1! \cdot 2! \cdot 3! \cdots 2^{2012}!},$$ but I didn't see how to convert this into binary.

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    The number of $0$'s is not tractable at all. The number of trailing $0$'s might be found with the help of [Kummer's theorem](https://en.wikipedia.org/wiki/Kummer%27s_theorem). – 2017-01-19
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    $N$ is square, so the number of 0's is even. – 2017-01-19
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    $N$ is not square, since ${2^{2012} \choose 2^{2011}}$ is not square. If $p$ is a prime between $2^{2011}$ and $2^{2012}$, then $p$ divides ${2^{2012} \choose 2^{2011}}$ but $p^2$ does not. – 2017-01-19
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    And squares don't always have an even number of $0$'s (e.g. $7^2 = 110001_2$ has $3$). – 2017-01-19

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