Integers between $1$ to $1000$ satisfying a condition that $1/2 * {2n \choose n}$ is even For how many integers $n,$ for $1 \leq n \leq 1000,$ is the number $\frac{1}{2}\left(\begin{array}{c}2 n \\ n\end{array}\right)$ even?
I'm thinking about expanding the top into $(2n \cdot (2n-1) \cdot ... \cdot (n+1))$ and the bottom is simply $n!$ and that expression would be $0 \mod 4$, but I'm not sure how to approach it after that point.
Here is my solution with $p$-adic valuation.
I claim $ \frac{1}{2} {2n\choose n}$ is odd if and only if $n = 2^{k}$ for some $k$.
This is true since $v_{2}(2^{k}!) = 2^{k} - 1$, which implies $n = 2^{k}$ as $v_{2}\left({2n\choose n}\right) = v_{2}(2n!) - 2v_{2}(n!)$. Now considering the binary representation of $n$, we see $v_{2}(n!)$ can be written as the sum of terms of the form $v_{2}(2^{i}!)$ from which the result follows.
By complementary counting, the answer is therefore $1000 - \lceil\log_2(1000)\rceil= 990$.
It is simpler to find $n$s where the term is not even.  A quick substitution of $n = 2^i$ for $i = 0, 1, 2, \ldots , 9$ shows this is the case.
If $n = 2^i$ we have $\frac{1}{2} {2 \cdot 2^i \choose 2^i} = \frac{2^i (2^{i+1}-1)(2^{i+1}-2) \cdots (2^{i+1}-2^i)}{2^{i-1}!}$.  The number of factors of $2$ in the numerator are the same as the number of factors of $2$ in the denominator, so the term is not even.
We need not consider any other numbers where there are no factors of $2$, as these simply cannot be even.
Thus the total number is $1000 - 10 = 990$.