Parity Devil

By binomial expansion, $(x+1)^{1000} = \dbinom{1000}0 x^{1000} + \dbinom{1000}1 x^{999} + \dbinom{1000}2 x^{998} + \cdots + \dbinom{1000}{1000} x^0$ .

So we want to find the number of odd numbers among the set of numbers, $\left \{\dbinom{1000}0 , \dbinom{1000}1 , \dbinom{1000}2 , \ldots, \dbinom{1000}{1000} \right \}$ .

Referencing Lucas' theorem :

In particular, $\binom{m}{n}$ is divisible by $p$ if and only if at least one of the base- $p$ digits of $n$ is greater than the corresponding base- $p$ digit of $m$ .

If we rewrite 1000 in base 2, we have $1000_{10} = 1111101000_2$ . Since there's six 1's in its binary representation, the answer is $2^6 = \boxed{64}$ .