AMC 12 Mock Exam 2 Problem 19

Such $n$ have at most $10$ binary digits; write $n=\sum_{k=0}^{9}d_k2^k$ with exactly five $d_k=1$. Since $2\equiv-1\pmod3$, we have $2^k\equiv1$ for even $k$ and $2^k\equiv-1$ for odd $k$. If $e$ of the five $1$'s are in the even positions $0,2,4,6,8$ (and $5-e$ in the odd positions $1,3,5,7,9$), then $$n\equiv e-(5-e)=2e-5\equiv 2e+1\pmod 3.$$ This is $0$ exactly when $2e\equiv2$, i.e., $e\equiv1\pmod3$, so $e=1$ or $e=4$. The count is $\binom51\binom54+\binom54\binom51=25+25=\boxed{50}$.

(Guessing that a third of the $\binom{10}{5}=252$ such numbers work gives the distractor $84$.)