Team Selection Test

The answer is $1024$. Let us call a partition of $\{1,2,\dots ,n\}$ into two sets nice partition for $n$, if none of the sets contains two distinct elements whose sum is a power of $2$. Let $p_n$ be the number of nice partitions for $n$. We observe that removing $n$ from a nice partition for $n$ gives a nice partition for $n-1$. Therefore, we can obtain the nice partitions for $n$ by adding $n$ to the nice partitions for $n-1$. If $2^m<n<2^{m+1}$ for some positive integer $m$, then as $2^m<n+1\le n+(n-1)<2^{m+2}$ and $1\le 2^{m+1}-n\le n-1$, we must add $n$ into the set not containing $2^{m+1}-n$. Therefore, $p_n=p_{n-1}$. If $n=2^m$ for some positive integer $m$, then since $2^m<n+1\le n+(n-1)<2^{m+1}$, we can add $n$ into any of the sets and hence $p_n=2\cdot p_{n-1}$. As $p_2=2$ and $2^{10}<2012<2^{11}$, we conclude that $p_{2012}=2^{10}=1024$.