The answer is $2$ Lemma. If an odd number is written on the board say $m.$ Then let the other two ( or more) numbers be of the form $2^{\alpha}-m$ and $2^{\beta}-m.$ But then adding these two numbers, we get $v_2( 2^{\alpha}-m+ 2^{\beta}-m)= 1. $ Hence these should be $1$. So board will be like $[m,1]$. $B:=\{a_1,a_2,\ldots, a_k\}$ consider $T=\min(v_2(a_i))$ now $B':=B/2^T$