Note that,except the first move where we only put some powers of $2$,in every other move we can make at most two different number equal.This is because of : 1) $2^x+2^x= 2^(x+1)$ 2)$2^m+2^n=2^p+2^q$ implies ${m,n}={p,q}$ [Indeed proof of (2) is simple.If $p$ is max of ${p,q,m,n}$ then $2^p+2^q>2^p - 1 > 1+ 2^2+...2^(p-1)>2^m+2^n$.contradiction.] So,$15-1=14$ numbers can be make equal to the other number in at least $14/2=7$ moves and with the first move at least 8 moves needed. For a example put ${1,2,2^2,2^3,2^4,2^5,2^6,2^7,0,0,0,0,0,0,0}$ In the second move we can get ${2,2^2,2^3,2^4 ,2^5,3^6,2^7,2^7,2^7,0,0,0,0,0,0}$ by putting $1$,$2$,$2^2$,$2^3$,$2^4$,$2^5$,$2^6$ in the first 7 box and $2^7$ in the 9th box. In the third move we can get ${2^2,2^3,2^4,2^5,2^6,2^7,2^7,2^7,2^7,2^7,0,0,0,0,0}$ by putting $2$,$2^2$,$2^3$,$2^4$,$2^5$,$2^6$ in the first 6 box and $2^7$ in the 10 box.So in this way at the end of $ 8$ moves we get all numbers equal to $2^7$.