Sol07052006

Notice:

$1=1/2+1/3+1/6;\ 1=2/3+1/3$

$1=1/2+1/3+1/6(2/3+1/3)=1/2+1/3+1/9+1/18\,$

$1=1/2+1/3+1/9+1/18(2/3+1/3)=1/2+1/3+1/9+1/27+1/54\,$

$\cdots\,$

$1 = 1/2+1/3^1+1/3^2+\cdots 1/3^n + 1/(2*3^n)\,$

Then with n=2004 we get 2006 terms. Our set is:

$\{2,3^1,3^2,\cdots,3^{2004},2*3^{2004}\}$