Solutions to the 66

A–1 We proceed by induction, with base case 1= 2030. Suppose all integers less than n− 1 can be represented. If n is even, then we can take a representation of n/2 and multiply each term by 2 to obtain a representation of n. If n is odd, put m = blog3 nc, so that 3m ≤ n < 3m+1. If 3m = n, we are done. Otherwise, choose a representation (n−3m)/2= s1+ · · ·+sk in the desired form. Then n = 3m +2s1 + · · ·+2sk, and clearly none of the 2si divide each other or 3m. Moreover, since 2si ≤ n− 3m < 3m+1 − 3m, we have si < 3m, so 3m cannot divide 2si either. Thus n has a representation of the desired form in all cases, completing the induction.