Show that every positive integer is a sum of one or more
numbers of the form 2^r*3^s, where r and s are nonnegative
integers and no summand divides another.

Remarks: This problem was originally created by Paul ErdÅ‘s.

Note that the representations need not be unique: for instance,**
11 = 2+9 = 3+8:**

This turns out to be a lot simpler than it seems. Armando is right with his even number idea: make set 2*n simply doubling everything from set n. Odd numbers n are almost as easy: pick the biggest power of three you can, call it 3^x, and then tack on the set for the even number n-3^x.

Then every set for each number n can be recursively built:

S_1 = {1}

S_2 = 2*S_1 = {2}

S_3 = 3^1 = {3}

S_4 = 2*S_2 = {4}

S_5 = 3^1 + S_2 = {3,2}

S_6 = 2*S_3 = {6}

S_7 = 3^1 + S_4 = {3,4}

S_8 = 2*S_4 = {8}

S_9 = 3^2 = {9}

S_10 = 2*S_5 = {6,4}

S_11 = 3^2 + S_2 = {9,2}

S_12 = 2*S_6 = {12}

S_13 = 3^2 + S_4 = {9,4}

S_14 = 2*S_7 = {6,8}

S_15 = 3^2 + S_6 = {9,6}

S_16 = 2*S_8 = {16}

S_17 = 3^2 + S_8 = {9,8}

S_18 = 2*S_9 = {18}

S_19 = 3^2 + S_10 = {9,6,4}

S_20 = 2*S_10 = {12,8}

S_21 = 3^2 + S_12 = {9,12}

S_22 = 2*S_11 = {18,4}

S_23 = 3^2 + S_14 = {9,6,8}

S_24 = 2*S_12 = {24}

S_25 = 3^2 + S_16 = {9,16}

S_26 = 2*S_13 = {18,8}

S_27 = 3^3 = {27}