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 Erdős.
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}
Solution
