The greatest integer on the "odd" side of the street is
3 + 6(20 - 1) = 3 + 6(19) = 3 + 114 = 117
2 houses have one digit each 3, 9
3 houses will have three digits each 105, 111, 117
So...15 houses will have two digits each
So...for the "odd" side he collects 2(1) + 15(2) + 3(3) = 2 + 30 + 9 = $ 41
For the "even" side, the greatest integer is 4 + 6(20-1) = 4 + 6(19) = 118
1 house has one digit
4 houses have three digits each [ 100, 106, 112, 118]
So.....15 houses wil have two digits each
So....for the "even" side he collects 1(1) + 15(2) + 4(3) = 1 + 30 + 12 = $43
So....the total he collects is $ [ 41 + 43 ] = $ 84