A street has 20 houses on each side, for a total of 40 houses. The addresses on the south side of the street form an arithmetic sequence, as do the addresses on the north side of the street. On the south side, the addresses are 4, 10, 16, etc., and on the north side they are 3, 9, 15, etc. A sign painter paints house numbers on a house for $1 per digit. If he paints the appropriate house number once on each of these 40 houses, how many dollars does he collect?

Guest Dec 8, 2018

#1**+2 **

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

CPhill Dec 8, 2018