Question 1208966
<pre>
Let's pretend they put plastic digits on ALL the houses on both sides of the
street. and that there are the same number of houses on both sides of the
street.

The number of digits in all the even integers 0,2,...,2n is the same as the
number of digits in all the odd integers 1,3,...,2n+1

It would have cost twice as much, $58.12x2 = $116.24, to put plastic digits on
all the houses on both the north and south sides of the street, that is, IF the
first even-numbered house were numbered 0.

[Even though 0 is not normally used as a house number, let's pretend it is
on this street anyway.]

If we consider 0 as a 1-digit number, then

there are 10 1-digit numbers which would cost $0.04x10=$0.40,

there are 100-10=90 2-digit numbers which would cost $0.04x2x90=$7.20,

and there are 1000-100=900 3-digit numbers which would cost $0.04x3x900=$108.

That would cost $0.40+$7.20+$108=$115.60

Since $116.24-115.60=$0.64, that means they placed $0.64/$0.04 = 16 more
plastic digits than they placed on houses numbered with 3 or fewer digits.

So 16 plastic digits were placed on 4 houses with 4-digit numbers.

These are the houses numbered 1000, 1001, 1002, and 1003.

So the even house numbers are 0,2,4,...,1002 and there are 1002/2=501 of them.

The odd house numbers are 1,3,5,...,1003 and there are also 501 of them. 

Answer: There are 501 odd numbered houses on the north side of the street.

Edwin</pre>