Answer
3200cents or $32
Explanation:
This can be solve using Arithmetic progression with the first term a =4 and common difference d =4 we need to know the number of term n that will give 120
the formular is
a+(n-1)d =nth term if you substitute the figures into the formular then
We will have 4 +( n-1) 4= 120
4 + 4n-4=120........evaluate like terms
4 - 4 +4n = 120
4n =120
n =120/4
n = 30 which means the total number of houses that were numbered are 30
2 of the houses will have single digit (I.e. 4 and 8)
6 of the houses will have 3 digits
( 100,104,108,112,116,120)
The rest will have 3 digits
2+6+ the rest = 30
the rest = 30-6-2
the rest = 22
Therefore total number of digits will be
1 x 2
+
2 x 22
+
3 x 6
=
2
+
44
+
18
=
64 digits
If single digit cost 50 cents
Then 64 digit will cost
50 cent x 64 = 3200 cents