The house numbers in a housing development are the multiples of four starting at 4 and ending at 120. If metal digits cost 50 cents each, what is the total cost of all of the metal digits that are needed to number the houses in this development?

Answer

3200cents or $32

Step-by-step 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