Determine the sum of all single-digit replacements for z such that the number 24, z38 is divisible by 6.

The sum of all single-digit replacements for z is 12

Step-by-step explanation:

* Lets explain how to solve the problem

- The number is divisible by 6 if it divisible by 2 and 3

- Any even number divisible by 2

- The number is divisible by 3 is the sum of its digits divisible by 3

* Now lets solve the problem

- The number 24, z38 is divisible by 6

- We need to find all the possible values of z which keep the number

divisible by 6

∵ Lets add the sum of the digits without z

∵ 2 + 4 + 3 + 8 = 17

∵ 18 is the nearest number to 17

∵ 18 is divisible by 3

∴ Add 17 by 1 to get 18

∴ z = 1

- Lets check the number

∵ The number is 24,138

∵ 24,138 : 6 = 4023

∴ The number is divisible by 6

∵ 21 is the next number after 18 and divisible by 3

∴ We must add 17 by 4 to get 21

∴ z = 4

- Lets check the number

∵ The number is 24,438

∵ 24,438 : 6 = 4073

∴ The number is divisible by 6

∵ 24 is the next number after 18 and divisible by 3

∴ We must add 17 by 7 to get 24

∴ z = 7

- Lets check the number

∵ The number is 24,738

∵ 24,738 : 6 = 4123

∴ The number is divisible by 6

- There is no other value for z because if we take the next number

of 24 divisible by 3 it will be 27, then we must add 17 by 10 but

10 not a single digit

∴ The possible values of z are 1, 4, 7

∴ The sum of them = 1 + 4 + 7 = 12

∴ The sum of all single-digit replacements for z is 12