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

12

Step-by-step explanation:

For a number to be divisible by 6, it must be divisible by 3 and 2. Since our number is obviously an even number, we only need to worry about whether it is divisible by 3. In order to check divisibility by 3, we find the sum of our digits: In order for to be divisible by 3, 17+z must be divisible by 3, meaning that z is 1, 4, or 7.

We can verify that  and are all divisible by 6, and therefore our answer is  

Hope this helped! :)

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step-by-step explanation: