Determine the sum of all single-digit replacements for $z$ such that the number ${24{,}z38}$ is divisible by 6.
For a number to be divisible by 6, it must be divisible by 3 and 2. Since our number ${24{,}z38}$ 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: $2 + 4 + z + 3 + 8 = 17 + z.$ In order for ${24{,}z38}$ to be divisible by 3, $17 + z$ must be divisible by 3, meaning that $z$ is $1,$ $4,$ or $7.$

We can verify that ${24{,}138},$ ${24{,}438},$ and ${24{,}738}$ are all divisible by 6, and therefore our answer is $1 + 4 + 7 = \boxed{12}.$
12
Level 3
Prealgebra
Problem: Determine the sum of all single-digit replacements for $z$ such that the number ${24{,}z38}$ is divisible by 6.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the sum of all single-digit replacements for $z$ such that the number ${24{,}z38}$ is divisible by 6, we need to apply the divisibility rules for 6. A number is divisible by 6 if it is divisible by both 2 and 3.

First, we check the condition for divisibility by 2. A number is divisible by 2 if its last digit is even. In the number ${24{,}z38}$, the last digit is 8, which is even. Therefore, the number is divisible by 2 regardless of the value of $z$.

Next, we check the condition for divisibility by 3. A number is divisible by 3 if the sum of its digits is divisible by 3. The digits of the number ${24{,}z38}$ are 2, 4, $z$, 3, and 8. We calculate the sum of the known digits:

2 + 4 + 3 + 8 = 17.

Now, we include $z$ in the sum:

Sum = 17 + $z$.

For ${24{,}z38}$ to be divisible by 3, the expression $17 + z$ must be divisible by 3. We can find the values of $z$ that satisfy this condition by evaluating $17 + z$ for each single-digit value of $z$ (0 through 9) and checking the divisibility by 3.

Calculating for each $z$:

- If $z = 0$: $17 + 0 = 17$ (not divisible by 3)
- If $z = 1$: $17 + 1 = 18$ (divisible by 3)
- If $z = 2$: $17 + 2 = 19$ (not divisible by 3)
- If $z = 3$: $17 + 3 = 20$ (not divisible by 3)
- If $z = 4$: $17 + 4 = 21$ (divisible by 3)
- If $z = 5$: $17 + 5 = 22$ (not divisible by 3)
- If $z = 6$: $17 + 6 = 23$ (not divisible by 3)
- If $z = 7$: $17 + 7 = 24$ (divisible by 3)
- If $z = 8$: $17 + 8 = 25$ (not divisible by 3)
- If $z = 9$: $17 + 9 = 26$ (not divisible by 3)

The values of $z$ that make $17 + z$ divisible by 3 are 1, 4, and 7. 

Now, we sum these valid values of $z$:

1 + 4 + 7 = 12.

Thus, the final answer is 12.<|END_OF_LATENT|> The final answer is $12$.