From left to right the first three digits of a 4 digits number add up to 8 which digits could be in the ones place if the 4 digit number is divisible by 6

The answer are

A. 5

B. 6

C. 0

D. 4

For any number to be divisible by 6, it must be divisible by both 2 and 3

To be divisible by 2, the unit digit must be even or zero, so that eliminates A - 5.

To be divisible by 3, the sum of the digits must be a multiple of 3 (divisible by 3)
Since we have a sum of 8 for the first 3 digits we need a sum of:
9, 12, 15, 18 etc
to get sum of 9, the last digit must be 1, but it can't be odd
to get a sum of 12, the last digit must be 4.
That could be a choice
to get a sum of 15, the last digit must be 7, no good
to get a sum of 18, our last digit must be 10, which is not single digit.
We can stop here.

The last digit has to be 4, which is choice D

Reiny can you help me with 5 questions

To solve this problem, we need to find the possible digits that can be in the ones place of a 4-digit number such that the number is divisible by 6.

First, let's consider the condition given which states that from left to right, the first three digits of the number add up to 8.

We'll calculate all the possible combinations of digits that add up to 8.

Since the first three digits add up to 8, the digits could be:

1 + 2 + 5 = 8
1 + 3 + 4 = 8
2 + 3 + 3 = 8

Next, we need to find the possible digits that could go in the ones place such that the whole number is divisible by 6.

In order for a number to be divisible by 6, it needs to be divisible by both 2 and 3.

To be divisible by 2, the number must end in an even digit (0, 2, 4, 6, or 8).

To be divisible by 3, the sum of all the digits of the number must add up to a multiple of 3.

Now, let's check which of the above combinations have digits in the ones place that satisfy these conditions.

For the combination 1 + 2 + 5 = 8, the possible digits in the ones place could be 0, 2, 4, 6, or 8.

For the combination 1 + 3 + 4 = 8, the possible digits in the ones place could be 0, 2, 4, 6, 8.

For the combination 2 + 3 + 3 = 8, the only possible digit in the ones place is 2.

Therefore, if the 4-digit number is divisible by 6 and the first three digits add up to 8, the possible digits that could be in the ones place are 0, 2, 4, 6, or 8.