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No.

To check for divisibility by 6, the number must be divisible by both 2 and 3:

To check divisibility by 2: if the number is even (last digit is one of 0, 2, 4, 6, 8) then the number is divisible by 2. Last digit of 56 is 6 which is even so 56 is divisible by 2.

To check divisibility by 3: if the sum of the digits is divisible by 3, then so is the original number; the test can be repeated on the sum, so if it is repeated until a single digit remains, the original number is divisible by 3 if this single digit is 3, 6 or 9.

For 56: 5 + 6 = 11 → 11: 1 + 1 = 2; 2 is not one of 3, 6 or 9, thus 56 is not divisible by 3.

As 56 is not divisible by both 2 and 3, it is not divisible by 6.
No.
We can ask ourselves what are the multiples of six?

6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, and so on.

We see that 56 is not in this set of numbers. So it's not a multiple of six. We can also see that each of these numbers can be divided by six. So a quicker way of deciding would be to try to divide 56 by 6. 56/6 = 9-4/6 or 9-2/3. It doesn't divide evenly so again, 56 is not a multiple of 6.
no
60 = 6 x 10
54 = 6 x 9
so 56 is not a multiple of 6

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No, 6 is not a multiple of 56.

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Q: Is 6 a multiple of 56?
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