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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Because 7 divides 56 evenly, 56 is the LCM of 7 and 56.
The multiplication problem 12 times 56 is approximately 784.
The least common multiple of the numbers 2,842 and 56 is 11,368.The least common multiple of the numbers 28, 42 and 56 is 168.
The Least Common Multiple (LCM) for 42 56 is 168.
Least Common Multiple (LCM) for 56 31 is 1,736