1, 2, 4, 7, 8, 14, 28, 56
Two is divisible by 56, but the number does not come out evenly. You get 0.0357 as your answer. 56 is divisible by 2 and it goes in evenly, however. (56/2 is 28).
56 is divisible by 1, 2, 4, 7, 8, 14, 28, and 56.
56 is divisible by: 1, 2, 4, 7, 8, 14, 28, 56.
56
5+6=11 which is not divisible by 3, then 56 is not divisible by 3
All multiples of 56, which is an infinite number.
No, 56/9 = 6.22222(recurring)
To show that the product of 56 and 39 is exactly divisible by 21, you can check if both numbers are divisible by the prime factors of 21, which are 3 and 7. Since 56 is divisible by 7 (56 ÷ 7 = 8) and 39 is divisible by 3 (39 ÷ 3 = 13), their product will also be divisible by both 3 and 7. Therefore, 56 x 39 is divisible by 21.
7
No.
Because 8 times 7 = 56
No - 56/45 = 1.24 recurring (that is, 1.24444..).