Because 8 times 7 = 56
56
56 is divisible by: 1, 2, 4, 7, 8, 14, 28, 56.
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.
All multiples of 56, which is an infinite number.
No. Both are divisible by 7.
56
If the last digit doubled subtracted from the rest is a multiple of 7, the whole number is divisible by 7. Example: 686 68 - 12 = 56 56 is divisible by 7. 686 is divisible by 7.
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.
7
All multiples of 56, which is an infinite number.
1, 2, 4, 7, 8, 14, 28, 56
8, 7, 2, 28
No. Both are divisible by 7.
7x8 is divisible by: 1 2 4 7 8 14 28 and 56.
2, 4, 7 and 14.
7 x 8 = 56