If you are wanting factors there is:
1, 3, 13, 39
No, it is divisible by 1, 3, 13, 39.
40 is nearest to 39 and is exactly by divisible 4
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.
No. 39 is odd but 6 is even and all multiples of 6 are even.
yes if you include decimal places.... any number is divisible except zero. The answer is 7.8
No, it is divisible by 1, 3, 13, 39.
40 is nearest to 39 and is exactly by divisible 4
Because it is divisible by numbers other than 1 and itself. 39 is divisible by 1, 3, 13, 39.
No - 39/4 = 9.75
No.
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.
Yes; if 39 is divided by 3, you get 13. Another way to check this is add 13+ 13+ 13, and you will get 39.
If its divisible by 5 AND 2 it must be divisible by 10 So you just have to pick the only number between 21 and 39 that's divisible by 10
All multiples of 39, which is an infinite number.
39 divided by 10 is 3.9 so 39 goes into 10 3times
3,2,5,6,9,10
Certainly I can: 100 ÷ 39 = 222/39 ≈ 2.564 However, 100 is not divisible by 39, that is 39 does not divide into 100 with no remainder.