Well, honey, let me break it down for you. 56 is divisible by 2 because it's even, by 4 because the last two digits form a number divisible by 4, and by 8 because the last three digits form a number divisible by 8. However, it's not divisible by 3, 5, 6, or 9. So, to answer your question, 56 is not divisible by each of those numbers.
56 is divisible by: 1, 2, 4, 7, 8, 14, 28, 56.
56 of them.
The numbers that are divisible by 28 are infinite. The first four are: 28, 56, 84, 112 . . .
1x56 2x56 3x56 etc.
All number that are completely divisible by 10 (with no remainder) end in 0 (zero). Dividing 565 by 10 yields 56 with a remainder of 5 (or 56.5 or 56½) - so - NO - 565 is NOT divisible by 10.
56 is divisible by: 1, 2, 4, 7, 8, 14, 28, 56.
56 of them.
The numbers that are divisible by 56 are in the form 56k where k is the integer constant.
The numbers that are divisible by 28 are infinite. The first four are: 28, 56, 84, 112 . . .
1x56 2x56 3x56 etc.
8, 7, 2, 28
All number that are completely divisible by 10 (with no remainder) end in 0 (zero). Dividing 565 by 10 yields 56 with a remainder of 5 (or 56.5 or 56½) - so - NO - 565 is NOT divisible by 10.
Well, honey, let me break it down for you. Yes, 56 is divisible by 2, 4, 7, and 8. But it ain't gonna fly with 3, 5, 6, 9, and 10. So, in a nutshell, 56 can be divided by some of those numbers, but not all of 'em. Hope that clears things up for ya!
There are 10 numbers between 10 and 80 that are divisible by 7 with no remainder: 14, 21, 28, 35, 42, 49, 56, 63, 70, and 77.
224 is evenly divisible by these numbers: 1, 2, 4, 7, 8, 14, 16, 28, 32, 56, 112 and 224.
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).
No number between 20 and 40 is divisible by all 4 numbers. The smallest number divisible by all 4 of those numbers is: 56