2, 4, 7 and 14.
7 x 29 = 203, 7 x 71 = 497. There are therefore 71 - 28 ie 43 multiples of 7 in the range.
8√7. Since the square root of 4 is 2 you just keep dividing the number by 4 and multiplying 2 outside the radical until you get to a number not divisible by 4. (√448 = √[112*4] = 2√112) then (2√112 = 2√[28*4] = 4√28) then (4√28 = 2√[7*4] = 8√7)
You can determine if any number is divisible by nine by adding its digits together: 2 + 1 + 2 +4 = 9 if the result is divisible by 9, then the original number is divisible by 9. In this case, the answer is divisible by 9 so, yes it is (in fact it is 236 x 9).
None. The smallest number divisible by 528 is 528.
this # is divisible by 3, 5, and 9 :-) ;-P
28 is not divisible by 3.
28 is divisible by 1, 2, 4, 7, 14, 28.
28 is divisible by: 1, 2, 4, 7, 14, 28.
28 and its multiples.
The numbers that are divisible by 28 are infinite. The first four are: 28, 56, 84, 112 . . .
28, 56, 84, and all other multiples of 28 that are factors of other numbers are also divisible by 28.
evenly divisible 28 times
yes 4x7=28
28 and 90 are both divisible by 2, as they are both even numbers. Additionally, 28 is divisible by 4 and 7, while 90 is divisible by 3, 5, 6, 9, 10, 15, 18, 30, 45, and 90.
No.
28 is a composite number as it is divisible by numbers other than itself and one. 28 is divisible by 1, 2, 4, 7, 14 and 28
yes as four sevens go in to twenty-eight * * * * * No, that answers the question: "Is 28 divisible by 7?" 7 is not divisibe by 28.