2, 4, 7 and 14.
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).
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)
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
No.
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.
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
GCF(28, 36) = 4