No.
evenly divisible 28 times
Yes with a remainder of 6 but it is not a factor of 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.
The factors of 18 are: 1, 2, 3, 6, 9, 18The factors of 28 are: 1, 2, 4, 7, 14, 28
28 is not divisible by 3.
A number that is divisible by 2 is not always evenly divisible by 6. It must also be divisible by 3, and must not be smaller than 6. The numbers 2, 4, 8, 10, 14, 16, 20, 22, 26, and 28 etc. are all divisible by 2 but not (evenly) by 6. The opposite is true, though. Any number divisible by 6 is also divisible by 2.
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 . . .
6 is not divisible by 162. 162 is divisible by 6.
28, 56, 84, and all other multiples of 28 that are factors of other numbers are also divisible by 28.