1827/6 = 609/2 will give 304 with remainder 1. Yes 1827 is divisible by 6 but not evenly (meaning does not leave a 0 remainder)
Any of its multiples
No. 1,827 is only divisible by 1, 3, 7, 9, 21, 29, 63, 87, 203, 261, 609, 1827.
No. 483 is not divisible by 6.A number is divisible by 6 if it is divisible by both 2 and 3.It is divisible by 2 if it is even and it is divisible by 3 if the sum of the digits is a multiple of 3.483 is not divisible by 6 since it is not divisible by 2 although it is divisible by 3.
If a number is divisible by 2 and 3, it is divisible by 6.
Since 3 is a factor of 6, any number divisible by 6 is also divisible by 3. But since 2 is also a factor of 6, then any number divisible by 6 must also be divisible by 2. This means that any number divisible by 6 is an even number. So if a number is odd and it is divisible by 3, then it is not divisible by 6. For example, 12 is divisible by 3, but since it is even, it is also divisible by 6. But 15 is divisible by 3, and it is odd, so it is not divisible by 6.
No, because when something is divisible by 6, it is divisible by 2 or 3. This is divisible by 3, but not 2.
3 9 0
Any of its multiples
1,827 is divisible by: 1, 3, 7, 9, 21, 29, 63, 87, 203, 261, 609, 1827.
No. 1,827 is only divisible by 1, 3, 7, 9, 21, 29, 63, 87, 203, 261, 609, 1827.
Because the final digital sum is 9 then it is divisible by 9 as follows 1827/9 = 203
Any of its factors: 1, 3, 7, 9, 21, 29, 63, 87, 203, 261, 609, 1827
6 is not divisible by 162. 162 is divisible by 6.
No. 483 is not divisible by 6.A number is divisible by 6 if it is divisible by both 2 and 3.It is divisible by 2 if it is even and it is divisible by 3 if the sum of the digits is a multiple of 3.483 is not divisible by 6 since it is not divisible by 2 although it is divisible by 3.
If it is divisible by 2 and 3, it is divisible by 6.
if a number is divisible by 2 and 3 then its divisible by 6
If a number is divisible by 2 and 3, it is divisible by 6.