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. Both are divisible by 5.No. Both are divisible by 5.No. Both are divisible by 5.No. Both are divisible by 5.
35 is divisible by 5 because any number that ends in 5 is divisible by 5.
Yes, 1255 is divisible by 5, and any number that ends in 5 is divisible by 5.
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)
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, because when something is divisible by 6, it is divisible by 2 or 3. This is divisible by 3, but not 2.
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
No. Both are divisible by 5.No. Both are divisible by 5.No. Both are divisible by 5.No. Both are divisible by 5.
YES. 285 is divisible by 5. A number is divisible by 5 if it ends with 0 or 5
35 is divisible by 5 because any number that ends in 5 is divisible by 5.
Yes, 1255 is divisible by 5, and any number that ends in 5 is divisible by 5.
It's false because we have numbers that is divisible by 10 but not divisible by 5 and vice versa, we have numbers that is divisible by 10 but not divisible by 5.