No. 1,827 is only divisible by 1, 3, 7, 9, 21, 29, 63, 87, 203, 261, 609, 1827.
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)
Any of its multiples
3 9 0
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, 1255 is divisible by 5, and any number that ends in 5 is 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.
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.