1, 3, 263, 789.
Nope, divisible by 3 (263 times)
2,367 is evenly divisible by 1, 3, 9, 263, 789 and 2367.
No.
no
No.
264
It can be simplified from 6/4734 to 1/789
5 789 is divisible by 7 so there is no remainder. The answer is 827.
0.0038
No. It isn't. If it doesn't have an even number in the last one of a number, it isn't
try 3. In looking for primes, a good place to start (after seeing if it even or not,) is to add the digits in the number together. In this case it would be 7+8+9 equals 24. If the sum is divisible by 3, then the original number is divisible by three.