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
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.
Zero out the Y to find X 3X - 2(0) = -789 3X = -789 X = -263 ------------- now to check you need to find Y, so zero out X 3(0) - 2Y = -789 - 2Y = - 789 Y = 394.5 0r 394 and 1/2 convert to fraction = 789/2 ------------ check 3(-263) - 2(789/2) = -789 -789 - 2(789/2) = -789 cancel one 2 and multiply through by that 2 -1578 + 789 = - 789 -789 = - 789 checks