What are the factors of -228?

Answer 1

I assume that we are speaking entirely of positive integers. There are six divisors of 228 that are less than 15: 1, 2, 3, 4, 6, and 12. There are six divisors of 228, greater than 15: 19, 38, 57, 76, 114, and 228 itself. There are several convenient ways to check that you have got them all, without the labour of dividing the first 228 integers into 228. One way is the following: Note, first, that 228 = 19 X 12 = 19 X 3 X 2 X 2. We may call that its 'prime factorisation'. Because every positive integer can be factorised into positive prime integers in only one way, we may safely conclude that every positive divisor of 228 must contain 19, either once or not at all; similarly, it must contain 3 once or not at all; and it must contain 2, twice, once, or not at all. That allows 2 X 2 X 3 = 12 choices for a divisor of 228; hence, there must be twelve divisors of 228, including, of course, 1 and 228 itself. Now, half of these divisors must be less than the square-root of 228, which is just a little bit more than 15; half must be more than 15. 228 is not a perfect square, so we may conclude that six divisors are less than 15. They are easy to identify, since their divisors, in turn, must contain some combination of 1's, 2's and 3's. The remaining six divisors, greater than 15, can be found by dividing the first six divisors into 228. And the job is done. If you want negative divisors, simply take the negatives of all the twelve positive divisors. canislunis

Answer 2

1, 2, 3, 4, 6, 12, 19, 38, 57, 76, 114, 228

-1, -2, -3, -4, -6, -12, -19, -38, -57, -76, -114, -228