Positive integers a, b and c are all less than 10. If the sum of all

----
What do we know to start?
----

We have the following six possible configurations, where each of a,b,c each are a digit of a three-digit number.
ABC
ACB
BAC
BCA
CAB
CBA

Then we know the sum of these numbers is 1998, i.e. ABC + ACB + BAC + BCA + CAB + CBA = 1998

So, knowing a number of the format XYZ = 100x + 10y + z, we can convert the above statement into something more workable:

100(a + a + b + b + c + c) + 10(b + c + a + c + a + b) + (c + b + c + a + b + a) = 1998
100(2a + 2b + 2c) + 10(2a + 2b + 2c) + (2a + 2b + 2c) = 1998
111(2a + 2b + 2c) = 1998
222(a + b + c) = 1998
a + b + c = 1998 / 222 = (999 * 2) / (2 * 111) = 999 / 111 = 9

So we know a + b + c = 9

Off the bat, we should recognize that knowing two of these numbers necessarily tells us the third number. So the answer can't be E. But let's see if we can do better.


---------------
What does A tell us?
---------------

c = 4

a + b + c = 9
a + b + 4 = 5
a + b = 5

Note that a + b is an odd number. There is no way we could get 5 from a + b if a = b. You could also figure this out by plugging in the possible combinations of a and b.

So Statement A is sufficient

---------------
What does B tell us?
---------------

a = 3

Let's pull back the statement we used earlier: a + b + c = 9

If a = 3, then 3 + b + c = 9

This means b + c = 6

Is it possible for b to be 3? Of course! Suppose a = b = c = 3. Then this checks out.

But it's not necessary for b to be 3. Consider the following counterexample: b = 2, c = 4. This works, but a != b.

So statement B is not sufficient

-------

The answer is A