The problem states that all 9 single digits in the problem are different; in other words, there are no repeated digits.
(1) SUFFICIENT: Given 3a = f = 6y, the only possible value for y = 1. Any greater value for y, such as y = 2, would make f greater than 9. Since y = 1, we know that f = 6 and a = 2.
We can now rewrite the problem as follows:
2 b c
+ d e 6
= x 1 z
In order to determine the possible values for z in this scenario, we need to rewrite the problem using place values as follows:
200 + 10b + c + 100d + 10e + 6 = 100x + 10 + z
This can be simplified as follows:
196 = 100(x - d) - 10(b + e) + 1(z - c)
Since our focus is on the units digit, notice that the units digit on the left side of the equation is 6 and the units digit on the right side of the equation is (z - c). Thus, we know that 6 = z - c.
Since z and c are single positive digits, let's list the possible solutions to this equation.
z = 9 and c = 3
z = 8 and c = 2
z = 7 and c = 1
However, the second and third solutions are NOT possible because the problem states that each digit in the problem is different. The second solution can be eliminated because c cannot be 2 (since a is already 2). The third solution can be eliminated because c cannot be 1 (since y is already 1). Thus, the only possible solution is the first one, and so z must equal 9.
(2) INSUFFICIENT: The statement f - c = 3 yields possible values of z. For example f might be 7 and c might be 4. This would mean that z = 1. Alternatively, f might be 6 and c might be 3. This would mean that z = 9.
The correct answer is A.
hope the OE helps
