Is x a multiple of 9? (1) x = (b!)(c), where b and c are distinct int

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Since we have 1 variable (x) and 0 equations, D is most likely to be the answer. So, we should consider each condition on its own first.

Condition 1)
If b = 3 and c =3, then x=(3!)*3 = 18 which is a multiple of 9 and the answer is 'yes'.
If b = 3 and c =4, then x=(3!)*4 = 24 which is not a multiple of 9 and the answer is 'no'.
Since condition 1) does not yield a unique answer, it is not sufficient.

Condition 2)
If c = 3 and x = 18, then x is a multiple of 9 and the answer is 'yes'.
If c = 3 and x = 12, then x is not a multiple of 9 and the answer is 'no'.
Since condition 2) does not yield a unique answer, it is not sufficient.

Conditions 1) & 2)
Since c is an integer with 2<c<4, we have c = 3.
Since b is an integer with b > 2 and b≠3=c we have b≥4 and b! is always a multiple of 3.
Thus, x = (b!)c is a multiple of 9 and the answer is 'yes', since both b! and c are multiples of 3.

Since both conditions together yield a unique answer, they are sufficient.

Therefore, C is the answer.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.