题目信息
If n > 4, what is the value of the integer n ?
A:Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B:Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C:BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D:EACH statement ALONE is sufficient.
E:Statements (1) and (2) TOGETHER are NOT sufficient.
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正确答案：
A:Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Algebra Simplifying algebraic expressions
Because the numerators of the two fractions have several common factors, and similarly for the denominators, a reasonable strategy is to begin by appropriately canceling these common factors.
The manipulations above show that n = 7.
Alternatively, we could begin by reducing each of the fractions to lowest terms by using identities such as n! = (n − 3)! × (n − 2)(n − 1)(n), and then performing operations on the resulting equation; SUFFICIENT.
For the same reason given in (1) above, we begin by canceling factors that are common on the left and right sides of the equality.
The manipulations above show that the original equation is identically true for all integers greater than 4, and thus n can be any integer greater than 4.
Alternatively, we could begin by reducing each of the fractions to lowest terms by using identities such as n! = (n − 3)! × (n − 2)(n − 1)(n), and then performing operations on the resulting equation; NOT sufficient.
The correct answer is A;statement 1 alone is sufficient.
Because the numerators of the two fractions have several common factors, and similarly for the denominators, a reasonable strategy is to begin by appropriately canceling these common factors.
The manipulations above show that n = 7.
Alternatively, we could begin by reducing each of the fractions to lowest terms by using identities such as n! = (n − 3)! × (n − 2)(n − 1)(n), and then performing operations on the resulting equation; SUFFICIENT.
For the same reason given in (1) above, we begin by canceling factors that are common on the left and right sides of the equality.
The manipulations above show that the original equation is identically true for all integers greater than 4, and thus n can be any integer greater than 4.
Alternatively, we could begin by reducing each of the fractions to lowest terms by using identities such as n! = (n − 3)! × (n − 2)(n − 1)(n), and then performing operations on the resulting equation; NOT sufficient.
The correct answer is A;statement 1 alone is sufficient.
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