GMAT备考的奇偶性： 奇数（odd），又称单数，是整数中不能被2整除的数，奇数的个位为1，3，5，7，9。可用2k+1表示，这里k就是整数。
偶数（even），又称双数，若某数是2的倍数，它就是偶数（双数），可表示为2k。 所有整数不是奇数（单数），就是偶数（双数）。
奇数个奇数相加减=奇数，偶数个奇数相加减=偶数，奇数和偶数相加减=奇数，任意个偶数相加减=偶数，相乘时有一个偶数时结果是偶数，只有全部是奇数相乘结果才是奇数。
例： If n is an integer, is n even?
(1) n^2-1 is an odd integer.
(2) 3n + 4 is an even integer.
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
【解析】D 条件1：n^2-1是奇数，n^2是偶数，n即为偶数（一个数的平方的奇偶性和这个数本身的奇偶性是一致的），充分； 条件2：3n + 4是偶数，3n是偶数，n即为偶数，充分。
If x、y is positive integer, is xy even?
(1)y=x+1
(2)y= x^2+1
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
【解析】D 条件1：y=x+1，x和y就是连续的两个整数，那么x和y一奇一偶，xy是偶数，充分； 条件2：y=x^2+1，y和x^2就是连续的两个整数，因为x^2和x的奇偶性一致，就能得到y和x还是一奇一偶，xy是偶数，充分。
If A and B are positive integers such that A-B and A / B are both even integers, which of the following must be an odd integer?
(A)A/2 (B)B/2 (C)(A+B)/2 (D)(A+2)/2 (E)(B+2)/2
【解析】D A-B是偶数：那么A和B都是奇数或都是偶数； A/B是偶数：那么A=B*偶数，A一定是偶数； 所以A,B都是偶数，假设B=2N,A=B*偶数=2N*偶数=2N*2M=4MN； 带入选项： A/2=2MN，偶数，排除； B/2=N，奇偶性不确定，排除； (A+B)/2=2MN+N，奇偶性不确定，排除； (A+2)/2 =2MN+1，一定是奇数，正解； (B+2)/2=N+1，奇偶性不确定，排除。
If x and y are integers, is y an even integer?
(1) 2y – x = x^2 – y^2 (2) x is an odd integer.
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.
【解析】A 条件1：y(y+2)=x(x+1)，x和x+1是连续的两个整数，所以x和x+1一奇一偶，x(x+1)是偶数，所以y(y+2)是偶数，y和y+2的奇偶性是一致的，所以只能是y和y+2同为偶数，充分； 条件2：x是奇数，x和y的奇偶性无关，y奇偶性未知，不充分。
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