GMAT備考|GMAT數(shù)學(xué)考點

GMAT備考的奇偶性: 奇數(shù)(odd),又稱單數(shù),是整數(shù)中不能被2整除的數(shù),奇數(shù)的個位為1,3,5,7,9??捎?k+1表示,這里k就是整數(shù)。

?偶數(shù)(even),又稱雙數(shù),若某數(shù)是2的倍數(shù),它就是偶數(shù)(雙數(shù)),可表示為2k。 所有整數(shù)不是奇數(shù)(單數(shù)),就是偶數(shù)(雙數(shù))。

?奇數(shù)個奇數(shù)相加減=奇數(shù),偶數(shù)個奇數(shù)相加減=偶數(shù),奇數(shù)和偶數(shù)相加減=奇數(shù),任意個偶數(shù)相加減=偶數(shù),相乘時有一個偶數(shù)時結(jié)果是偶數(shù),只有全部是奇數(shù)相乘結(jié)果才是奇數(shù)。

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 條件1:n^2-1是奇數(shù),n^2是偶數(shù),n即為偶數(shù)(一個數(shù)的平方的奇偶性和這個數(shù)本身的奇偶性是一致的),充分; 條件2:3n + 4是偶數(shù),3n是偶數(shù),n即為偶數(shù),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 條件1:y=x+1,x和y就是連續(xù)的兩個整數(shù),那么x和y一奇一偶,xy是偶數(shù),充分; 條件2:y=x^2+1,y和x^2就是連續(xù)的兩個整數(shù),因為x^2和x的奇偶性一致,就能得到y(tǒng)和x還是一奇一偶,xy是偶數(shù),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是偶數(shù):那么A和B都是奇數(shù)或都是偶數(shù); A/B是偶數(shù):那么A=B*偶數(shù),A一定是偶數(shù); 所以A,B都是偶數(shù),假設(shè)B=2N,A=B*偶數(shù)=2N*偶數(shù)=2N*2M=4MN; 帶入選項: A/2=2MN,偶數(shù),排除; B/2=N,奇偶性不確定,排除; (A+B)/2=2MN+N,奇偶性不確定,排除; (A+2)/2 =2MN+1,一定是奇數(shù),正解; (B+2)/2=N+1,奇偶性不確定,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 條件1:y(y+2)=x(x+1),x和x+1是連續(xù)的兩個整數(shù),所以x和x+1一奇一偶,x(x+1)是偶數(shù),所以y(y+2)是偶數(shù),y和y+2的奇偶性是一致的,所以只能是y和y+2同為偶數(shù),充分; 條件2:x是奇數(shù),x和y的奇偶性無關(guān),y奇偶性未知,不充分。

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