题目内容

12.设集合S={A0,A1,A2,A3},在S上定义运算为:AiAj =Ak,其中ki+j被4除的余数,i,j=0,1,2,3.则满足关系式(xx)  A2=A0x(xS)的个数为(  )

(A)1                        (B)2                               (C)3                        (D )4

答案:B

解析:∵xS,∴下面验证x的取值.

(A0A0)A2=A0A2=A2A0,

(A1A1)A2=A2A2=A0,

(A2A2)A2=A0A2=A2A0,

(A3A3)A2=A2A2=A0,

x的个数有2个.

 


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