irrigationman
幼苗
共回答了19个问题采纳率:78.9% 举报
设集合
![](https://img.yulucn.com/upload/3/36/33608131f64a6a3da6edcf80ec04cbc9_thumb.jpg)
,在S上定义运算
![](https://img.yulucn.com/upload/f/8c/f8cb9f3826151e43cfa6d4dd31a13f72_thumb.jpg)
,其中k为i+j被4除的余数,
![](https://img.yulucn.com/upload/9/23/9235ad7a6cde2a3289efd9c20af6755f_thumb.jpg)
,则使关系式
![](https://img.yulucn.com/upload/4/39/4398f510d33c21512507442c5c4f66e0_thumb.jpg)
成立的有序数对(i,j)的组数为()
A.4 B.3 C.2 D.1
A
分析:由已知中集合S={A
0 ,A
1 ,A
2 ,A
3 },在S上定义运算⊕:A
i ⊕A
j =A
k ,其中k为i+j被4除的余数,i,j=0,1,2,3,分别分析A
i 取A
0 ,A
1 ,A
2 ,A
3 时,式子的值,并与A
0 进行比照,从而可得到答案.
当A
i =A
0 时,(A
i ⊕A
i )⊕A
j =(A
0 ⊕A
0 )⊕A
j =A
0 ⊕A
j =A
j =A
0 ,∴j=0
当A
i =A
1 时,(A
i ⊕A
i )⊕A
j =(A
1 ⊕A
1 )⊕A
j =A
2 ⊕A
j =A
0 ,∴j=2
当A
i =A
2 时(A
i ⊕A
i )⊕A
j =(A
2 ⊕A
2 )⊕A
j =A
0 ⊕A
j =A
0 ,∴j=4
当A
i =A
3 时(A
i ⊕A
i )⊕A
j =(A
3 ⊕A
3 )⊕A
j =A
2 ⊕A
j =A
0 =,∴j=2
∴使关系式(A
i ⊕A
i )⊕A
j =A
0 成立的有序数对(i,j)的组数为4组.
故选A.
1年前
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