题目内容
下面是用字母表示的三种形式的六位数(x、y、z表示三个数字,且x不等于0),形如( )的六位数,一定能被3整除.
A、xxyyzz | B、xyxyxy | C、xyyxzy |
分析:用十进制表示出结果,提公因式分组分解,找出公有的因数,再进一步分解质因数即可求得问题的答案.
解答:解:A、xxyyzz,
=100000x+10000x+1000y+100y+10z+z,
=110000x+1100y+11z,
=11(10000x+100y+z),
此数一定能被11整除,但不能被3整除;
B、xyxyxy,
=100000x+10000y+1000x+100y+10x+y,
=101010x+10101y,
=10101(10x+y),
因为10101=3×7×13×37,
所以此数一定能被3整除;
C、xyyxzy,
=100000x+10000y+1000y+100x+10z+y,
=100100x+11001y+10z,
此数不能被3整除.
故选:B.
=100000x+10000x+1000y+100y+10z+z,
=110000x+1100y+11z,
=11(10000x+100y+z),
此数一定能被11整除,但不能被3整除;
B、xyxyxy,
=100000x+10000y+1000x+100y+10x+y,
=101010x+10101y,
=10101(10x+y),
因为10101=3×7×13×37,
所以此数一定能被3整除;
C、xyyxzy,
=100000x+10000y+1000y+100x+10z+y,
=100100x+11001y+10z,
此数不能被3整除.
故选:B.
点评:此题主要考查利用十进制、分组分解因式以及分解质因数研究数的整除性.
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