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⇔ 100a+10b + c − 100c − 10b − a=99a − 99c = 99 (a−c)
=> abc - cba chia hết cho 99
Ta có : \(\overline{abc}-\overline{cba}=\left(100a+10b+c\right)-\left(100c+10b+a\right)\)
\(=\left(100a-a\right)+\left(10b-10b\right)-\left(100c-c\right)\)
\(=99a-99c=99\left(a-c\right)\) chia hết cho 99
![](https://rs.olm.vn/images/avt/0.png?1311)
A, ab + bc chia het cho 11
Ta có : 10 a +b +10b +a
=11a +11b
=11 (a+b) chia het cho 11
B, abc - cba chia het cho 99
Ta có :( 100a +b +c ) - ( 100c +b+a )
=99a - 99c
=99 (a-b) chia het cho 99
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\overline{abcabc}=100000a+10000b+1000c+100a+10b+c\) \(=100100a+10010b+1001c\) \(=1001\left(100a+10b+c\right)=7\cdot11\cdot13\left(100a+10b+c\right)⋮7,11,13\)
b) Ta có: \(\overline{ab}-\overline{ba}=10a+b-10b-a=9a-9b\) \(=9\left(a-b\right)⋮9\)
c) Ta có: \(\overline{abc}-\overline{cba}=100a+10b+c-100c-10b-a=99a-99c=99\left(a-c\right)⋮99\)
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theo đề bài ta có aba - cba =(100a + 10b + c) - (100c + 10b + c) =(100a - a) + (10b - 10b) + (100c - 100c) =99a + 99c =99.(a + c) chia hết cho 99 => abc - cba chia hết cho 99
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ab + ba = (10a + b) + (10b + a) = 11a + 11b = 11.(a + b) chia hết cho 11
b) abc - cba = (100a + 10b + c) - (100c + 10b + a) = 99a - 99c = 99.(a - c) chia hết cho 99
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
abc - cba = 100a + 10b + c - 100c - 10b - a = (100a - a) + (10b - 10b) - (100c - c) = 99a - 99c = 99(a - c) chia hết cho 99
Ta có:
abc-bca=100a+10b+c-100c-10b-a=( 100a-a )+(10b-10b)-(100c-c)=99a-99c=99,(a-c)chia hết cho 99
abc-cba
=100a+10b+c-100c-10b-a
=99a-99c chia hết cho 99