ta có:abcabc=abc.1001=abc.7.11.13 chia hết cho 7,11 và 13

Ta có: \(\overline{abcabc}=\overline{abc}\times1001\)

Mà: \(1001=7\times11\times13\)

\(\Rightarrow\overline{abcabc}=\overline{abc}\times7\times13\times11\) ⋮ 7, 13, 11 (đpcm) 

abcabc = 100000a+10000b+1000c+100a+10b+c 
ababab = 100000a+10000b+1000a+100b+10a+b 
-->(abcabc +ababab ) =201110a+20111b+1001c 
=91(2210a+221b+11c) 

= 7.13 (2210a+221b+11c) chia hết cho 7

Giải:
Ta có:

abcabc = 100000.a + b.10000 + c.1000 + a.100 + b.10 +c

ababab = 100000.a + b.10000 + a.1000 + b.100 + a.10 + b

\(\Rightarrow\) abcabc + ababab = 201110.a + 20111.b + 1001.c = 91.( 2210.a + 221.b + 11.c ) chia hết cho 7 ( vì 91 = 13.7 chia hết cho 7 )

\(\Rightarrowđpcm\)

1) abcd = ab x 100 + cd 

= ab x 99 + ab + cd

Vậy nếu ab + cd chia hết cho 11 

Thì abcd chia hết cho 11 

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: \(\overline{abc}-\overline{cba}=100a+10b+c-100c-10b-a=99a-99c=99\left(a-c\right)⋮99\)