`AAa,b∈NN` , ta có:
`1)\overline{abab}=1000a+100b+10a+b=(1000a+10a)+(100b+b)=1010a+101b=1012a+99b-2a+2b=11(92a+9b)-2(a-b)`
Có: `11⋮11⇒11(92a+9b)⋮11`
Vậy để `\overline{abab}⋮11⇔a-b⋮11.`
`2)` $\overline{aaabbb}$`=100000a+10000a+1000a+100b+10b+b=111000a+111b=111.(1000a+b)`
Có: `111=3.37⇒111⋮37⇒111.(1000a+b)⋮37`.
Vậy $\overline{aaabbb}$`⋮ 37.`
