Theo bài ra ta có :
\(\overline{abcabc}\)
\(=\overline{abc}.1000+\overline{abc}.1\)
\(=\overline{abc}.\left(1000+1\right)\)
\(=\overline{abc}.1001\)
\(=\overline{abc}.11.91\)
\(=\left(\overline{abc}.91\right).11\)
\(\Rightarrow\overline{abcabc}⋮11\left(đpcm\right)\)
Ta có:
\(\overline{abcabc}=1001\overline{abc}=11.91\overline{abc}\)
Vì \(11.91\overline{abc}\) \(⋮\) 11 nên \(\overline{abcabc}\) \(⋮\) 11
\(\Rightarrow\) ĐPCM(điều phải chứng minh)
abcabc \(⋮\) 11 vì:
abcabc = abc . 1000 + abc
abcabc = abc . ( 1000 + 1 )
abcabc = abc . 1001
abcabc = abc . 11 . 91
Mà 11 \(⋮\) 11 \(\Rightarrow\) abc . 11 . 91 \(⋮\) 11
Vậy abcabc \(⋮\) 11 ( đpcm )
