ab+cd+eg chia hết cho 11
Mà 9999ab = 99.11.ab chia hết cho 11 và 99cd = 9.11.cd chia hết cho 11
=> 9999ab+99cd+ab+cd+eg chia hết cho 11
=> 10000ab+100cd+eg chia hết cho 11
=> ab0000+cd00+eg chia hết cho 11
=> abcdeg chia hết cho 11
=> ĐPCM
Tk mk nha
Ta có: \(\overline{abcdeg}=10000\overline{ab}+100\overline{cd}+\overline{eg}=9999\overline{ab}+99\overline{cd}+\left(\overline{ab}+\overline{cd}+\overline{eg}\right)\)
Mà \(999\overline{ab}⋮11;99\overline{cd}⋮11;\left(\overline{ab}+\overline{cd}+\overline{eg}\right)⋮11\)
\(\Rightarrow9999\overline{ab}+99\overline{cd}+\left(\overline{ab}+\overline{cd}+\overline{eg}\right)⋮11\)
Vậy...
abcdeg=10000ab+100cd+eg=9999ab+99cd+(ab+cd+eg)
Mà ab + cd + eg chia hết cho 11
Suy ra abcdeg chia hết cho 11 khi ab + cd + eg chia hết cho 11 ( do 9999ab+99cd chia hết cho 11)
Tk mình đi!
ta có : \(\overline{abcdeg}=\overline{ab}.10000+\overline{cd}.100+\overline{eg}\)
\(=9999.\overline{ab}+\overline{ab}+99.\overline{cd}+\overline{cd}+\overline{eg}\)
\(=\left(9999.\overline{ab}+99.\overline{ab}\right)+\left(\overline{ab}+\overline{cd}+\overline{eg}\right)\)
\(=11.\left(909.\overline{ab}+9.\overline{cd}\right)+\left(\overline{ab}+\overline{cd}+\overline{eg}\right)\)
ta có \(11.\left(909.\overline{ab}+9.\overline{cd}\right)⋮11\) ; mà \(\left(\overline{ab}+\overline{cd}+\overline{eg}\right)⋮11\)
\(\Rightarrow\overline{abcdeg}⋮11\)