Question

\(\overline{ab}+\overline{cd}+\overline{eg}⋮11\) thì \(\overline{abcdeg}⋮11\)

Answer 1

Ta có: \(\overline{abcdeg}\) = 10000.\(\overline{ab}\) + 100.\(\overline{cd}\) + \(\overline{eg}\)

= (9999.\(\overline{ab}\) + 99.\(\overline{cd}\) ) + ( \(\overline{ab}\) + \(\overline{cd}\) + \(\overline{eg}\))

Theo bài ra, ta có: \(\overline{ab}\) + \(\overline{cd}\) + \(\overline{eg}\) \(⋮\) 11

Vì 9999.\(\overline{ab}\) + 99.\(\overline{cd}\) \(⋮\) 11 và \(\overline{ab}\) + \(\overline{cd}\) + \(\overline{eg}\) \(⋮\) 11

nên (9999.\(\overline{ab}\) + 99.\(\overline{cd}\) ) + ( \(\overline{ab}\) + \(\overline{cd}\) + \(\overline{eg}\)) \(⋮\) 11

Vậy \(\overline{abcdeg}\) \(⋮\) 11