Question

Chứng minh rằng nếu : (\(\overline{ab}\)+\(\overline{cd}\)+\(\overline{eg}\)) \(⋮\) 11 thì \(\overline{abcdeg}\) \(⋮\) 11

Mọi người giúp mình với , mai kiểm tra rùi

Answer 1

Ta có : \(\overline{abcdeg}=\overline{ab}.1000+\overline{cd}.100+\overline{eg}\)

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

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

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

\(\Rightarrow\overline{abcdeg}⋮11\left(đpcm\right)\)

Answer 2

Ta có:

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

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

\(=\overline{ab}.11.909+\overline{cd}.11.9+\left(\overline{ab}+\overline{cd}+\overline{eg}\right)\)

\(=11\left(\overline{ab}.909+\overline{cd}.9\right)+\left(\overline{ab}+\overline{cd}+\overline{eg}\right)\)

Vì \(11\left(\overline{ab}.909+\overline{cd}.9\right)⋮11\) và \(\overline{ab}+\overline{cd}+\overline{eg}⋮11\)

nên \(\overline{abcdeg}⋮11\)

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

Answer 3

ta có \(\overline{abcdeg}\)=\(\overline{ab}\).1000+\(\overline{cd}\).100+\(\overline{eg}\).1

=9999.\(\overline{ab}\)+99.\(\overline{cd}\)+\(\left(\overline{ab}+\overline{cd}+\overline{eg}\right)⋮11\)

\(\Rightarrow\overline{abcdeg}⋮11\)

Answer 4

Ta có :

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

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

= \(\overline{ab}\) .909.11 + \(\overline{cd}\) . 9.11 +( \(\overline{cd}\) + \(\overline{eg}\) + \(\overline{ab}\) )

= 11.( \(\overline{ab}\) .909 + \(\overline{cd}\) . 9) + ( \(\overline{cd}\) + \(\overline{eg}\) + \(\overline{ab}\) )

Vì 11 \(⋮\) 11 \(\Rightarrow\) 11.( \(\overline{ab}\) .909 + \(\overline{cd}\) . 9) \(⋮\) 11 (1)

Theo bài ra , ( \(\overline{cd}\) + \(\overline{eg}\) + \(\overline{ab}\) ) \(⋮\) 11 (2)

Từ (1) và (2) \(\Rightarrow\) 11.( \(\overline{ab}\) .909 + \(\overline{cd}\) . 9) + ( \(\overline{cd}\) + \(\overline{eg}\) + \(\overline{ab}\) ) \(⋮\) 11

Hay \(\overline{abcdeg}\) \(⋮\) 11

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