Question

\(CMR:2a-5b+6c⋮17\)nếu \(a-11b+3c⋮17\left(a,b,c\in Z\right)\)

 

Answer 1

Nếu \(a-11b+3c⋮17\Rightarrow2\left(a-11b+3c\right)⋮17\)

\(\Rightarrow2a-22b+6c⋮17\Rightarrow\left(2a-5b+6c\right)-17b⋮17\)

Vì\(17b⋮17\Rightarrow2a-5b+3c⋮17\)

Answer 2

Vì \(a-11b+3c\) chia hết cho 17 => \(2\left(a-11b+3c\right)\)chia hết cho 17 =>      \(2a-22b+6c\)

Ta có:     \(\left(2a-22b+6c\right)-\left(2a-5b+6c\right)=17b\)chia hết  cho 17

Mà 2a - 22b + 6c chia hết cho 17 nên => 2a - 5b + 6c chia hết cho 17

Vậy 2a - 5b + 6c chia hết cho 17.

Answer 3

\(a-11b+3c⋮17\)\(\Rightarrow2\left(a-11b+3c\right)⋮17\)\(\Rightarrow2a-22b+6c⋮17\)

mà \(17b⋮17\)\(\Rightarrow\left(2a-22b+6c\right)+17b⋮17\)

\(\Rightarrow2a-22b+6c+17b⋮17\)\(\Rightarrow2a-5b+6c⋮17\left(đpcm\right)\)