Ta có : \(\left(a;b\in N\right)\)
\(a+16b⋮6\)
Xét tổng :
\(\left(a+16b\right)+\left(11a+8b\right)=12a+24b=12\left(a+2b\right)⋮6\)
\(\Rightarrow11a+8b=12\left(a+2b\right)-\left(a+16b\right)\)
Mà \(\left\{{}\begin{matrix}a+16b⋮6\\12\left(a+2b\right)⋮6\end{matrix}\right.\)
\(\Rightarrow12\left(a+2b\right)-\left(a+16b\right)=11a+8b⋮6\)
Vậy \(a+16b⋮6\Rightarrow11a+8b⋮6\rightarrowđpcm\)
Chỉ làm 1 câu thôi câu sau làm tương tự:
Ta có:Nếu
\(a+16b⋮6\)
\(\Leftrightarrow6\left(a+16b\right)⋮6\)
\(\Leftrightarrow6a+96b⋮6\)
\(6a+6.\left(16b\right)⋮6\)
Vì 6a chia hết cho 6;6.(16b) cũng chia hết cho 6
\(\Leftrightarrow6a+96b⋮6\)
\(\Leftrightarrow a+16b⋮6\left(đpcm\right)\)
