thôi bạn ơi

trả lời giùm

\(\overline{abcd}=1000a+100b+10c+d=\)

\(=\left(986a+87b\right)+\left(14a+13b+10c+d\right)=\)

\(=\left(34.29.a+3.29.b\right)+\left(14a+13b+10c+d\right)=\)

\(=29\left(34a+3b\right)+\left(14a+13b+10c+d\right)⋮29\)

Mà \(29\left(34a+3b\right)⋮29\Rightarrow14a+3b+10c+d⋮29\)

\(\Rightarrow2\left(14a+13b+10c+d\right)=28a+26b+20c+2d⋮29\)

\(\Rightarrow28a+26b+20c+2d-29\left(a+b+c+d\right)=\)

\(=-3a-3b-9c-27d=-\left(a+30+9c+27d\right)⋮29\)

\(\Rightarrow a+3b+9c+27d⋮29\)

+ Ta có

n=abcd=1000a+100b+10c+d=986a+87b+14a+13b+10c+d=29(34a+3b)+(14a+13b+10c+d) chia hết cho 29

Mà 29(34a+3b) chia hết cho 29 nên (14a+13b+10c+d) cũng chia hết cho 29

+ Ta lại có

a+3b+9c+27d=29(a+b+c+d)-(28a+26b+20c+2d)=29(a+b+c+d)-2(14a+13b+10c+d)

Mà 29(a+b+c+d) chia hết cho 29 và (14a+13b+10c+d) cũng chia hết cho 29 nên 2(14a+13b+10c+d) chia hết cho 29

=> a+3b+9c+27d chia hết cho 29

thanks

như bên dưới là dc rùi...

chắc chắn là đúng

\(\overline{abcd}=1000a+100b+10c+d⋮29\)

\(\Leftrightarrow1000a-29.34a+100b-29.3b+10c+d⋮29\)

\(\Leftrightarrow14a+13b+10c+d⋮29\)

\(\Leftrightarrow28a+26b+20c+2d⋮29\)(vì \(\left(2,29\right)=1\))

\(\Leftrightarrow29a-28a+29b-26b+29c-20c+29d-2d⋮29\)

\(\Leftrightarrow a+3b+9c+27d⋮29\).

Ta có đpcm.