Ta có:

abcd=1000a+100b+10c+d=986a+87b+14a+13b+10c+d=29.(34a+3b)+(14a+13b+10c+d)

=>14a+13b+10c+d 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)

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

=>a+3b+9c+27d chia hết cho 29 (ĐPCM)

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

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

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

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

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

Vậy ...

thôi bạn ơi