a) Có: \(100a+8b=4\left(25a+2b\right)⋮4\)
mà \(2b+c⋮4\)
\(\Rightarrow100a+\left(2b+8b\right)+c=100a+10b+c=\overline{abc}⋮4\left(đpcm\right)\)
b) Có: \(1000a+96b+8c=8\left(125a+12b+c\right)⋮8\)
mà \(4b+2c+d⋮8\)
\(\Rightarrow1000a+\left(96b+4b\right)+\left(8c+2c\right)+d=1000a+100b+10c+d=\overline{abcd}⋮8\left(đpcm\right)\)