Đáp án :

`a)M \vdots 8`

`b)M \vdots 24`

Giải thích các bước giải :

`a)M=32+10^(2015)+10^(2016)+10^(2017)+10^(2018)`

`<=>M=32+10^(2013).(10^2+10^3+10^4+10^5)`

`<=>M=32+10^(3.671).(10^2+10^3+10^4+10^5)`

`<=>M=32+(10^3)^(617).(10^2+10^3+10^4+10^5)`

`<=>M=32+1000^(617).(10^2+10^3+10^4+10^5)`

Vì `1000 \vdots 8`

`=>1000^(617) \vdots 8`

`=>1000^(617).(10^2+10^3+10^4+10^5) \vdots 8`

Mà `32 \vdots 8`

`=>32+1000^(617).(10^2+10^3+10^4+10^5) \vdots 8`

`=>M \vdots 8`

`b)M=32+10^(2015)+10^(2016)+10^(2017)+10^(2018)`

`<=>M=32+\bar{100...0}+\bar{100...0}+\bar{100...0}+\bar{100...0}`

`<=>M=\bar{111100...032}`

Tổng các chữ số của `M` là :

`1+1+1+1+0+0+...+0+3+2=9 \vdots 3`

`=>M \vdots 3`

Mà `M \vdots 8`

`=>M \vdots 24`

Vậy : `M \vdots 24`

Tham khảo

 `a) M=32+10^{2015}+10^{2016}+10^{2017}+10^{2018}`

`⇒M=32+10^{2013}×(10^2+10^3+10^4+10^5)`

`⇒M=32+(10^3)^{671}×(10^2+10^3+10^4+10^5)`

`⇒M=32+1000^{671} ×(10^2+10^3+10^4+10^5)`

Vì `1000 \vdots 8⇒1000^{671} ×(10^2+10^3+10^4+10^5) \vdots 8` và `32 \vdots 8`

`⇒M \vdots 8`

`b) M=32+10^{2015}+10^{2016}+10^{2017}+10^{2018}`

`⇒M=32+10...0+100....0+1000....0+10000.....0`

`⇒M=11110000...32`

`⇒M=(1+1+1+1+0+0+...+0+3+2)=9 \vdots 3`

Có \begin{cases}M\vdots 8\\M \vdots 3\\\end{cases}

`⇒M \vdots 24`

`⇒M` chia `24` dư `0`