Giải thích các bước giải:
Bài 4:
Để $\overline{825ab}\quad\vdots\quad 2,5$
$\to \overline{825ab}\quad\vdots\quad 10$
$\to b=0$
$\to \overline{825a0}\quad\vdots\quad 9$
$\to 8+2+5+a+0\quad\vdots\quad 9$
$\to 15+a\quad\vdots\quad 9$
$\to a=3$
Bài 8:
Ta có :
$A=(1+2+2^2)+(2^3+2^4+2^5)+..+(2^{2016}+2^{2017})+2^{2018})+2^{2019}+2^{2020}$
$\to A=(1+2+2^2)+2^3(1+2+2^2)+..+2^{2016}(1+2+2^2)+2^{2019}(1+2)$
$\to A=(1+2^3+..+2^{2016})(1+2+2^2)+2^{2019}(1+2)$
$\to A=7.(1+2^3+..+2^{2016})+3.(2^3)^{673}$
$\to A=7.(1+2^3+..+2^{2016})+3.8^{673}$
Ta có $8\equiv 1(mod 7)\to 3.8^{673}\equiv 3.1=3(mod 7)$
$\to A$ chia 7 dư 3
