Bài 1:
a) 13. 12 + 26. 27 - 13. 0
= 13. 12 + 13. 2. 27 - 0
= 13. 12 + 13. 54
= 13. (12 + 54)
= 13. 66 $\vdots$ 31 vì 66 $\vdots$ 33
Vậy 13. 12 + 26. 27 - 13. 0 $\vdots$ 33 (đpcm).
b) $6^{5}. 5$ - $3^{5}$
= $(2. 3)^{5}. 5$ - $3^{5}$
= $2^{5}$. $3^{5}.5$ - $3^{5}$
= $3^{5}$. ($2^{5}.5 - 1$)
= $3^{5}. 159$
= $3^{5}. 53.$ $3$ $\vdots$ 53
Vậy $3^{5}. 53.$ $3$ $\vdots$ 53 (đpcm).
c) $3^{4n+1}$ + $2^{4n+1}$
=$3^{4n}. 3$ + $2^{4n}. 2$
= $81^{n}. 3$ + $16^{n}. 2$
$(\overline{...1}). 3$ + $(\overline{...6}). 2$ (vì số có tận cùng là 0; 1; 5; 6 thì mũ bao nhiêu lên vẫn có tận cùng là chính nó
= $(\overline{...3})$ + $(\overline{...2})$
= $(\overline{...5})$ $\vdots$ 5
Vậy $3^{4n+1}$ + $2^{4n+1}$ $\vdots$ 5 (đpcm).
d) $2$ + $2^{2}$ + $2^{3}$ + $2^{4}$ + ... + $2^{20}$
= ($2$ + $2^{2}$ + $2^{3}$ + $2^{4}$ + $2^{5}$) + ($2^{6}$ + $2^{7}$ + $2^{8}$ + $2^{9}$ + $2^{10}$) + ($2^{11}$ + $2^{12}$ + $2^{13}$ + $2^{14}$ + $2^{15}$) + ($2^{16}$ + $2^{17}$ + $2^{18}$ + $2^{19}$ + $2^{20}$)
= $2$. ($2$ + $2^{6}$ + $2^{11}$ + $2^{16}$) . ($1$ + $2$ + $2^{2}$ + $2^{3}$ + $2^{4}$)+ $2^{6}$. ($1$ + $2$ + $2^{2}$ + $2^{3}$ + $2^{4}$) + $2^{11}$. ($1$ + $2$ + $2^{2}$ + $2^{3}$ + $2^{4}$) + $2^{16}$. ($1$ + $2$ + $2^{2}$ + $2^{3}$ + $2^{4}$)
= ($2$ + $2^{6}$ + $2^{11}$ + $2^{16}$). ($1$ + $2$ + $2^{2}$ + $2^{3}$ + $2^{4}$)
= ($2$ + $2^{6}$ + $2^{11}$ + $2^{16}$). $31$ $\vdots$ $31$
Vậy $2$ + $2^{2}$ + $2^{3}$ + $2^{4}$ + ... + $2^{20}$ $\vdots$ $31$ (đpcm).
