Question

chứng minh rằng A=1+2^2+2^4+2^6....+2^20+2^22 chia hết cho 119

giúp tui nha ,cầu đó :((

Answer 1

Ta có:

$A=1+2^2+2^4+2^6+...+2^{20}+2^{22}$

$=(1+2^2+2^4)+(2^6+2^8+2^{10})+(2^{12}+2^{14}+2^{16})+(2^{18}+2^{20}+2^{22})$

$=21+2^6\cdot(1+2^2+2^4)+2^{12}\cdot(1+2^2+2^4)+2^{18}\cdot(1+2^2+2^4)$

$=21+2^6\cdot21+2^{12}\cdot21+2^{18}\cdot21$

$=21\cdot(1+2^6+2^{12}+2^{18})$

Vì $21\vdots7$

nên $21\cdot(1+2^6+2^{12}+2^{18})\vdots7$

hay $A\vdots7$ (1)

Lại có:

$A=1+2^2+2^4+2^6+...+2^{20}+2^{22}$

$=(1+2^2+2^4+2^6)+(2^8+2^{10}+2^{12}+2^{14})+(2^{16}+2^{18}+2^{20}+2^{22})$

$=85+2^8\cdot(1+2^2+2^4+2^6)+2^{16}\cdot(1+2^2+2^4+2^6)$

$=85+2^8\cdot85+2^{16}\cdot85$

$=85\cdot(1+2^8+2^{16})$

Vì $85\vdots17$

nên $85\cdot(1+2^8+2^{16})\vdots17$

hay $A\vdots17$ (2)

Mặt khác: $(7,17)=1$ (3)

Từ (1); (2) và (3) $\Rightarrow A\vdots 7\cdot17=119$

$\text{#}Toru$