Question

[Coin Hunter]

Giúp mình với:

1. CMR:

a) (1991¹⁹⁹⁷-1997¹⁹⁹⁶) \(⋮\) 10

b) (2⁹+2⁹⁹) \(⋮\) 100

2. CMR:

Nếu x \(⋮̸\)3 thì x² \(\equiv\) 1 (\(mod3\))

* Dùng đồng dư thức nhé

Answer 1

[Akai Haruma]

Bài 1: 

a. $1991\equiv 1\pmod {10}$
$\Rightarrow 1991^{1997}\equiv 1^{1997}\equiv 1\pmod {10}$
$1997^2\equiv -1\pmod {10}$

$\Rightarrow 1997^{1996}=(1997^2)^{998}\equiv (-1)^{998}\equiv 1\pmod {10}$

Do đó:

$1991^{1997}-1997^{1996}\equiv 1-1\equiv 0\pmod {10}$
Vậy $1991^{1997}-1997^{1996}\vdots 10$

b.

$2^9+2^{99}=2^9(1+2^{90})$

Hiển nhiên $2^9(1+2^{90}\vdots 4$ nên để cm nó chia hết cho 100 thì ta chỉ cần cm $2^{90}+1\vdots 25$

Có:

$2^{10}\equiv -1\pmod {25}$

$\Rightarrow 2^{90}=(2^{10})^9\equiv (-1)^9\equiv -1\pmod {25}$

$\Rightarrow 2^{90}+1\equiv -1+1\equiv 0\pmod {25}$

Vậy $2^{90}+1\vdots 25$

$\Rightarrow 2^9+2^{99}=2^9(2^{90}+1)\vdots 100$

Answer 2

[Akai Haruma]

Bài 2:

$x\not\vdots 3$

Tức là $x\equiv \pm 1\pmod 3$

$\Rightarrow x^2\equiv (\pm 1)^2\equiv 1\pmod 3$