A= (1+3^2+3^4)+.......+(2^2002+2^2004+2^2006)

 = 91+......+ 2^2002.(1+3^2+3^4)

= 91+.+ 2^2002.91 chia hết cho 91 (đpcm)

b, Ta có: 9A= 3^2+3^4+....+3^2008

               9A-A= 3^2008-1 => 8A= 3^2008-1 => 8A+1= 3^2008

 Thay vào ta có 27^263x.9^5y = 3^2008 => 9^263x.3^263x.9^5y= 3^2008 => 9^( 263x+5y).3^263x= 3^2008

 => 3^263x= 3^2008-9^( 263x+5y) => 3^263x= 9^1004-9^( 263x+5y) => 3^263x= 3^{2.(1004-263x-5y)}

=>  263x= 2008-2.263x-10.y => 263x+2.263.x+ 10y= 2008

=> 789x + 10y= 2008 . Vì 10y chia hết cho 2; 2008 chia hết cho 2 => 789x chia hết cho 2.

 Mà (789; 2)=1 => x chia hết cho 2 . Do x là số nguyên tố nên x= 2 => y = 43.

 Vậy (x; y)= (2; 43)

 Không biết đúng không ^o^

a: (x-3)(y+1)=15

=>\(\left(x-3\right)\left(y+1\right)=1\cdot15=15\cdot1=\left(-1\right)\cdot\left(-15\right)=\left(-15\right)\cdot\left(-1\right)=3\cdot5=5\cdot3=\left(-3\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-3\right)\)

=>(x-3;y+1)\(\in\){(1;15);(15;1);(-1;-15);(-15;-1);(3;5);(5;3);(-3;-5);(-5;-3)}

=>(x,y)\(\in\){(4;14);(18;0);(2;-16);(-12;-2);(6;4);(8;2);(0;-6);(-2;-4)}

b: Sửa đề:\(m=1+3+3^2+3^3+...+3^{99}+3^{100}\)

\(m=1+3+\left(3^2+3^3+3^4\right)+\left(3^5+3^6+3^7\right)+...+\left(3^{98}+3^{99}+3^{100}\right)\)

\(=4+3^2\left(1+3+3^2\right)+3^5\left(1+3+3^2\right)+...+3^{98}\left(1+3+3^2\right)\)

\(=4+13\left(3^2+3^5+...+3^{98}\right)\)

=>m chia 13 dư 4

\(m=1+3+3^2+...+3^{99}+3^{100}\)

\(=1+\left(3+3^2+3^3+3^4\right)+...+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)

\(=1+3\left(1+3+3^2+3^3\right)+3^5\left(1+3+3^2+3^3\right)+...+3^{97}\left(1+3+3^2+3^3\right)\)

\(=1+40\left(3+3^5+...+3^{97}\right)\)

=>m chia 40 dư 1

mình mới học lớp 5