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![](https://rs.olm.vn/images/avt/0.png?1311)
\(^{A=1+3+3^2+3^3+...+3^{119}}\)
\(A=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{117}+3^{118}+3^{119}\right)\)
\(^{=13+3^3\left(1+3+3^2\right)+...+3^{117}\left(1+3+3^2\right)}\)
\(^{=13+3^3.13+...+3^{117}.13}\)
\(13\left(1+3^3+...+3^{117}\right)\)chia hết cho 13
=>A chia hết cho 13
![](https://rs.olm.vn/images/avt/0.png?1311)
E = 1 + 3 + 32 + 33 +.....+3119
E = ( 1 + 3 + 32) +....+ ( 3117 + 3118+ 3119)
E = 13 + ......+ 3117.( 1 + 3 + 32)
E = 13 +.....+ 3117 . 13
E = 13. ( 30 + ....+ 3117)
13 ⋮ 13 ⇒ 13. (30 +....+3117) ⋮ 13 ⇒ E = 1 +3+32+ ....+3119⋮13(đpcm)
=\(\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{117}+3^{118}+3^{119}\right)\)
= \(13+3^3\left(1+3+3^2\right)+...+3^{117}\left(1+3+3^2\right)\)
=\(13+3^3.13+...+3^{117}.13\)
=\(13.\left(1+3^2+...+3^{117}\right)\) chia hết cho 13
![](https://rs.olm.vn/images/avt/0.png?1311)
M=1+3+3^2+3^3+^3+...+3^118+3^119
=(1+3+3^2)+(3^3+3^4+3^5)+...+(3^117+3^118+3^119)
=13+3^3(1+3+3^2)+...+3^117(1+3+3^2)
=13+3^3.13+..+3^117.13
=13(1+3^3+...+3^117) chia hết cho 13
Vậy Mchia hết cho 13
ai chơi truy kích thì kết bạn vs mình nha
rồi khi nào tạo phòng solo đao phong chibi với nhau 1 ván
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=2+2^2+2^3+...+2^{60}\)
\(=2\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(=7\cdot\left(2+...+2^{58}\right)⋮7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=> A=(1+3+32)+(33+34+35)+...+(3117+3118+3119)
=> A=13+33(1+3+32)+...+32017(1+3+32)
=> A=13+33.13+...+32017.13
=> A=13(1+33+...+32017) chia hết cho 13
![](https://rs.olm.vn/images/avt/0.png?1311)
\(D=1+3+3^2+...+3^{119}\)
\(=\left(1+3+3^2\right)+...+3^{117}\left(1+3+3^2\right)\)
\(=13\left(1+...+3^{117}\right)⋮13\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=3+3^2+3^3+...+3^{118}+3^{119}+3^{120}\\ =\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{118}+3^{119}+3^{120}\right)\\ =3.\left(1+3+3^2\right)+3^4.\left(1+3+3^2\right)+...+3^{118}\left(1+3+3^2\right)\\ =\left(3+3^4+...+3^{118}\right).\left(1+3+3^2\right)\\ =\left(3+3^4+...+3^{118}\right).13⋮13\left(ĐPCM\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, mình nghĩ là \(16^5+2^{15}\)
ta có : \(16^5=2^{20}\)
=>\(16^5+2^{15}=2^{20}+2^{15}\)
=\(2^{15}.2^5+2^{15}\)
\(=2^{15}.\left(2^5+1\right)\)
\(=2^{15}.33\)
mà \(2^{15}.33⋮33\)
\(=>16^5+2^{15}⋮33\)
Hứa r phải làm thôi:
Đặt:
\(A=1+3+3^2+.....+3^{119}\)
\(A=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+.....+\left(3^{117}+3^{118}+3^{119}\right)\)
\(A=1\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+....+3^{117}\left(1+3+3^2\right)\)
\(A=1.13+3^3.13+....+3^{117}.13\)
\(A=\left(1+3^3+....+3^{117}\right).13\)
\(A⋮13\rightarrowđpcm\)
Đặt \(A=1+3+3^2+3^3+...+3^{119}\)
\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{117}+3^{118}+3^{119}\right)\)
\(=13+\left(1.3^3+3.3^3+3^2.3^3\right)+...+\left(1.3^{117}+3.3^{117}+3^2.3^{117}\right)\)
\(=13+3^3\left(1+3+3^2\right)+...+3^{117}\left(1+3+3^2\right)\)
\(=13+3^3.13+...+3^{117}.13\)
\(=13.\left(1+3^3+...+3^{117}\right)⋮13\)
Vậy \(A⋮13\)