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a,
`3A=3+3^3+3^3+...+3^{53}`
`3A-A=(3+3^3+3^3+...+3^{53})-(1+3+3^3+3^3+...+3^{52})`
`2A=3^{53}-1`
`A=(3^{53}-1)/2`
b,
`A=1+3+3^3+3^3+...+3^{52}`
`A=(1+3+3^2)+(3^3+3^4+3^5)+....+(3^{50}+3^{51}+3^{52})`
`A=(1+3+3^2)+3^3*(1+3+3^2)+....+3^{50}*(1+3+3^2)`
`A=(1+3+3^2)*(1+3^3+....+3^{50})`
`A=13*(1+3^3+....+3^{50})`
Do `13 \vdots 13 => A=13*(1+3^3+....+3^{50})\vdots 13 `
Vậy `A \vdots 13 `
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=3+3^2+3^3+3...+3^{31}\\ =\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{28}+3^{29}+3^{30}\right)+3^{31}\\ =\left(3+3^2+3^3\right)+3^3\left(3+3^2+3^3\right)+...+3^{27}\left(3+3^2+3^3\right)+3^{31}\\ =39+3^3.39+...+3^{27}.39+3^{31}\\ =39.\left(1+3^3+...+3^{27}\right)+3^{31}\\ Mà:39.\left(1+3^3+...+3^{27}\right)⋮13\left(Do:39⋮13\right)\\ Mà:3^{31}:13\left(dư:3\right)\\ Vậy:39.\left(1+3^3+...+3^{27}\right)+3^{31}:13\left(dư:3\right)\\ \Rightarrow A:13\left(dư:3\right)\)
A = 3 + 3² + 3³ + ... + 3³¹
= 3 + 3² + 3³ + 3⁴ + 3⁵ + 3⁶ + 3⁷ + ... + 3²⁹ + 3³⁰ + 3³¹
= 3 + (3² + 3³ + 3⁴) + (3⁵ + 3⁶ + 3⁷) + ... + (3²⁹ + 3³⁰ + 3³¹)
= 3 + 3².(1 + 3 + 3²) + 3⁵.(1 + 3 + 3²) + ... + 3²⁹.(1 + 3 + 3²)
= 3 + 3².13 + 3⁵.13 + ... + 3²⁹.13
= 3 + 13.(3² + 3⁵ + ... + 3³¹)
Do 13.(3² + 3⁵ + ... + 3³¹) ⋮ 13
⇒ 3 + 13.(3² + 3⁵ + ... + 3³¹) chia 13 dư 3
Vậy A chia 13 dư 3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=3+3^2+3^3+...+3^{100}\)
\(=3+\left(3^2+3^3+3^4\right)+...+\left(3^{98}+3^{99}+3^{100}\right)\)
\(=3+3^2\left(1+3+3^2\right)+...+3^{98}\left(1+3+3^2\right)\)
\(=3+3^2.13+...+3^{98}.13\)
\(=3+13\left(3^2+...+3^{98}\right)\)
\(\Rightarrow B⋮̸13\)
\(\Rightarrow B:13\) dư 3.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1+3+3^2+3^3+...+3^{2022}\)
\(=1+\left(3+3^2+3^3\right)+...+\left(3^{2020}+3^{2021}+3^{2022}\right)\)
\(=1+3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{2020}\left(1+3+3^2\right)\)
\(=1+13\left(3+3^4+...+3^{2020}\right)\)
=>A chia 13 dư 1
Bạn ơi, bạn cũng xem lại giúp mình luôn nha
2020 đâu có chia hết cho 3
Với lại dãy này có 2023 số đó bạn, 2023 cũng đâu chia hết cho 3 đâu
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$A=1+(3+3^2+3^3)+(3^4+3^5+3^6)+....+(3^{2014}+3^{2015}+3^{2016})$
$=1+3(1+3+3^2)+3^4(1+3+3^2)+...+3^{2014}(1+3+3^2)$
$=1+3.13+3^4.13+....+3^{2014}.13$
$=1+13(3+3^4+...+3^{2014})$
$\Rightarrow A-1\vdots 13(1)$
Mặt khác:
$A=1+(3+3^2+3^3+3^4)+....+(3^{2013}+3^{2014}+3^{2015}+3^{2016})$
$=1+3(1+3+3^2+3^3)+....+3^{2013}(1+3+3^2+3^3)$
$=1+(3+...+3^{2013})(1+3+3^2+3^3)$
$=1+40(3+....+3^{2013})$
$\Rightarrow A-1\vdots 5(2)$
Từ $(1); (2)$ mà $(5,13)=1$ nên $A-1\vdots (5.13)$ hay $A-1\vdots 65$
$\Rightarrow A$ chia $65$ dư $1$
![](https://rs.olm.vn/images/avt/0.png?1311)
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
\(+\)Ta thấy A có số số hạng là: \(\left(2021-1\right);1+1=2021\)(số)
\(+\)Ta nhóm \(3\)số hạng liên tiếp vào \(1\)nhóm, ta được: \(2021:3=673\)dư \(2\)số
\(\Rightarrow A=\left(3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{2019}+3^{2020}+3^{2021}\right)\)
\(\Rightarrow A=\left(3+3^2\right)+\left(3^3\cdot1+3^3\cdot3+3^3\cdot3^2\right)+...+\left(3^{2019}\cdot1+3^{2019}\cdot3+3^{2019}\cdot3^2\right)\)
\(\Rightarrow A=\left(3+3^2\right)+3^3\cdot\left(1+3+3^2\right)+...+3^{2019}\cdot\left(1+3+3^2\right)\)
\(\Rightarrow A=12+3^3\cdot13+...+3^{2019}\cdot13\)
\(\Rightarrow A=12+13\cdot\left(3^3+3^6+3^9+...+^{2019}\right)\)
Vì\(\hept{\begin{cases}12:13=0dư12\\13\cdot\left(3^3+3^6+3^9+...+3^{2019}\right)⋮13\end{cases}}\)
\(\Rightarrow A:13dư12\)
Vậy \(A:13dư12\)
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