a/ \(A=\left(3+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{57}+3^{58}+3^{59}+3^{60}\right)\)

\(A=3\left(1+3+3^2+3^3\right)+3^5\left(1+3+3^2+3^3\right)+...+3^{57}\left(1+3+3^2+3^3\right)\)

\(A=40\left(3+3^5+3^9+...+3^{53}+3^{57}\right)\)Chia hết cho 4; 5

Ta cũng có

\(A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\)

\(A=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\)

\(A=13\left(3+3^4+3^7+...+3^{55}+3^{58}\right)\) chia hết cho 13

b/ \(3A=3^2+3^3+3^4+...+3^{61}\)

\(A=\frac{3A-A}{2}=\frac{3^{61}-3}{2}< 3^{61}\)

a/ \(A=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{59}+3^{60}\right)\)

\(A=12+3^2\left(3+3^2\right)+3^{58}\left(3+3^2\right)=12\left(1+3^2+3^4+...+3^{56}+3^{58}\right)\) chia hết cho 12

c/ \(A=3+\left(3^2+3^3+3^4+...+3^{60}\right)\)

\(A=3+3^2\left(1+3+3^2+...+3^{58}\right)\)

Ta có \(3^2\left(1+3+3^2+...+3^{58}\right)\) chia hết cho 9 => A chia 9 dư 3

d/ Từ câu A ta có

\(A=40\left(3+3^5+3^9+...+3^{53}+3^{57}\right)\)=> chữ số tận cùng của A là 0

61380

a) 3114 chia hết cho 3

b) 71 244 chia hết cho cả 2 và 3

c) 9999 chia hết cho 9

d) 61 380 chia hết cho 2,3,5 và 9

a/ 3114-3414-3714

b/ 71244

c/9999

d/61380

giúp mình nhanh lên với mai mình đi học rùi

b)=3^1+(3^2+3^3+3^4)+(3^5+3^6+3^7)+....+(3^58+3^59+3^60)

=3^1+(3^2.1+3^2.3+3^2.9)+(3^5.1+3^5.3+3^5.9)+......+(3^58.1+3^58.3+3^58.9)

=3^1+3^2.(1+3+9)+3^5.(1+3+9)+.....+3^58.(1+3+9)

=3+3^2.13+3^5.13+.........+3^58.13

=3.13.(3^2+3^5+....+3^58)

vi tich tren co thua so 13 nen tich do chia het cho 13

=

bai1

a) A=(3¹+3²)+(3³+3⁴)+...+(3⁵⁹+3⁶⁰)

=(3^1.1+3^1.3)+...+(3^59.1+3^59.2)

=3^1.(1+3)+...+3^59.(1+3)

=3^1.4+....+3^59.4

=4.(3^1+...+3^59)

vi tich tren co thua so 4 nen tich do chia het cho 4

Bài 2:(12a + 36b) = (12a + 12 x 3 x b) = 12( a + 3b)chia hết cho 12