a) Đặt A = \(6^5.5-3^5\)

\(=\left(2.3\right)^5.5-3^5\)

\(=2^5.3^5.5-3^5\)

\(=3^5.\left(2^5.5-1\right)\)

\(=3^5.\left(32.5-1\right)\)

\(=3^5.159\)

\(=3^5.3.53⋮53\)

Vậy \(A⋮53\)

b) Đặt \(B=2+2^2+2^3+...+2^{120}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{119}+2^{120}\right)\)

\(=2.\left(1+2\right)+2^3.\left(1+2\right)+...+2^{119}.\left(1+2\right)\)

\(=2.3+2^3.3+...+2^{119}.3\)

\(=3.\left(2+2^3+...+2^{59}\right)⋮3\)

Vậy \(B⋮3\)

\(B=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{118}+2^{119}+2^{120}\right)\)

\(=2.\left(1+2+2^2\right)+3^4.\left(1+2+2^2\right)+...+2^{118}.\left(1+2+2^2\right)\)

\(=2.7+2^4.7+...+2^{118}.7\)

\(=7.\left(2+2^4+...+2^{118}\right)⋮7\)

Vậy \(B⋮7\)

\(B=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)\)

\(+...+\left(2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)

\(=2.\left(1+2+2^2+2^3+2^4\right)+2^6.\left(1+2+2^2+2^3+2^4\right)\)

\(+2^{116}.\left(1+2+2^2+2^3+2^4\right)\)

\(=2.31+2^6.31+...+2^{116}.31\)

\(=31.\left(2+2^6+...+2^{116}\right)⋮31\)

Vậy \(B⋮31\)

\(B=\left(2+2^2+2^3+2^4+2^5+2^6+2^7+2^8\right)+\left(2^9+2^{10}+2^{11}+2^{12}+2^{13}+2^{14}+2^{15}+2^{16}\right)\)

\(+...+\left(2^{113}+2^{114}+2^{115}+2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)

\(=2.\left(1+2+2^2+2^3+2^4+2^5+2^6+2^7\right)+2^9.\left(1+2+2^2+2^3+2^4+2^5+2^6+2^7\right)\)

\(+...+2^{113}.\left(1+2+2^2+2^3+2^4+2^5+2^6+2^7\right)\)

\(=2.255+2^9.255+...+2^{113}.255\)

\(=255.\left(2+2^9+...+2^{113}\right)\)

\(=17.15.\left(2+2^9+...+2^{113}\right)⋮17\)

Vậy \(B⋮17\)

c) Đặt C = \(3^{4n+1}+2^{4n+1}\)

Ta có:

\(3^{4n+1}=\left(3^4\right)^n.3\)

\(2^{4n}=\left(2^4\right)^n.2\)

\(3^4\equiv1\left(mod10\right)\)

\(\Rightarrow\left(3^4\right)^n\equiv1^n\left(mod10\right)\equiv1\left(mod10\right)\)

\(\Rightarrow3^{4n+1}\equiv\left(3^4\right)^n.3\left(mod10\right)\equiv1.3\left(mod10\right)\equiv3\left(mod10\right)\)

\(\Rightarrow\) Chữ số tận cùng của \(3^{4n+1}\) là \(3\)

\(2^4\equiv6\left(mod10\right)\)

\(\Rightarrow\left(2^4\right)^n\equiv6^n\left(mod10\right)\equiv6\left(mod10\right)\)

\(\Rightarrow2^{4n+1}\equiv\left(2^4\right)^n.2\left(mod10\right)\equiv6.2\left(mod10\right)\equiv2\left(mod10\right)\)

\(\Rightarrow\) Chữ số tận cùng của \(2^{4n+1}\) là \(2\)

\(\Rightarrow\) Chữ số tận cùng của C là 5

\(\Rightarrow C⋮5\)

d) Đặt \(D=75+\left(4^{2006}+4^{2005}+4^{2004}+...+1\right).25\)

Đặt \(E=4^{2006}+4^{2005}+4^{2004}+...+1\)

\(\Rightarrow4E=4^{2007}+4^{2006}+4^{2005}+...+4\)

\(\Rightarrow3E=4E-E\)

\(=\left(4^{2007}+4^{2006}+4^{2005}+...+4\right)-\left(4^{2006}+4^{2005}+4^{2004}+...+1\right)\)

\(=4^{2007}-1\)

\(\Rightarrow E=\dfrac{\left(4^{2007}-1\right)}{3}\)

\(\Rightarrow D=75+\dfrac{4^{2007}-1}{3}.25\)

Ta có:

\(4^{2007}=\left(4^2\right)^{1003}.4\)

\(4^2\equiv6\left(mod10\right)\)

\(\left(4^2\right)^{1003}\equiv6^{1003}\left(mod10\right)\equiv6\left(mod10\right)\)

\(\Rightarrow4^{2007}\equiv\left(4^2\right)^{1003}.4\left(mod10\right)\equiv6.4\left(mod10\right)\equiv4\left(mod10\right)\)

\(\Rightarrow\) Chữ số tận cùng của \(4^{2007}\) là 4

\(A=\left(2+2^2\right)+...+\left(2^{99}+2^{100}\right)\)

\(A=2\cdot\left(1+2\right)+...+2^{99}\cdot\left(1+2\right)\)

\(A=2\cdot3+...+2^{99}\cdot3\)

\(A=3\cdot\left(2+...+2^{99}\right)⋮3\left(đpcm\right)\)

2 ý kia tương tự

Giải:

Đặt S=(2+2^2+2^3+...+2^100)

=2.(1+2+2^2+2^3+2^4)+2^6.(1+2+2^2+2^3+2^4)+...+(1+2+2^2+2^3+2^4).296

=2.31+26.31+...+296.31

=31.(2+26+...+296)\(⋮\)31

Ta có :

\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)

=> \(A=(2+2^2)+(2^3+2^4)+...+(2^{99}+2^{100})\)

=> \(A=2(1+2)+2^3(1+2)+...+2^{99}(1+2)\)

=> \(A=2.3+2^3.3+...+2^{99}.3\)

=> \(A=(2+2^3+...+2^{99}).3\)chia hết cho 3             ( 1 )

Ta lại có :

\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)

=> \(A=2(1+2+2^2+2^3+...+2^{98}+2^{99})\)chia hết cho 2       ( 2 )

Từ ( 1 ) và ( 2 ) ta có :

A chia hết cho 2 . 3 hay A chia hết cho 6

Ta có :

\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)

=> ​\(A=\left(2+2^2+2^3+2^4+2^5\right)+....\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)​

=> \(A=2\left(1+2+2^2+2^3+2^4\right)+...+2^{96}\left(1+2+2^2+2^3+2^4\right)\)

=> \(A=2.31+...+2^{96}.31\)

=> \(A=\left(2+...+2^{96}\right)31\)chia hết cho 31

\(A=2+2^2+2^3+2^4+...+2^{99}+2^{100}\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{99}\left(1+2\right)\)

\(=3\left(2+2^3+...+2^{99}\right)\) ⋮ 3 \(\left(đpcm\right)\)

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

\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{99}\left(1+2\right)=3\left(2+2^3+...+2^{99}\right)⋮3\)

\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{99}\left(1+2\right)\)

\(=3\left(2+...+2^{99}\right)⋮3\)

\(=>A=2\cdot\left(2+1\right)+2^3\cdot\left(2+1\right)+.....+2^{99}\left(2+1\right)\)

\(=>A=3.\left(2+2^3+....+2^{99}\right)⋮3\)