1. \(A=2^{2016}-1\)

\(2\equiv-1\left(mod3\right)\\ \Rightarrow2^{2016}\equiv1\left(mod3\right)\\ \Rightarrow2^{2016}-1\equiv0\left(mod3\right)\\ \Rightarrow A⋮3\)

\(2^{2016}=\left(2^4\right)^{504}=16^{504}\)

16 chia 5 dư 1 nên 16^504 chia 5 dư 1

=> 16^504-1 chia hết cho 5

hay A chia hết cho 5

\(2^{2016}-1=\left(2^3\right)^{672}-1=8^{672}-1⋮7\)

lý luận TT trg hợp A chia hết cho 5

(3;5;7)=1 = > A chia hết cho 105

2;3;4 TT ạ !!

Tui biet nhung ko tra loi dc

Ta có: \(A=2+2^2+2^3+...+2^{120}\)

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

\(=14+2^3\cdot14+...+2^{117}\cdot14\)

\(=14\cdot\left(1+2^3+...+2^{117}\right)⋮7\)

Ta có: \(A=2+2^2+2^3+...+2^{120}\)

\(=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)+...+\left(2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)

\(=62+2^5\cdot62+...+2^{115}\cdot62\)

\(=62\cdot\left(1+2^5+...+2^{115}\right)⋮31\)

Ta có: \(A=2+2^2+2^3+...+2^{120}\)

\(=\left(2+2^2+2^3+2^4+2^5+2^6\right)+\left(2^7+2^8+2^9+2^{10}+2^{11}+2^{12}\right)+...+\left(2^{115}+2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)

\(=126+126\cdot2^6+...+126\cdot2^{114}\)

\(=126\cdot\left(1+2^6+...+2^{114}\right)⋮21\)

??!!?

Đặt biểu thức trên là A.

Ta có: A=2^2008-8

            A=(2^4+2^5+....+2^2008)-(8+2^4+....+2^2007)

            A=2x(8+2^4+....+2^2007)-(8+2^4+....+2^2007)

       A=8+2^4+2^5+2^6+2^7+2^8+2^9+2^10+2^11+2^12+....+2^2003+2^2004+2^2005+2^2006+2^2007(có 2005 số hạng)

A=(8+2^4+2^5+2^6+2^7)+                                                                                                       (2^8+2^9+2^10+2^11+2^12)+....+(2^2003+2^2004+2^2005+2^2006+2^2007)(có 401 nhóm)

A=8x(1+2+4+8+16)+2^8x(1+2+4+8+16)+.....+2^2003x(1+2+4+8+16)

A=8x31+2^8x31+....+2^2003x31

A=31x(8+2^8+...+2^2003)

A là tích có thừa số 31 nên A chia hết cho 31(đpcm)

thiệt chớ tao hiểu tao chết liền

\(16^5-2^{15}.\)

\(=\left(2^4\right)^5-2^{15}.\)

\(=2^{20}-2^{15.}\)

\(=2^{15}\left(2^5-1\right).\)

\(=2^{15}\left(32-1\right).\)

\(=2^{15}.31⋮31\left(đpcm\right).\)

Bài 3: 

a) Ta có: \(C=2+2^2+2^3+...+2^{99}+2^{100}\)

\(=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)

\(=2\left(1+2+2^2+2^3+2^4\right)+2^6\left(1+2+2^2+2^3+2^4\right)+...+2^{96}\left(1+2+2^2+2^3+2^4\right)\)

\(=31\cdot\left(2+2^6+...+2^{96}\right)⋮31\)(đpcm)

Bài 1: 

Ta có: \(A=3^{n+2}-2^{n+2}+3^n-2^n\)

\(=3^n\cdot9-2^n\cdot4+3^n-2^n\)

\(=3^n\left(9+1\right)-2^n\left(4+1\right)\)

\(=10\left(3^n-2^{n-1}\right)⋮10\)

Vậy: A có chữ số tận cùng là 0

Bài 2: 

Ta có: \(abcd=1000\cdot a+100\cdot b+10\cdot c+d\)

\(\Leftrightarrow abcd=1000\cdot a+96\cdot b+8c+2c+4b+d\)

\(\Leftrightarrow abcd=8\left(125a+12b+c\right)+\left(2c+4b+d\right)\)

mà \(8\left(125a+12b+c\right)⋮8\)

và \(2c+4b+d⋮8\)

nên \(abcd⋮8\)(đpcm)

\(8^{30}+8^{31}+8^{32}\)

\(=8^{30}.1+8^{30}.8+8^{30}.8^2\)

\(=8^{30}.1+8^{30}.8+8^{30}.64\)

\(=8^{30}\left(1+8+64\right)\)

\(=8^{30}.73\)

\(=\left(2^3\right)^{30}.73\)

\(=2^{90}.73\)

\(=2^{89}.146⋮146\rightarrowđpcm\)

\(4^{25}+4^{26}+4^{27}+4^{28}+4^{29}+4^{30}\)

\(=4^{25}.1+4^{25}.4+4^{25}.4^2+4^{25}.4^3+4^{25}.4^4+4^{25}.4^5\)

\(=4^{25}.1+4^{25}.4+4^{25}.16+4^{25}.64+4^{25}.256+4^{25}.1024\)

\(=4^{25}\left(1+4+16+64+256+1024\right)\)

\(=4^{25}.1365\)

\(=4^{25}.195.7⋮7\rightarrowđpcm\)

à há, giờ mới biết mi làm sao biết đc cách giải BTVN

1.

a) Ta có :

\(8^{30}+8^{31}+8^{32}=8^{30}\left(8+8^2+8^3\right)\)

\(=8^{30}.584\)

* Do \(584⋮146\Rightarrow584.8^{30}⋮146\)

Vậy \(8^{30}+8^{31}+8^{32}⋮146\)

b) \(4^{25}+4^{26}+4^{27}+4^{28}+4^{29}+4^{30}\)

\(=\left(4^{25}+4^{26}\right)+\left(4^{27}+4^{28}\right)+\left(4^{29}+4^{30}\right)\)\(=4^{25}\left(1+4^2\right)+4^{27}\left(1+4^2\right)+4^{29}\left(1+4^2\right)\)\(=\left(1+4^2\right)\left(4^{25}+4^{27}+4^{29}\right)\)

\(=17\left(4^{25}+4^{27}+4^{29}\right)⋮17\)

\(\RightarrowĐpcm\)

tik mik nha !!!