Q= 2+22+.....+29+210
Chứng minh rằng Q chia hết cho 31
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NT
26 tháng 12 2017
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 !!
NB
0
21 tháng 2 2021
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: \(A=2+2^2+2^3+...+2^{120}\)
\(=\left(2+2^2+2^3+2^4+2^5+2^6\right)+\left(2^7+2^8+2^9+2^{10}+2^{11}+2^{12}\right)+...+\left(2^{115}+2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)
\(=126+126\cdot2^6+...+126\cdot2^{114}\)
\(=126\cdot\left(1+2^6+...+2^{114}\right)⋮21\)
HH
4
9 tháng 4 2016
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)
TP
0
HA
1
NN
28 tháng 12 2017
\(16^5-2^{15}.\)
\(=\left(2^4\right)^5-2^{15}.\)
\(=2^{20}-2^{15.}\)
\(=2^{15}\left(2^5-1\right).\)
\(=2^{15}\left(32-1\right).\)
\(=2^{15}.31⋮31\left(đpcm\right).\)
13 tháng 1 2021
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)
13 tháng 1 2021
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)
4 tháng 8 2017
\(8^{30}+8^{31}+8^{32}\)
\(=8^{30}.1+8^{30}.8+8^{30}.8^2\)
\(=8^{30}.1+8^{30}.8+8^{30}.64\)
\(=8^{30}\left(1+8+64\right)\)
\(=8^{30}.73\)
\(=\left(2^3\right)^{30}.73\)
\(=2^{90}.73\)
\(=2^{89}.146⋮146\rightarrowđpcm\)
\(4^{25}+4^{26}+4^{27}+4^{28}+4^{29}+4^{30}\)
\(=4^{25}.1+4^{25}.4+4^{25}.4^2+4^{25}.4^3+4^{25}.4^4+4^{25}.4^5\)
\(=4^{25}.1+4^{25}.4+4^{25}.16+4^{25}.64+4^{25}.256+4^{25}.1024\)
\(=4^{25}\left(1+4+16+64+256+1024\right)\)
\(=4^{25}.1365\)
\(=4^{25}.195.7⋮7\rightarrowđpcm\)
ta có
\(Q=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)\)
\(Q=2\left(1+2+2^2+2^3+2^4\right)+2^6\left(1+2+2^2+2^3+2^4\right)\)
\(Q=2.31+2^6.31=31\left(2+2^6\right)\)
vậy Q chia hết cho 31