Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
3^6 chia 7 dư 1
3^96 chia 7 dư 1
3^4 chia 7 dư 4
3^100 chia 7 dư 4
b)8.7.6.5.4.3.2.1=(8.7)(6.2)(4.3).=(55+1)(11+1)(11+1).5 chia 11 dư 1.1.5=5
8! chia 11 dư 5
Bạn alibaba nguyễn sai rồi nên mình sửa lại rồi bạn xem nhé :
Lời giải :
Ta có : \(331\equiv1\left(mod15\right)\)
\(\Rightarrow331^{332}\equiv1^{332}\equiv1\left(mod15\right)\left(1\right)\)
Ta có : \(2^4\equiv1\left(mod15\right)\)
\(\Rightarrow2^{333}=\left(2^4\right)^{83}.2\equiv2\left(mod15\right)\)
\(\Rightarrow332^{333}\equiv2^{333}\equiv2\left(mod15\right)\left(2\right)\)
Ta có : \(3^5\equiv3\left(mod15\right)\)
\(\Rightarrow3^{334}=3^{5.66}.3^4\equiv3^{66}.3^4\equiv3^{70}\equiv\left(3^5\right)^{14}\equiv3^{14}\equiv\left(3^5\right)^2.3^4\equiv3^2.3^4\equiv3^6\equiv9\left(mod15\right)\)
\(\Rightarrow333^{334}\equiv3^{334}\equiv9\left(mod15\right)\left(3\right)\)
Từ ( 1 ) , ( 2 ) , ( 3 ) suy ra : \(A\equiv\left(1+2+9\right)\equiv12\left(mod15\right)\)
Vậy A chia cho 15 dư 12
A = (tự chép lại đề)
\(\Leftrightarrow A=\left(330+1\right)^{332}+\left(333-1\right)^{333}+\left(332+1\right)^{334}\)
\(\Leftrightarrow A=\left(330+1+333-1+332+1\right)+\left(x\right)^{332+333+334}\)
\(\Rightarrow A=996\)
\(\Rightarrow A\)chia 15 dư : \(996:15=66\) dư 6
=> A chia 15 dư 6
a) Ta có:
\(A=2+2^2+2^3+...+2^{24}\)
\(\Rightarrow A=\left(2+2^2+2^3\right)+...+\left(2^{22}+2^{23}+2^{24}\right)\)
\(\Rightarrow A=14+...+2^{21}.\left(2+2^2+2^3\right)\)
\(\Rightarrow A=14+...+2^{21}.14\)
\(\Rightarrow A=\left(1+...+2^{21}\right).14⋮14\)( đpcm )
\(A=2+2^2+2^3+...+2^{24}\)
\(\Rightarrow A=\left(2+2^2+2^3+2^4\right)+...+\left(2^{21}+2^{22}+2^{23}+2^{24}\right)\)
\(\Rightarrow A=2\left(1+2+2^2+2^3\right)+...+2^{21}\left(1+2+2^2+2^3\right)\)
\(\Rightarrow A=2.15+...+2^{21}.15\)
\(\Rightarrow A=15\left(2+...+2^{21}\right)⋮15\left(đpcm\right)\)
b) Mk sửa đề chút là A chia 16 dư 15 nhé
Ta có:
\(A=2+2^2+2^3+...+2^{24}\)
\(\Rightarrow A=\left(2+2^2+2^3+2^4+2^5\right)+...+\left(2^{20}+2^{21}+2^{22}+2^{23}+2^{24}\right)\)
\(\Rightarrow A=2\left(1+2+2^2+2^3+2^4\right)+...+2^{20}\left(1+2+2^2+2^3+2^4\right)\)
\(\Rightarrow A=2.31+...+2^{20}.31\)
\(\Rightarrow A=\left(2+2^{20}\right).31\)
Vì 31 chia 16 dư 15 nên suy ra đpcm
1. \(\left\{{}\begin{matrix}a+495⋮a\\195-a⋮a\end{matrix}\right.\)
\(\Rightarrow\left(a+495\right)+\left(195-a\right)⋮a\)
\(\Leftrightarrow690⋮a\)
\(\Rightarrow a\in\left\{1,2,3,.....,345,690\right\}\)
Mà : \(a\) lớn nhất, \(a\in N\)
\(\Rightarrow a=690\)
Vậy : \(a=690\)
Tổng A có 2012 số hạng. Nhóm 4 số thành 1 nhóm. Ta có:
A = (2+22+23+24)+(25+26+27+28)+.......+(22009+22010+22011+22012)
A = 2(1+2+22+23)+25(1+2+22+23)+.....+22009(1+2+22+23)
A = 2.15 + 25.15 +.....+22009.15
A = 15 (2+25+.....+22009) chia hết cho 15
=> A chia 15 dư 0