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10^6 - 5^7
= (2^6 x 5^6) - 5^7
= 5^6 x (2^6 - 5)
= 5^6 x 59
vậy nó chia hết cho 59.
10^6-5^7
=5^6.2^6-5^7
=5^6.2^6-5^6.5
=5^6.(2^6-5)
=5^6.59 chia hết cho 59
\(A=3^{2022}-2^{2022}+3^{2020}-2^{2020}\\=(3^{2022}+3^{2020})-(2^{2022}+2^{2020})\\=3^{2020}\cdot(3^2+1)-2^{2020}\cdot(2^2+1)\\=3^{2020}\cdot10-2^{2019}\cdot2\cdot5\\=3^{2020}\cdot10-2^{2019}\cdot10\)
Ta có: \(\left\{{}\begin{matrix}3^{2020}\cdot10⋮10\\2^{2019}\cdot10⋮10\end{matrix}\right.\)
\(\Rightarrow3^{2020}\cdot10-2^{2019}\cdot10⋮10\)
hay \(A⋮10\) (đpcm)
\(\text{#}Toru\)
\(\left(27^{21}-9^{31}-3^{60}\right)\)
\(=\left[\left(3^3\right)^{21}-\left(3^2\right)^{31}-3^{60}\right]\)
\(=\left(3^{63}-3^{62}-3^{60}\right)\)
\(=3^{60}\left(3^3-3^2-3\right)\)
\(=3^{60}.17\)
\(\Rightarrow\left(27^{21}-9^{31}-3^{60}\right)⋮17\)
\(\RightarrowĐPCM\)
\(\left(27^{21}-9^{31}-3^{60}\right)\)
\(=\left(3^3\right)^{21}-\left(3^2\right)^{31}-3^{60}\)
\(=\left(3^{63}-3^{62}-3^{60}\right)\)
\(=3^{60}\left(3^3-3^3-3\right)\)
\(=3^{60}.17\)
\(\Rightarrow\left(27^{21}-9^{31}-3^{60}\right)⋮17\)
Vậy (2721 - 931 - 360 ) \(⋮\)17
Ta có: \(5+5^2+5^3+....+5^{12}\)
\(=\left(5+5^2\right)+\left(5^3+5^4\right)+.......+\left(5^{11}+5^{12}\right)\)
\(=\left(5+5^2\right)+5^2\left(5+5^2\right)+........+5^{10}\left(5+5^2\right)\)
\(=\left(5+5^2\right).\left(1+5^2+.......+5^{10}\right)\)
\(=30.\left(1+5^2+......+5^{10}\right)⋮30\)(1)
Ta lại có: \(5+5^2+5^3+......+5^{12}\)
\(=\left(5+5^2+5^3\right)+\left(5^4+5^5+5^6\right)+.......+\left(5^{10}+5^{11}+5^{12}\right)\)
\(=5\left(1+5+5^2\right)+5^4\left(1+5+5^2\right)+........+5^{10}\left(1+5+5^2\right)\)
\(=5.31+5^4.31+......+5^{10}.31\)
\(=31\left(5+5^4+......+5^{10}\right)⋮31\)(2)
Từ (1) và (2) \(\Rightarrowđpcm\)
Xem cách làm câu (b);(c);(d)
Bạn tham khảo:
Câu hỏi của Nguyễn Ngọc Thảo My - Toán lớp 7 - Học toán với OnlineMath
các bạn giúp mik nha
Cho A bằng 5^2021+1 phần 5^2022+1 ; B bằng 5^2020+1 phần 5^2021+1. Hãy so sánh A và B
a) Lập bảng
| n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | ... |
| 7n | 7 | 9 | 3 | 1 | 7 | 9 | 3 | 1 | ... |
| 9n | 9 | 1 | 9 | 1 | 9 | 1 | 9 | 1 | ... |
Ta có: 2018 : 4 = 504 (dư 2)
Suy ra \(2017^{2018}+2019^{2018}= \overline{...9}+\overline{...1}=\overline{...0}\)
Vậy 20172018 + 20192018 chia hết cho 10
b) Làm tương tự như câu a)
\(=5^{20}+\left(5^2\right)^{11}+\left(5^{ }^3\right)^7\)
=\(5^{^{ }20}+5^{22}+5^{21}\)
\(=5^{20}\cdot\left(1+5^2+5^1\right)\)
=\(5^{20}\cdot\left(1+25+5\right)\)
=\(5^{20}\cdot31\)
Vì 31 chia hết chó 31 nên
\(5^{20}+25^{^{ }11}+125^7\)chia hết cho 31
\(^{5^{20}+25^{11}+125^7}\)=\(1.5^{20}+25.25^{10}+\left(5^3\right)^7\)=\(1.5^{20}+25.\left(5^2\right)^{10}+5^{21}\)=\(1.5^{20}+25.5^{20}+5.5^{20}\)
=\(^{5^{20}.\left(1+25+5\right)}\)=\(5^{20}.31\)chia hết cho 31
Vậy \(5^{20}+25^{11}+125^7\)chia hết cho 31
Ta có:
\(92^3\equiv2\left(mod6\right)\)
\(\Rightarrow92^{30}\equiv\left(92^3\right)^{10}\left(mod6\right)\equiv2^{10}\left(mod6\right)\equiv4\left(mod6\right)\)
\(\Rightarrow92^{90}\equiv\left(92^{30}\right)^3\left(mod6\right)\equiv4^3\left(mod6\right)\equiv4\left(mod6\right)\)
\(\Rightarrow92^{93}\equiv92^{90}.92^3\left(mod6\right)\equiv4.2\left(mod6\right)\equiv2\left(mod6\right)\)
\(139^2\equiv1\left(mod6\right)\)
\(\Rightarrow139^{20}\equiv\left(139^2\right)^{10}\left(mod6\right)\equiv1^{10}\left(mod6\right)\equiv1\left(mod6\right)\)
\(\Rightarrow92^{93}+139^{20}+3\equiv2+1+3\left(mod6\right)\equiv6\left(mod6\right)\equiv0\left(mod6\right)\)
Vậy \(\left(92^{93}+139^{20}+3\right)⋮6\)