CMR: E=2+22+23+...+2120 chia hết cho 3, chia hết cho 7
F=3+33+35+...+31991 Chia hết cho 13, chia hết cho 41
Mấy vị tiền bối làm ơn giúp em với!!!
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10 tháng 10 2021
\(B=2+2^2+2^3+...+2^{60}\)
\(=2\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(=7\cdot\left(2+...+2^{58}\right)⋮7\)
21 tháng 10 2023
a: \(G=8^8+2^{20}\)
\(=2^{24}+2^{20}\)
\(=2^{20}\left(2^4+1\right)=2^{20}\cdot17⋮17\)
b: Sửa đề: \(H=2+2^2+2^3+...+2^{60}\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)\)
\(=3\left(2+2^3+...+2^{59}\right)⋮3\)
\(H=2+2^2+2^3+...+2^{60}\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(=7\left(2+2^4+...+2^{58}\right)⋮7\)
\(H=2+2^2+2^3+...+2^{60}\)
\(=\left(2+2^2+2^3+2^4\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(=2\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)\)
\(=15\left(2+2^5+...+2^{57}\right)⋮15\)
c: \(E=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{1989}\left(1+3+3^2\right)\)
\(=13\left(1+3^3+...+3^{1989}\right)⋮13\)
\(E=1+3+3^2+3^3+...+3^{1991}\)
\(=\left(1+3+3^2+3^3+3^4+3^5\right)+\left(3^6+3^7+3^8+3^9+3^{10}+3^{11}\right)+...+3^{1986}+3^{1987}+3^{1988}+3^{1989}+3^{1990}+3^{1991}\)
\(=364\left(1+3^6+...+3^{1986}\right)⋮14\)
KV
23 tháng 12 2023
A = 8⁸ + 2²⁰
= (2³)⁸ + 2²⁰
= 2²⁴ + 2²⁰
= 2²⁰.(2⁴ + 1)
= 2²⁰.17 ⋮ 17
Vậy A ⋮ 17
22 tháng 10 2021
\(C=\left(3+3^3+3^5\right)+\left(3^7+3^9+3^{11}\right)+...+\left(3^{1987}+3^{1989}+3^{1991}\right)\\ C=\left(3+3^3+3^5\right)+3^6\left(3+3^3+3^5\right)+...+3^{1986}\left(3+3^3+3^5\right)\\ C=\left(3+3^3+3^5\right)\left(1+3^6+...+3^{1986}\right)\\ C=273\left(1+3^6+...+3^{1986}\right)\\ C=13\cdot21\left(1+3^6+...+3^{1986}\right)⋮13\\ C=\left(3+3^3+3^5+3^7\right)+\left(3^9+3^{11}+3^{13}+3^{15}\right)+...+\left(3^{1985}+3^{1987}+3^{1989}+3^{1991}\right)\\ C=\left(3+3^3+3^5+3^7\right)+3^8\left(3+3^3+3^5+3^7\right)+...+3^{1984}\left(3+3^3+3^5+3^7\right)\\ C=\left(3+3^3+3^5+3^7\right)\left(1+3^8+...+3^{1984}\right)\\ C=2460\left(1+3^8+...+3^{1984}\right)\\ C=41\cdot60\left(1+3^8+...+3^{1984}\right)⋮41\)
LM
25 tháng 10 2021
a, 6100 - 1 = (6 . 6 . 6 ..... 6) - 1 = [(...6) . (...6) . (...6) ..... (...6)] - 1 = (...6) - 1 = ...5 \(⋮\) 5
LM
25 tháng 10 2021
b, 2120 - 1110 = (21 . 21 . 21 . 21 . 21..... 21) - (11 . 11 . 11 . 11 ..... 11) = [(...1) . (...1) . (...1) . (...1).....(...1)] - [(...1) . (...1) . (...1) . (...1).....(...1)] = (...1) - (...1) = ....0 \(⋮\) 2; \(⋮\) 5
PT
1
22 tháng 12 2023
b: \(B=16^5+2^{15}\)
\(=\left(2^4\right)^5+2^{15}\)
\(=2^{20}+2^{15}\)
\(=2^{15}\left(2^5+1\right)=2^{15}\cdot33⋮33\)
c: \(45⋮9;99⋮9;180⋮9\)
Do đó: \(45+99+180⋮9\)
=>\(C⋮9\)
d: \(D=2+2^2+2^3+...+2^{60}\)
\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)
\(=2\left(1+2+2^2\right)+2^4\cdot\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(=7\left(2+2^4+...+2^{58}\right)⋮7\)
\(D=2+2^2+2^3+...+2^{60}\)
\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)\)
\(=15\left(2+2^5+...+2^{57}\right)\)
=>D chia hết cho cả 3 và 5