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Giải:
a) \(M=21^9+21^8+21^7+...+21+1\)
Do \(21^n\) luôn có tận cùng là 1
\(\Rightarrow M=21^9+21^8+21^7+...+21+1\)
Tân cùng của M là:
\(1+1+1+1+1+1+1+1+1+1=10\) tận cùng là 0
\(\Rightarrow M⋮10\)
\(\Leftrightarrow M⋮2;5\)
b) \(N=6+6^2+6^3+...+6^{2020}\)
\(N=6.\left(1+6\right)+6^3.\left(1+6\right)+...+6^{2019}.\left(1+6\right)\)
\(N=6.7+6^3.7+...+6^{2019}.7\)
\(N=7.\left(6+6^3+...+6^{2019}\right)⋮7\)
\(\Rightarrow N⋮7\)
Ta thấy: \(N=6+6^2+6^3+...+6^{2020}⋮6\)
Mà \(6⋮̸9\)
\(\Rightarrow N⋮̸9\)
c) \(P=4+4^2+4^3+...+4^{23}+4^{24}\)
\(P=1.\left(4+4^2\right)+4^2.\left(4+4^2\right)+...+4^{20}.\left(4+4^2\right)+4^{22}.\left(4+4^2\right)\)
\(P=1.20+4^2.20+...+4^{20}.20+4^{22}.20\)
\(P=20.\left(1+4^2+...+4^{20}+4^{22}\right)⋮20\)
\(\Rightarrow P⋮20\)
\(P=4+4^2+4^3+...+4^{23}+4^{24}\)
\(P=4.\left(1+4+4^2\right)+...+4^{22}.\left(1+4+4^2\right)\)
\(P=4.21+...+4^{22}.21\)
\(P=21.\left(4+...+4^{22}\right)⋮21\)
\(\Rightarrow P⋮21\)
d) \(Q=6+6^2+6^3+...+6^{99}\)
\(Q=6.\left(1+6+6^2\right)+...+6^{97}.\left(1+6+6^2\right)\)
\(Q=6.43+...+6^{97}.43\)
\(Q=43.\left(6+...+6^{97}\right)⋮43\)
\(\Rightarrow Q⋮43\)
Chúc bạn học tốt!
Lời giải:
$A=(4+4^2)+(4^3+4^4)+....+(4^{23}+4^{24})$
$=(4+4^2)+4^2(4+4^2)+....+4^{22}(4+4^2)$
$=(4+4^2)(1+4^2+...+4^{22})$
$=20(1+4^2+...+4^{22})\vdots 20$
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$A=(4+4^2+4^3)+(4^4+4^5+4^6)+....+(4^{22}+4^{23}+4^{24})$
$=4(1+4+4^2)+4^4(1+4+4^2)+....+4^{22}(1+4+4^2)$
$=(1+4+4^2)(4+4^4+...+4^{22})$
$=21(4+4^4+....+4^{22})\vdots 21$
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Vậy $A\vdots 20; A\vdots 21$. Mà $(20,21)=1$ nên $A\vdots (20.21)$ hay $A\vdots 420$
A = (4 + 4^2 + 4^3 + 4^4 + 4^5 + 4^6) + (4^7 + 4^8 + 4^9 + 4^10 + 4^11 + 4^12) + (4^13 + 4^14 + 4^15 + 4^16 + 4^17 + 4^18) + (4^19 + 4^20 + 4^21 + 4^22 + 4^23 + 4^24)
A = (4 + 4^2 + 4^3 + 4^4 + 4^5 + 4^6) + 4^6(4 + 4^2 + 4^3 + 4^4 + 4^5 + 4^6) + 4^12(4 + 4^2 + 4^3 + 4^4 + 4^5 + 4^6) + 4^18(4 + 4^2 + 4^3 + 4^4 + 4^5 + 4^6)
A = (4 + 4^2 + 4^3 + 4^4 + 4^5 + 4^6).(1+4^6+4^12+4^18)
A = 5460.(1+4^6+4^12+4^18)
A = 420 . 13(1+4^6+4^12+4^18) => A chia hết cho 420
A = 20.21.13(1+4^6+4^12+4^18) => A chia hết cho 20 ; 21
A=(13+132)+(133+134)+.......................+(1399+13100)
A=1.(13+132)+132.(13+132)+..............+1398.(13+132)
A=1.182+132.182+..........................+1398.182
A+182.(1+132+..............+1398) Chia hết cho 182
--> A chia hết cho 182
Sửa đề: A=4+4^2+4^3+...+4^23+4^24
A=4(1+4+4^2)+...+4^22(1+4+4^2)
=21(4+...+4^22) chia hết cho 21
A=(4+4^2)+4^2(4+4^2)+...+4^22(4+4^2)
=20(1+4^2+...+4^22) chia hết cho 20
Câu 1:
$A=(2+2^2)+(2^3+2^4)+(2^5+2^6)+....+(2^{2019}+2^{2020})$
$=2(1+2)+2^3(1+2)+2^5(1+2)+....+2^{2019}(1+2)$
$=(1+2)(2+2^3+2^5+...+2^{2019})=3(2+2^3+2^5+...+2^{2019})\vdots 3$
-----------------
$A=2+(2^2+2^3+2^4)+(2^5+2^6+2^7)+....+(2^{2018}+2^{2019}+2^{2020})$
$=2+2^2(1+2+2^2)+2^5(1+2+2^2)+....+2^{2018}(1+2+2^2)$
$=2+(1+2+2^2)(2^2+2^5+....+2^{2018})$
$=2+7(2^2+2^5+...+2^{2018})$
$\Rightarrow A$ chia $7$ dư $2$.
Câu 2:
$B=(3+3^2)+(3^3+3^4)+....+(3^{2021}+3^{2022})$
$=3(1+3)+3^3(1+3)+...+3^{2021}(1+3)$
$=(1+3)(3+3^3+...+3^{2021})=4(3+3^3+....+3^{2021})\vdots 4$
-------------------
$B=(3+3^2+3^3)+(3^4+3^5+3^6)+...+(3^{2020}+3^{2021}+3^{2022})$
$=3(1+3+3^2)+3^4(1+3+3^2)+....+3^{2020}(1+3+3^2)$
$=(1+3+3^2)(3+3^4+...+3^{2020})=13(3+3^4+...+3^{2020})\vdots 13$ (đpcm)
\(C=1+4+4^2+4^3+4^4+....+4^{20}\)
\(C=\left(1+4+4^2\right)+\left(4^3+4^4+4^5\right)+...+\left(4^{18}+4^{19}+4^{20}\right)\)
\(C=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...+4^{18}\left(1+4+4^2\right)\)
\(C=\left(1+4+4^2\right)\left(1+4^3+...+4^{18}\right)\)
\(C=21.\left(1+4^3+...+4^{18}\right)\)
Vì 21 chia hết cho 21 nên \(21.\left(1+4^3+...+4^{18}\right)\) chia hết cho 21(đpcm)
Chúc bạn học tốt!!!
\(C=1+4+4^1+4^2+4^3+4^4+...+4^{20}\)
\(C=\left(1+4+4^2\right)+\left(4^2+4^3+4^4\right)+...+\left(4^{18}+4^{19}+4^{20}\right)\) \(C=\left(1+4+4^2\right)+4^3.\left(1+4+4^2\right)+...+4^{18}.\left(1+4+4^2\right)\)
\(C=\left(1+4+4^2\right).\left(1+4^3+...+4^{18}\right)\)
\(C=21.\left(1+4^3+...+4^{18}\right)\)
Vì \(21⋮21\) \(\Rightarrow21.\left(1+4^3+...+4^{18}\right)\)
Vậy \(C⋮21\)