Chứng tỏ rằng: (5 + 5^2 + 5^3 + 5^4 + 5^5 + ... + 5^29 + 5^30) chia hết cho 6
(a + a^2 + a^3 + a^4 + ... + a^29 + a^30) chia hết cho (a + 1),(a thuộc N)
(3 + 3^2 + 3^3 + 3^4 + 3^5 + ... + 3^29 + 3^30) chia hết cho 4
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2.Gọi số cần tìm là \(x\left(x\ne0,x>9\right)\)
Ta có:
\(53=mx+2\left(m\in N\right)\\ \Rightarrow51=mx\\ \Rightarrow x\inƯ\left(51\right)\left(1\right)\\ 77=nx+9\left(n\in N\right)\\ \Rightarrow68=nx\\ \Rightarrow x\inƯ\left(68\right)\left(2\right)\)
Từ (1) và (2) ta có:
\(x\inƯC\left(51,68\right)\)
\(51=3\cdot17\\ 68=2^2\cdot17\\ \Rightarrow\text{ƯCLN}\left(51,68\right)=17\\ ƯC\left(51,68\right)=Ư\left(17\right)=\left\{1;17\right\}\)
Vì x > 9 nên x = 17
Vậy số chia là 17
3. Làm câu b trước, các câu kia trả lời tương tự hoặc áp dụng điều đã chứng minh
b,
\(a+a^2+a^3+a^4+...+a^{29}+a^{30}\\ =\left(a+a^2\right)+\left(a^3+a^4\right)+...+\left(a^{29}+a^{30}\right)\\ =a\left(1+a\right)+a^3\left(1+a\right)+...+a^{29}\left(1+a\right)\\ =\left(1+a\right)\left(a+a^3+...+a^{29}\right)⋮a+1\)
Vậy \(a+a^2+a^3+a^4+...+a^{29}+a^{30}⋮a+1\) với a thuộc N
A=3+32+33...+329+330
A=(3+32+33)+...+(328+329+330)
A=3.(1+3+32)+...+328.(1+3+32)
A=3.13+...+328.13
A=13.(3+...+328) chia hết cho 13
A= 3(1+3+3^2)+3^4(1+3+3^2)+3^7(1+3+3^2)+...+3^28.(1+3+3^2)
A=(1+3+3^2)(3+3^4+3^7+...+3^25+3^28)
=13.(3+3^4+3^7+...3^28) vậy A chia hết cho 13
1):
Ta có: 51 chia hết cho 3
120 chia hết cho 3
453 chia hết cho 3
=>51a+120b+453c chia hết cho 3
2):
Ta có:
A=5+52+53+...+530
=>A=(5+52)+52(5.52)+...+528(5+52)
=>A=(5+52).(52+54+...+528)
Vì 5+52=30 chia hết cho 6
=>A chia hết cho 6
Ta có : A = (51+52+53)+(54+55+56)+...+(528+529530)
A = 155 + 53.(51+52+53)+...+527.(51+52+53)
A = 155 + 53. 155+...+527.155
A = 155.(1+53+...+527) chia hết cho 155
Vậy A chia hết cho 155
(5+52+53)+(54+55+56)+...+(528+529+530)
= 155 +53(5+52+53)+...+527(5+52+53)
=155+53.155+...+527.155
=155(1+53+..+527) chia hết cho 155
Ta có : A = 5 + 52 + 53 + ..... + 58
=> A = (5 + 52) + (53 + 54) + ..... + (57 + 58)
=> A = (5 + 52) + 52(5 + 52) + ..... + 56(5 + 52)
=> A = 30 + 52.30 + .... + 56.30
=> A = 30(1 + 52 + .... + 56)
Vì (1 + 52 + .... + 56) là số nguyên
Vậy A = 30(1 + 52 + .... + 56) chia hết cho 30
A=5+5^2+5^3+...+5^20
=(5+5^2)+(5^3+5^4)+...+(5^19+5^20)
=(5+5^2)+5^2(5+5^2)+...5^18(5+5^2)
=30+5^2.30+5^4.30+5^6.30+..+5^18.30
=30(1+5^2+5^4+5^6+..+5^18)(chia hết cho 30)
Vậy A là bội của 30
a)A=5+52+53+...+58
A= (5+52)+(53+54) + ... + (57+58)
A= 5( 1+5) + 52(5+52)+... + 56(5+52)
A= 30 + 52 . 30 + ... +56.30
A = 30 ( 1 + 52+...+56) chia hết cho 30
=> A chia hết cho 30
b)B=3+33+35+37+...+329
B = (3 + 33 + 35) + (37+39+311) + ... + ( 327+328+329)
B = 273 + 36 (3 + 33 + 35) + ... + 326 (3 + 33 + 35)
B = 273 + 36.273 + ... + 326.273
B = 273 ( 1 + 36+...326) chia hết cho 273
=> B chia hết cho 273
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