Cho A = 1 + 5 + 52 + 53 +...+ 559
a, Chứng tỏ A ⋮ 31
b, So sánh A và B = 560 : 4
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KJ
2
30 tháng 11 2021
a: \(A=\left(1+5+5^2\right)+...+5^{57}\left(1+5+5^2\right)\)
\(=31\left(1+...+5^{57}\right)⋮31\)
AH
Akai Haruma
Giáo viên
30 tháng 11 2021
Lời giải:
a.
$A=1+5+5^2+5^3+...+5^{59}$
$= (1+5+5^2)+(5^3+5^4+5^5)+....+(5^{57}+5^{58}+5^{59})$
$=(1+5+5^2)+5^3(1+5+5^2)+....+5^{57}(1+5+5^2)$
$=31+5^3,31+,,,,,+5^{57}.31$
$=31(1+5^3+...+5^{57})\vdots 31$ (đpcm)
b.
$A=1+5+5^2+...+5^{59}$
$5A=5+5^2+5^3+...+5^{60}$
$\Rightarrow 4A=5A-A=5^{60}-1< 5^{60}$
$\Rightarrow A< \frac{5^{60}}{4}=B$
AH
Akai Haruma
Giáo viên
18 tháng 11 2021
Lời giải:
a. Ta thấy:
$3+3^2+3^3+...+3^{99}\vdots 3$
$1\not\vdots 3$
$\Rightarrow A=1+3+3^2+...+3^{99}\not\vdots 3$
$\Rightarrow A\not\vdots 9$
b.
$A=(5+5^2)+(5^3+5^4)+...+(5^{39}+5^{40})$
$=5(1+5)+5^3(1+5)+...+5^{39}(1+5)$
$=5.6+5^3.6+....+5^{39}.6$
$=6(5+5^3+...+5^{39})$
$=2.3.(5+5^3+...+5^{39})$
$\Rightarrow A\vdots 2$ và $A\vdots 3$
![YTP] Huấn Hoa Hồng 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)
PT
24 tháng 1 2021
a. S = 5 + 52 + 53 + 54 + 55 + 56 +...+ 52012.
S = (5 + 52 + 53 + 54) + 55(5 + 52 + 53 + 54)+....+ 52009(5 + 52 + 53 + 54)
Vì (5 + 52 + 53 + 54) = 780 chia hết cho 65
Vậy S chia hết cho 65
b. Gọi số cần tìm là a ta có: (a - 6) chia hết cho 11; (a - 1) chia hết cho 4; (a - 11) chia hết cho 19.
(a - 6 + 33) chia hết cho 11; (a - 1 + 28) chia hết cho 4; (a - 11 + 38) chia hết cho 19.
(a + 27) chia hết cho 11; (a + 27) chia hết cho 4; (a + 27) chia hết cho 19.
Do a là số tự nhiên nhỏ nhất nên a + 27 nhỏ nhất
Suy ra: a + 27 = BCNN (4;11; 19).
Từ đó tìm được: a = 809
A = 10n + 18n - 1 = 10n - 1 - 9n + 27n
Ta biết số n và số có tổng các chữ số bằng n có cùng số dư khi chia cho 9 do đó 
nên 
* Vậy A chia hết cho 27
a. A = 1 + 5 + 52 + 53 + .... + 559
A = ( 1 + 5 + 52) + (53 + 54 + 55) +.....+ (557 + 558 + 559)
A = (1 + 5 + 52) + 53(1 + 5 + 52) + ..... + 557( 1 + 5 + 52)
A = (1 + 5 + 52)( 1 + 53 +......+ 557)
A = 31(1 + 53+.....+ 557)
Vì có một thừa số 31 nên A ⋮ 31
\(a,A=\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{57}+5^{58}+5^{59}\right)\\ A=\left(1+5+5^2\right)+5^3\left(1+5+5^2\right)+...+5^{57}\left(1+5+5^2\right)\\ A=\left(1+5+5^2\right)\left(1+5^3+...+5^{57}\right)\\ A=31\left(1+5^3+...+5^{57}\right)⋮31\\ b,5A=5+5^2+5^3+...+5^{60}\\ \Rightarrow5A-A=4A=5^{60}-1\\ \Rightarrow A=\dfrac{5^{60}-1}{4}=\dfrac{5^{60}}{4}-\dfrac{1}{4}< \dfrac{5^{60}}{4}=B\)