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4 tháng 11 2017
\(A=1+3+3^2+..........+3^{11}\)
\(\Leftrightarrow A=\left(1+3\right)+\left(3^2+3^3\right)+.........+\left(3^{10}+3^{11}\right)\)
\(\Leftrightarrow A=1\left(1+3\right)+3^2\left(1+3\right)+.........+3^{10}\left(1+3\right)\)
\(\Leftrightarrow A=1.4+3^2.4+.......+3^{10}.4\)
\(\Leftrightarrow A=4\left(1+3^2+..........+3^{10}\right)⋮4\left(đpcm\right)\)
HP
0
7 tháng 7 2017
1) ta có : 180= 9 x 5 x 4
Mà 1494 chia hết cho 9; 1495 chia hết cho 5 và 1496 chia hết cho 4.
=> 1494 x 1495 x 1496 chia hết cho 180.
2)Ta có 495= 9 x 5 x 11
Mà 1494 chia hết cho 9; 1495 chia hết cho 5 và 1496 chia hết cho 11.
=> 1494 x 1495 x 1496 chia hết cho 495.
k mình nha !!!
16 tháng 10 2022
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KX
0
NT
1
H
4 tháng 4 2018
\(A=n\left(n-1\right)\left(n+1\right)\left(n^2-4+5\right)\)
=> \(A=n\left(n+1\right)\left(n-1\right)\left(n^2-4\right)+n\left(n+1\right)\left(n-1\right)\cdot5\)
=> \(A=n\left(n+1\right)\left(n-1\right)\left(n-2\right)\left(n+2\right)+n\left(n-1\right)\left(n+1\right).5\)
Vì \(n\left(n+1\right)\left(n+2\right)\left(n-1\right)\left(n-2\right)\)là tích 5 số nguyên liên tiếp nên chia hết cho 2 và 5 => chia hết cho 2.5=10
n(n+1)(n-1) là tích 3 số nguyên liên tiếp nên chia hết cho 2 => 5n(n+1)(n-1) chia hết cho 2.5=10
=> A chia hết cho 10 đpcm
LT
1
BG
0
4 tháng 4 2017
Không thực hiện phép tính hãy cho biết 3 chữ số tận cùng của tích 12 chữ số nguyên dương đầu tiên
10 tháng 8 2016
Ta có: 2000a = 24.53.a
Ta đã biết số chính phương chỉ chứa các thừa số nguyên tố với số mũ chẵn, không chứa các thừa số nguyên tố với số mũ lẻ nên để 2000a là số chính phương thì a = 5.k2 (k thuộc N*)
Do a chia hết cho 6 => 5.k2 chia hết cho 6
Mà (5;6)=1 => k2 chia hết cho 6 => k chia hết cho 6
Mà a nhỏ nhất => k nhỏ nhất => k = 6
=> a = 5.62 = 5.36 = 180
Vậy số cần tìm là 180
x là 0, số 0 chia hết cho tất cả các số tự nhiên N*. và 4x+a=0, 2x +1=1,
0:1=0 :/
Đề bài là?