Tim so tu nhien m,n sao cho 2^m -2^n=1984
a, Cho ( a,b) = 1 . Chung minh rang (a.b, a+b)=1
b, Cho (a,b)= 1. Tim UCLN (11a+2b , 18a +5b)
C,, Cho A = m+n ; B=m^2+n^2.Trong do m va n la cac so tu nhien nguyen to cung nhau. Tim UCLN (A,B)
d, Tim cac so tu nhien n sao cho n^3 - n^2 + n-1la so nguyen to
Tim so tu nhien m,n sao cho:\(m^2+n^2+2mn+m+3n+2\)chinh phuong
tim tat ca so tu nhien m,n sao cho 2m+2015=\(_{||n-2016|+n-2016}\)
a) Chứng minh cac tich sau day la so chan:
( n+7 ) * ( n+10 ) va m*n*( m-n ) trong do m , n la so tu nhien ( vi ko co dau nhan nen minh viet dau sao )
b) Chưng minh rang voi n la so tu nhien thi B = n2 + 1 ko chia het cho 3
c)Tim so tu nhien n khi n2 chia het cho 3
bai 1 tim so tu nhien co 4 chu so ab cd biet abcd+abc+ab+a=4321.tim abcd
bai 2 cho m , n la cac so tu nhien va p la so nguyen to thoa man 9/m-1=m+n/p. tinh A= p^2-n
bai 3 tim so co 3 chu so biet abc 1000/a+b+c=abc.
Bài 1: Ký hiệu (abcd) là số tự nhiên có 4 chữ số.
(abcd) + (abc) + (ab) + (a) = 1111.a + 111.b + 11.c + d
Vậy 1111.a + 111.b + 11.c + d = 4321
+ Nếu a < 3 => 111.b + 11.c + d > 2098 (vô lý vì b, c, d < 10)
+ Nếu a > 3 => vế trái > 4321
Vậy a = 3 => 111.b + 11.c + d = 988
+ Nếu b < 8 => 11.c + d > 210 (vô lý vì c, d < 10)
+ Nếu b > 8 => vế trái > 988
Vậy b = 8 => 11.c + d = 100
+ Nếu c < 9 => d > 11 (vô lý)
Vậy c = 9; d = 1
=> (abcd) = 3891
cho M = 3 + 3^2 + 2^3 +3^4 + ... + 3^100 Hoi : a) M co chia het cho 4 cho 12 khong vi sao ? b) tim so tu nhien n biet rang 2M + 3 = 3^n
tim cac so tu nhien x , y biet :2^m+2^n=2^m+n
\(2^m+2^n=2^{m+n}\)
\(\Rightarrow2^{m+n}-2^m-2^n=0\)
\(\Rightarrow\left(2^{m+n}-2^m\right)-\left(2^n-1\right)=1\)
\(\Rightarrow2^m\left(2^n-1\right)-\left(2^n-1\right)=1\)
\(\Rightarrow\left(2^m-1\right)\left(2^{n-1}\right)=1\)
\(\Rightarrow\left\{\begin{matrix}2^n-1=1\\2^m-1=1\end{matrix}\right.\)\(\Rightarrow m=n=1\)
tim so tu nhien n sao cho n^2+2 chia het cho n+2
\(n^2+2⋮n+2\)
Có: \(2\left(n+2\right)⋮n+2\)
=> \(2n+4⋮n+2\)
=> \(\left(n^2+2\right)+\left(2n+4\right)⋮n+2\)
=> \(n^2+2+2n+4⋮n+2\)
=> \(n^2+2n+6⋮n+2\)
=> \(n\left(n+2\right)+6⋮n+2\)
Mà \(n\left(n+2\right)⋮n+2\)
=> \(6⋮n+2\)
=> \(n+2\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)
Lập bảng:
n+2 | -6 | -3 | -2 | -1 | 1 | 2 | 3 | 6 |
n | -8 | -5 | -4 | -3 | -1 | 0 | 1 | 4 |
tim so tu nhien n sao cho 3.(n+2) chia het cho n-2