\(2^n-1⋮7\Rightarrow2^n-1=7k\left(k\in N\right)\)
\(\Rightarrow2^n=7k+1\)
Vì \(7k+1\) luôn lẻ với mọi k Để \(2^n=7k+1\Leftrightarrow n=0\)
Với \(n=0\) thì \(2^0-1=1-1=0⋮7\)
Vậy \(n=0\)
tim tat ca cac so nguyen duong n sao cho:2^n-1 chia het cho 7
tim tat ca cac so nguyen duong n sao cho 2^n- chiahet cho 7
tim cac cap so nguyen duong nguyen to cung nhau sao cho: (x+y)/(x^2+y^2)=7/25
B1. Cho f(1)=1008,f(1)+f(2)+...+f(n)=n^2.f(n) f(n)la ham so duong
tinh f(2015)
B2 Cho cac so nguyen to q,p,r,s sao cho cac so sau deu la so nguyen to p^s+s^q,q^s+s^r,r^s+s^p
tim cac so p,q,r,s
B3 co a,b,c duong a mu 3+bmu3=c^3
so sanh a^2015+b^2015 va c^2015
tim cac cap so nguyen duong nguyen to cung nhau sao cho: (x+y)/(x^2+y^2)=7/25
mk can gap lam
tim k sao cho 3^(6n-1)-k.3^(3n-2)+1 chia het cho 7 voi moi so nguyen duong n
bai 1 tim cac stn x y z sao cho x+y+z=xyz
bai 2cmr neu m^2+mn+n^2chia het cho 9 voi m ,n la cac stn thi m,n chia het cho 3
bai 3 tim stn n nho nhat de cac phan so sau toi gian
7/n+9.8/n+10,...,31/n+33
bai 4chung to rang phan so m^3+2m/m^4+3m^2 +1 toi gian
bai 5 a,cmr khong ton tai so chinh phuong nao co dang abab
b,cho 51 so nguyen duong khong vuot qua 100 cmr ton tai 2 so trong 51 so noi tren ma 1 so chia het cho so con lai
c,tim cac so co 3 chu so sao cho hieu cua so ay va gom 3 chu so ay viet theo thu tu nguoc lai la mot so chinh phuong
bai 6cmr A=2x(x+y-1)+y^2+1 luon nhan gia tri khong am voi moi x,y
bai 7 tim hai so nguyen duong sao cho tong ,hieu thuong (so lon chia cho so nho )cua chung cong lai duoc 38
bai 8 cho ba so a,b,c thoa man a/2009=b/2010=c/2011
tinh gia tri cua bieu thuc M=4(a-b)(b-c)-(c-a)^2
tim cac so nguyen n sao cho P=2n -1 / n-1 la so nguyen
Tim a,b la cac so nguyen duong sao cho a+b^2 chia het cho a^2b-1
