This collection isn’t mine, I am mere who keep from Diverse groups for leading to share with men that Like mathematics. And if there is someone can show One, or two, or,…, I am pleased that see our harmony. But if it isn’t, I will require to show in next week. 1 let n be positive integer, show that 1^n+8^n-3^n-6^n is divisible by 10 2 for all real numbers x, define f(1-x)+xf(1+x)=x^3+4x^2-x+2 Find f(999) 3 for a,b,c≥0,and a+b+c=3,prove that √(a/(1+b+bc))+√(b/(1+c+ca))+√(c/(1+a+ab))≥√3 4 for all positive real numbers a,b and ab(a+b)=2000, what is minimum value of 1/a+1/b+1/(a+b) 5 if a,b,and c be positive real numbers,then Prove that a^a b^b c^c≥〖abc〗^(((a+b+c)/3) ) Solution 1 consider the following congruent 1^4n≡1^((4n+1))≡1^((4n+2))≡1^((4n+3))≡1(mod10) 8^4n≡6(mod10), 8^((4n+1))≡8(mod10) 〖 8〗^((4n+2))≡4(mod10), 8^((4n+3))≡2(mod10) 3^4n≡1(mod10), 3^((4n+1))≡3(mod10) 〖 3〗^((4n+2))≡9(mod10), 3^((4n+3))≡7(mod10) 6^4n≡6(mod10), 6^((4n+1))≡6(mod10) 〖 6〗^((4n+2))≡6(mod10), 3^((4n+3))≡6(mod10) (All equations above we ought to show) It follows that 1^4n+8^4n-3^4n-6^4n≡0(mod10) 1^((4n+1))+8^((4n+1))-3^((4n+1))-6^((4n+1))≡0(mod10) 1^((4n+2))+8^((4n+2))-3^((4n+2))-6^((4n+2))≡-10(mod10) 1^((4n+3))+8^((4n+3))-3^((4n+3))-6^((4n+3))≡-10(mod10) And as ∀a∈z, and 4, we have a=4k,4k+1,4k+2, Or 4k+3, (Division Algorithm) Therefore, we can write 10⁄((1^n+8^n-3^n-6^n ) ) 2 from f(1-x)+xf(1+x)=x^3+4x^2-x+2 When x=-3,f(4)-3f(-2)=14 When x=-2,f(3)-2f(-1)=12 When x=-1,f(2)-f(0)=6 When x=0, f(1)=2 When x=1,f(0)+f(2)=6 When x=2,f(-1)+2f(3)=24 When x=3,f(-2)+3f(4)=62 Hence, f(0)=0,f(-1)=0,f(1)=2,f(2)=6 f(-2)=2,f(3)=12, f(4)=20 Assume that f(x)=ax^3+bx^2+cx+d We have 0=f(0)=d=0 0=f(-1)=-a+b-c+d 2=f(1)=a+b+c+d 6=f(2)=8a+4b+2c+d 2=f(-2)=-8a+4b-2c+d Hence, -a+b-c=0….(1) a+b+c=2….(2) 4a+2b+c=3….(3) From (1),(2) will get b=1 From (2),(3) will get c=1 That is, a=0 Therefore, it’s equation is f(x)=a_n x^n+a_((n-1)) x^(n-1)+⋯+a_2 x^2+a_1 x+a_0 Where, a_n,a_((n-1)),⋯,a_3,and a_0 are to be zero And a_2,a_1 are to be 1 And f(999)=x(x+1)=999(1000)=999000 3 from a,b,c≥0,and a+b+c=3, we have 3.1 a=3,b=c=0 3.2 a,b≠0,c=0, and 3.3 a,b,c≠0 3.1 yes is ok (obvious) 3.2 we will get a+b=3, and 0
Posted on: Mon, 01 Sep 2014 03:15:04 +0000
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