Recent content by dr hannibal

16/64 is an unusual fraction such that when you cancel 6's from the top and bottom youre left with correct answer of 1/4. I need to Find a way for finding all other unusal fractions for a fraction n/m such that n>=11 , and m <=99. I have tried writing it as (10a+b)/(10b+c)=a/c ,but stuck...

thanks:)

so I have 2^{1990}=(199k+2)^{10} expanding I have. 2^{1990}=2^{10}+10.2^9. (199k)+\frac{10.9}{1.2} 2^8.(199k)^2+...+10.2. (199k)^9+(199K)^{10}-(1) now its clear 199|2^{1990}-2^{10} since I can take 199 out of the RHS. but the book seems to imply that the above equation(1) says...

Yup that's what I meant for your hint how would I use notation to represent it ..just one line would be enough c, its just I have been grappling with this question for far too long and have not made any headway.. Thanks

it means 3 can be written as 1+1+1 , 2+1, 1+2, 3 so 4 different ways..

sorry for the many threads Let S_n denote the number of ways of expressing n as positive integrs.. S_1=1 , s_2=2, s_3=4 .. Prove that S_n=S_{n-1}+S_{n-2} ---S_1+1 no idea to prove that :

thanks

thanks :) , one more small question when they have summed the above where is -2a_ib_ja_jb_i coming from?

Need help proving Cauchy Schwarz inequality ... the first method I know is pretty easy \displaystyle\sum_{i=1}^n (a_ix-b_i)^2 \geq 0 expanding this and using the discriminatant quickly establishes the inequality..The 2nd method I know is I think a easier one , but I don't have a clue about...