1708(1) is it b-a = 7
(2) total = 2012 pairs
Q1. We have T_{n}=(n^2+1)n! and S_{n}=T_{1}+T_{2}+....+T_{n}, If \frac{T_{n}}{S_{n}} =\frac{a}{b} where gcd(a,b)=1,, then find b-a.
Q2. No of ordered pairs (a,b) such that (a+ib)^{2010}=a-ib is
Q3. Let A,B,C be three subsets of X = \left\{ 1,2,3,....,n\right\} such that A∩B∩B=Null , A∩B≠φ , B∩C≠φ , the number of such (A,B,C) is :
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21Let |A| = r, Then A can be chosen in \binom{n}{r} ways , 1\le r \le n. Let |A\cap B|= t. then B can be chosen in \binom{r}{t}\cdot \binom{n-r}{p}\; \; , \; 1 \le t\le r , 1\le p \le n-r ways. Given A\cap B\cap C = \phi. Let |B\cap C| = k. Then C can be chosen in \binom{p}{k}\cdot 2^{r-t}\cdot 2^{n-r-p} = \binom{p}{k}\cdot 2^{n-p-t} ways. So total number of ways of choosing A,B,C is ------
\sum_{r=1}^{n}\sum_{t=1}^{r}\sum_{p=1}^{n-r}\sum_{k=1}^{p}\left (\binom{n}{r}\cdot \binom{r}{t}\cdot \binom{n-r}{p}\cdot \binom{p}{k}\cdot 2^{n-p-t}\right )
Which after computations (which only requires binomial theorem) , reduces (indeed reduces ,i have done the computation by hand) to \boxed{7^n + 5^n - 2\times 6^n}.
Notify if you need help with the computation. Basically sum up wrt. k,p,t,r in order.
Btw i know of a simpler solution...see here http://olympiads.hbcse.tifr.res.in/uploads/rmo-2004
211)
\sum_{r=1}^{n}(r^2+1)r! = \sum_{r=1}^{n}(r(r+1)-(r-1))r!= \sum_{r=1}^{n}(r+1)!r - r!(r-1)= n\cdot (n+1)!
So \frac{T_n}{S_n}= \frac{(n^2+1)n!}{n(n+1)!}= \frac{n^2+1}{n^2+n}
\gcd(n^2+1,n^2+n)= \gcd(n^2+1,n-1)=1 \; \; \text{if n is even and = 2 , if n is odd.}
So b- a= n-1 if n is even, and \frac{n-1}{2} if n is odd.
71@Aditya.. I didn't give any value of n, so how did you reached the answer as 49? Subhodip is correct.
@Subhodip
\gcd(n^2+1,n^2+n)= \gcd(n^2+1,n-1)=1 \; \; \text{if n is even and = 2 , if n is odd.}
Explain this Please! You know why!! :(
"Notify if you need help with the computation. "
yes, Please If you can!
"Then C can be chosen in"
Unclear for this! Help!
1708(2)
$\hspace{-16}Let $\mathbf{a+ib=z}$ and $\mathbf{a-ib=\bar{z}}$\\\\ Then $\mathbf{(a+ib)^{2010}=(a-ib)\Leftrightarrow z^{2010}=\bar{z}}$\\\\ $\mathbf{\mid z \mid^{2010}=\mid z\mid }$\\\\ $\mathbf{\mid z\mid .\left(\mid z \mid ^{2009}-1\right)=0}$\\\\ $\mathbf{\mid z \mid =1}$\\\\ Now Given that $\mathbf{z^{2010}=\bar{z}\Leftrightarrow z^{2011}=z.\bar{z}=\mid z\mid ^2}$\\\\ So $\mathbf{z^{2011}=1\Leftrightarrow z=1^{\frac{1}{2011}}=\cos\left(\frac{2k\pi}{2011}\right)+\sin\left(\frac{2k\pi}{2011}\right) }$\\\\ So $\mathbf{z=e^{\frac{2k\pi}{2011}}}$\\\\ Where $\mathbf{k\in \left\{0,1,2,........2010\right\}$\\\\ So Total $\mathbf{2011}$ solution\\\\ Now $\mathbf}z=0$ is also a solution \\\\ So Total $\mathbf{2011+1=2012}$ Solution
1708Thanks Shubhodip for very Nice answer.
I also want some Light on that line which is mentioned by vivek
Thanks
21I will show you the proof of the fact that :
if t= mq+ r, gcd(t,m) = gcd(m,r)
note that if k|a and k|b, then k≤ gcd(a,b)
Let gcd(t,m) = c
so c|t , and c|m, and you can see that c|r
since c|m and c|r, c= gcd(t,m)≤ gcd(m,r) (*)
Also gcd(m,r)|m and gcd(m,r)|r so gcd(m,r)|t
hence gcd(m,r)≤ gcd(m,t) (**)
combine (*) and (**) to get that gcd(m,r) = gcd(m,t). qed
For more see any elementary number theory book.
Here n2+ n = 1*(n2+ 1) + n-1, so by the above fact gcd(n2+ n,n2+1)= gcd(n2+1, n-1)
and n2+1 = (n+1)(n-1)+ 2
so gcd(n2+1,n-1) = gcd(n-1,2)
Now you can easily see that gcd(n-1,2) = 1, if n is even and = 2, if n is odd...
21"Then C can be chosen in"
Unclear for this! Help!
Sorry for not explaining
Follow my notations
There are t elements in common between A and B. So those must not be in C. Apart from those t elements there are B has got p elements. Since you want something common in between B and C, you must choose some k elements from t. Hence C(p,k). Also nothing said about A ∩ C. A has r elements , of them B has got t. So those t are strict NO for C. But you can choose any number of elements from r-t elements. So 2r-t. You can just as well select any number of elements from.. And there are n-r elements that are not in A. of those n-r, we have taken p for B. There are n-r-p left. You can choose any number of elements from them. so 2n-r-p .
Now multiply them...
As for the computations, I wont be able to upload it...It's two page long...See the link to the official solution given by me...I was just looking for a direct and straightforward way to solve it...which resulted in what i posted above..
71@subho.. Thanks.
But the rmo solution is really miraculous!! Very Nice it is!