Let a,b,c,d be real numbers such that $a+b+c+d=10$  then the minimum value of $a^2\cot 9^o+b^2\cot {27}^o+c^2\cot {63}^o+d^2\cot {81}^o$  is $5\sqrt{n}(n\in N)$  then n =

If $a_1,a_2,\mathrm{......},a_{n}and{b}_1,b_2,\mathrm{......},b_n$  are 2n real numbers, then    ${(a_1{b}_1+a_2{b}_2+\mathrm{......}{a}_n{b}_n)}^2$$\le (a_1^2+a_2^2+\mathrm{......}+a_n^2)(b_1^2+b_2^2+\mathrm{.......}+b_n^2)\mathrm{.........}\left(1\right)$
Let     $a_1=a\sqrt{\cot 9^o},a_2=b\sqrt{\cot {27}^o},a_3=c\sqrt{\cot {63}^o},$

$a_4=d\sqrt{\cot {81}^o}and{b}_1=\sqrt{\tan 9^o},b_2=\sqrt{\tan {27}^o},$

$b_3=\sqrt{\tan {63}^o},b_4=\sqrt{\tan {81}^o}$
Now using(1), we get 
 ${(a+b+c+d)}^2\le (a^2\cot 9^o+b^2\cot {27}^o+c^2\cot {63}^o+d^2\cot {81}^o)$

$(\tan 9^o+\tan {27}^o+\tan {63}^o+\tan {81}^o)\mathrm{.....}\left(2\right)$     
But a+b+c+d=10 (Given)
And  $(\tan 9^o+\tan {81}^o)+(\tan {27}^o+\tan {63}^o)$
= $\frac{1}{\sin 9^o\cos 9^o}+\frac{1}{\sin {27}^o\cos {27}^o}=\frac{2}{\sin {18}^o}+\frac{2}{\sin {54}^o}$
=  $\frac{2}{\frac{\sqrt{5}-1}{4}}+\frac{2}{\frac{\sqrt{5}+1}{4}}$

$=8\left[\frac{(\sqrt{5}+1)+(\sqrt{5}-1)}{4}\right]=4\sqrt{5}$   $\therefore$$From\left(2\right)$ we get
 $100\le 4\sqrt{5}(a^2\cot 9^o+b^2\cot {27}^o+c^2\cot {63}^o+d^2\cot {81}^o)$
 $\Rightarrow a^2\cot 9^o+b^2\cot {27}^o+c^2\cot {63}^o+d^2\cot {81}^o\ge \frac{25}{\sqrt{5}}$
 $=5\sqrt{5}=5\sqrt{n}\Rightarrow n=5$