Question

If $x+\frac{1}{x}=1$ and $p=x^{4000}+\frac{1}{x^{4000}}$ and $q$ be the digit at unit place in the number $2^{2^n}+1,n\in N$ and $n>1$, then the value of $p+q=$

Answer 1

$x+\frac{1}{x}=1$
$\Rightarrow x^2-x+1=0$
$\therefore x=-\omega ,-{\omega }^2$
Now for $x=-\omega ,p={\omega }^{4000}+\frac{1}{{\omega }^{4000}}=\omega +\frac{1}{\omega }=-1$
Similarly for $x=-{\omega }^2$ also $p=-1$
For $n>1,2^n=4k$
$\therefore 2^{2^n}=2^{4k}$
${16}^k=$ a number with last digit $=6$
$\Rightarrow q=6+1=7$
Hence $p+q=6$