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Q.
If $x+\frac{1}{x}=1$ and $p=x^{1000}+\frac{1}{x^{1000}}$ and $q$ be the digit at
unit place in the number $2^{2^{n}}+1, n \in N$ and $n>1$, then $p+q=$
Binomial Theorem
Solution:
$x+\frac{1}{x}=1 \Rightarrow x^{2}-x+1=0 \Rightarrow x=\frac{1 \pm \sqrt{3} i}{2}$
$\Rightarrow x=-\omega,-\omega^{2}$
Now, $p=\omega^{1000}+\frac{1}{\omega^{1000}}=\left(\omega^{3}\right)^{333} \cdot \omega+\frac{1}{\left(\omega^{3}\right)^{333} \cdot \omega}$
$=\omega+\frac{1}{\omega}=\omega+\omega^{2}=-1$
Similarly, for $x=-\omega^{2}$, also $p=-1$
For $n>1,2^{n}=4 k, k \in N$
$\therefore 2^{2^{n}}=2^{4 k}=(16)^{k}=$ a number with last digit $=6$
$\therefore q=\left(\right.$ the digit at unit place in $\left.2^{2^{n}}\right)+1=6+1=7$
$\therefore p+q=7+(-1)=6$