Question

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 $p+q=$

• A. 8
• B. 6
• C. 7
• D. 0

Answer 1

Answer: (B)

$x+\frac{1}{x}=1\Rightarrow x^2-x+1=0\Rightarrow x=\frac{1\pm \sqrt{3}i}{2}$
$\Rightarrow x=-\omega ,-{\omega }^2$
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 $n>1,2^n=4k,k\in N$
$\therefore 2^{2^n}=2^{4k}=(16{)}^k=$ a number with last digit $=6$
$\therefore q=($ the digit at unit place in $2^{2^n})+1=6+1=7$
$\therefore p+q=7+(-1)=6$