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 $2^{2^n}+1,n\in N$, then value of $p+q$ is

• A. $8$
• B. $6$
• C. $7$
• D. none of these

Answer 1

The correct option is B $6$

Let $x=-w$ where $w$ is the cube root of unity.
We know that $w^3=1$ and $1+w+w^2=0$
Now $-(w+\frac{1}{w})$
$=\frac{-(w^2+1)}{w}=\frac{-(-w)}{w}$
$=1$
Hence $x^{4000}+\frac{1}{x^{4000}}$
$=w^{3999}.w+\frac{1}{w^{3999}.w}$
$=(w^3{)}^{1333}.w=\frac{1}{(w^3{)}^{1333}.w}$
$=w+\frac{1}{w}$
$=-1=p$
And $z=2^{2^n}+1$
Let $n=2$
We get $z=16+1=17$
Let $n=3$
We get $2^8+1$
$256+1$
$=257$
Hence unit's digit will be $7$.
Thus $q=7.$
Hence $p+q=7-1=6$