Question

If x + $\frac{1}{x}$= 1 and p = x⁴⁰⁰⁰ + $\frac{1}{x^{4000}}$and q be the digit at unit place in the number $2^{2^{n}}$+ 1, nĪN& n > 1, then p + q =

• A. 8
• B. 6
• C. 7
• D. None of these

Answer 1

Answer: (B)

6

Answer 2

x + $\frac{1}{x}$= 1 Ž x² – x + 1 = 0

Ž x = $\frac{1 \pm \sqrt{3}i}{2}$Ž x = –w, –w²

Ž Now, p = w¹⁰⁰⁰ + $\frac{1}{\omega^{1000}}$

= (w³) ³³³. w + $\frac{1}{(\omega^{3})^{333}.\omega}$

= w + $\frac{1}{\omega}$= w + w² = –1

Similarly, for x = w², also p = –1

For n > 1, 2ⁿ = 4k, k Ī N

\ $2^{2^{n}}$= 2^4k = (16)^k = a number with last digit = 6

\ q = (the digit at unit place in $2^{2^{n}}$) + 1 = 6 + 1 = 7

\ p + q = 7 + (–1) = 6