Question

urgent please solve this trick question on number system​

Answer 1

From the assumed equation given in the question, we can derive,

$N-1=(2^1+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)$

Now we have to find a general form for the RHS of this equation.

Take the first two terms of the RHS.

$(2^1+1)(2^2+1)=3\times 5=15=2^4-1$

Now take the first three terms.

$(2^1+1)(2^2+1)(2^4+1)=(2^4-1)(2^4+1)=2^8-1$

Now take the first 4 terms.

$(2^1+1)(2^2+1)(2^4+1)(2^8+1)=(2^8-1)(2^8+1)=2^{16}-1$

So we get a general formula.

$\displaystyle \prod_{i=0}^{n}2^{2^i}+1=2^{2^{n+1}}-1$

Let me prove this formula by induction.

Since we found P(1), P(2) and P(3) successively, we can assume that P(k) is true too.

$\textsf{Let\ \ \displaystyle \prod_{i=0}^{k}2^{2^i}+1=2^{2^{k+1}}-1. }$

Consider P(k + 1).

$\displaystyle \prod_{i=0}^{k+1}2^{2^i}+1\ =\ \left(\prod_{i=0}^{k}2^{2^{i}}+1\right)(2^{2^{k+1}}+1)\\ \\ \\ =\ \left(2^{2^{k+1}}-1\right)\left(2^{2^{k+1}}+1\right)\\ \\ \\ =\ \left(2^{2^{k+1}}\right)^2-1\\ \\ \\ =\ 2^{2^{k+1}\cdot 2}-1\\ \\ \\ =\ 2^{2^{k+1+1}}-1\\ \\ \\ =\ 2^{2^{n+1}}-1$

Hence Proved!

Now, come back to the equation derived by us earlier!

$\begin{aligned}&N-1&=&\ \ (2^1+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)\\ \\ &N-1&=&\ \ 2^{64}-1\\ \\ &N&=&\ \ 2^{64}\\ \\ &2^{\lambda}&=&\ \ 2^{64}\\ \\ &\lambda&=&\ \ \mathbf{64}\end{aligned}$

Hence  64  is the answer.