Power of two problem

Number Theory Level 2

Is there any number $n$ such that $\begin{array}{l} {2^n} = 3 \times k\\ \end{array}$ where $n \in \mathbb{Z}^+,k \in \mathbb{N}?$

It is immposible to tell. No. Yes.

1 solution

Vinayak Srivastava
Jun 30, 2020

$3\times k$ must be odd natural number and $\geq 3$ if $k\in \mathbb{N}$

$2^{n}$ must be even if $n\geq 1$

If $n=0, 2^{n}=1,$ which is not in the form $3\times k$ .

If $n <0 , 2^{n}$ will not be an integer.

So, in all cases where $k\in \mathbb{N}$ and $n\in \mathbb{Z^{+}}$ , it is impossible.