What is the dimension of a hypercube with 524,288 edges?
I know that for $Q_n$, the number of edges is $(\frac{1}{2})n2^n = n2^{n-1}$. So I wrote this down in my book: $$(\frac{1}{2})n2^n = n2^{n-1} = 524288$$
My question is how do you solve for n? Do I multiply 524288 by 2 then solve for n? Like this: $$n2^n = 1048576$$ Or is it just a big guessing game of trial and error when you solve for n?
$1048576=2^{20 }=n2^n $
So let $n=2^a $ and $2^a2^{2^a}=2^{a+2^a}=2^{20} $
$a+2^a=20$
As $2^a>a$ then $2^a $ must be the leading power of $2$ in the binary representation of $20$ and it must be such that $a $ is what's left.
$20=16 +4$ so if this hyper cube is actually possible, it must be that $16=2^4$. Which it does.
So $n=2^a=16$.
So $a=4$ and $n=16$.