Let’s $n=p_1...p_m$, where $p_1\leq p_2\leq ...\leq p_m$. Then $n=t\cdot p_m$. Let’s consider numbers $t\cdot (p_m+1), t\cdot (p_m+2), ... , t\cdot (p_m+p_m-1)$. They all are between $n$ and $2n$. If $p_m>2$, then $p_m+1$ has complexity more than one, that is $t\cdot (p_m+1)$ has complexity more than $n$. So, satisfying $n$ is a power of two. Let’s prove that any power of two satisfies the condition(a). Suppose it’s wrong. That is, there exists a number $2^{n-1}<k=p_1...p_m<2^{n}$, where $m\geq n$ But $k$ can’t be equal to power of two, so $2^n>k>2^m\geq 2^n$ — contradiction. And let’s prove that any power of two doesn’t satisfy the condition (b). We just can consider the number $2^n<2^{n-1}\cdot 3< 2^{n+1}.

It's known that there is a power of $2$ in any interval $[n,2n]$ , Which implies that for $(a)$ : $n$ can be any power of $2$. But, in $(b)$ : We can find an integer $a=2^{k-1}\times 3$ in the interval $(2^k,2^{k+1})$ which it's complexity equal to $2^k$ Therefore, there is no such $n$ satisfying $(b)$.

I haven't read the solution. I don't understand what "the number of factors in the prime decomposition of an integer $n > 1$ the complexity of $n$. For example, complexity of numbers $4$ and $6$ is equal to $2$" mean. How does it differ from the number of factors? ...like $6$ have $(1+1)(1+1)=4$ factors