Bases, Digits, and Digits of Bases

When $n$ is written in base 2, it has $a$ digits, therefore:

$2^{a-1} \leq n < 2^a$

When $n$ is written in base $a$ , it has 5 digits, therefore:

$a^{4} \leq n < a^5$

For there to be an $n$ that exists in both of these ranges, we must have $2^a > a^4$ and $2^{a-1} < a^5$ .

Since we are concerned with the digits of $a$ in base 2, let $a=2^k$ .

When $2^a > a^4$ :

$\begin{aligned} 2^{(2^k)} &> 2^{4k} \\ 2^k &> 4k &\text{ (since we are working with positive integers)} \\ k &> 4 &\text{ (by observation)} \\ \implies a &> 2^4 \end{aligned}$

When $2^{a-1} < a^5$ :

$\begin{aligned} 2^{(2^k-1)} &< 2^{5k} \\ 2^k-1 &< 5k &\text{ (since we are working with positive integers)} \\ k &< 5 &\text{ (by observation)} \\ \implies a &< 2^5 \end{aligned}$

We now know that $a$ must at least exist within the range $2^4 < a < 2^5$ . All numbers in this range have 5 digits when written in base 2,

$\therefore a$ has $\boxed{5}$ digits when written in base 2.

One is looking for an integer whose fourth power binary expansion has at least 16 bits as $16_{10}=10000_2$ ,, the smallest 5-bit unsigned binary integer. That implies the integer is at least 17 as 16 will not do as $16^4=65536=\text{D5D1}_{17}$ . Fortunately 17 works: $17^4=83521=\text{10000}_{17}$ . $17_{10}=10001_2$ , which is 5 bits long. I must have spent too much time around binary computers!