5. Let $a$ and $n$ be natural numbers, and it is known that $a^{n}-2014$ is a number with $2014$ digits. Find the smallest natural number $k$ such that $a$ cannot be a $k$-digit number.
Solution. Let $a$ be a $k$-digit number, then $10^{k-1} \leq a2013$, i.e., $k>45$.

We will check for $k$ starting from 46, whether the condition of the absence of an integer in the specified interval is met. We will find that

(for $k=46$) $2013 / 462013$ );

(for $k=47$) $2013 / 472013$ );

(for $k=48$) $2013 / 482013$ );

but (for $k=49$) $2013 / 49 > 41$ (since $\quad 41 \cdot 49 = 45^2 - 1^2 = 2025 - 16 < 2013$).

Thus, the interval $[2013 / 49, 2013 / 48)$ does not contain any integers, and for $k=49$, there are no suitable $n$.

## Answer: $\mathbf{k}=49$.

## Criteria.

The problem is complex, so don't be stingy with points for reasonable thoughts. Any answer in the vicinity of $\sqrt{2014}$, which is preceded by something reasonable, is worth points. A point is deducted for the absence of what is in small print.