Peter wrote: The number $21982145917308330487013369$ is the thirteenth power of a positive integer. Which positive integer? So, we have that $A=21982145917308330487013369=B^{13}$. The first thing to notice is that this number has 26 digits=> $10^{25}<A<10^{26}$=>$B<100$ Next, it is not hard to show that $80^{13}<10^{25}$=> $80<B<100$. One more thing is that the last digit of $B$ should be $9$(=>B=89 or B=99) and that the sume of digits of $A$ does not dividisible by 9 =>B=89 .

Nice!

Sorry, but could you show me how to decide 80^13 < 10^25? I am a beginner, though I understand the steps you are going. Thanks.

Go by small steps. $ 80^{13} < 10^{25}$ is equivalent to $ \left ( \frac {80} {100} \right )^{13} < \frac {1} {10}$, then $ \left ( \frac {4} {5} \right )^{13} < \frac {1} {10}$, then $ 5^{12} > 2^{27}$. But $ 5^4 = 625 > 512 = 2^9$, so $ 5^{12} = (5^4)^3 > (2^9)^3 = 2^{27}$.

Let $a^{13}$ be the large number. Note that $a<100.$ By Fermat's Little Theorem, $$a^{13}\equiv a\equiv 11 \pmod{13}.$$Also observe that $$a^{13}\equiv a\equiv 8\pmod{9}.$$Since it has to be less than 100 and $$89\equiv 8\pmod{9},$$we are done.