The year is $ 183$, for sure.... But I have no reasons for it... Anyway...is it a problem of inequality?? I am very eager to see a solution then...

easy one. put the number$ A$. we have $ 17^2 < 300$ so $ 170^{13} < 17*3^6*10^{25} < A$ and its easy to prove$ A < 200$ $ a^{13} = A$ so we have $ (a,9) = 3$ and$ (a,5) = 1$ and$ a = 8k - 1$ so $ a = 183$

Good one Sumita....

another solution : $$x^{13} \equiv x (mod 10) \Rightarrow x\equiv 3 (mod 10)$$. $$x^{13}< 8\times 10^{29} \Rightarrow x^{13}< (200)^{13}\Rightarrow x< 200$$. $$x = \overline{1a3}$$. $$ x=10a+3 $$ so : $$ (10a+3)^{13}\equiv (10a)\binom{13}{1}3^{12} + 3^{13}\equiv 30 a + 23 \equiv 63 (mod 100) $$ so $$ 30a\equiv 40 (mod 100) \Rightarrow a \equiv 8 (mod 10) $$ . so $a$ must be $18$ so we have $ x=183 $

Hi,see here for another problem like this. http://www.artofproblemsolving.com/community/c146h150612p849619