In an old script found in ruins of Perspolis is written: This script has been finished in a year whose 13th power is 258145266804692077858261512663 You should know that if you are skilled in Arithmetics you will know the year this script is finished easily. Find the year the script is finished. Give a reason for your answer.
Problem
Source: Iranian National Olympiad (3rd Round) 2008
Tags: inequalities, number theory proposed, number theory
20.09.2008 13:40
The year is $ 183$, for sure.... But I have no reasons for it... Anyway...is it a problem of inequality?? I am very eager to see a solution then...
20.09.2008 15:57
easy one. put the number$ 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$
20.09.2008 19:13
Good one Sumita....
11.09.2015 13:51
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 $
13.09.2016 09:47
Hi,see here for another problem like this. http://www.artofproblemsolving.com/community/c146h150612p849619