I ran into a nice question from one book in Discrete Mathematics. I want to someone lean me how solve such a problem, because I prepare for entrance exam.
if the time is "Wednesday 4 afternoon", after $47^{74}$ hours, we are in what hours? and what day?
Thanks to all.
If now it's “Wednesday, 4pm”, then $16$ hours ago it was “Wednesday, 0:00”.
Thus the problem is to know what hour it is $47^{74}+16$ hours after “Wed, 0:00”. This is obviously solved by computing the remainder of $47^{74}+16$ divided by $24$; since $47$ is coprime with $24$ and $\varphi(24)=8$, from Fermat-Euler we can say $$ 47^{8}\equiv 1\pmod{24} $$ hence $47^{74}\equiv 47^2=2209\equiv 1\pmod{24}$. Therefore, adding back $16$, we know that we'll be at 17:00, that is, 5pm.
In order to know what day it will be, compute the remainder of $47^{74}+16$ modulo $7\cdot 24$; recall that $\varphi(7\cdot24)=6\cdot 8=48$.
By Fermat-Euler, you need to compute the remainder of $$ 47^{74-48}=47^{26}=47^2\cdot47^8\cdot47^{16} $$ Now (all congruences are modulo $168$) \begin{align} 47^2&=2209\equiv25\\ 47^3&\equiv25\cdot47=1175\equiv167\equiv-1\\ \end{align} So $47^8=(47^3)^2\cdot47^2\equiv25$ and $47^{16}=(47^3)^5\cdot47\equiv-47\equiv121$.
Therefore $$ 47^{74}\equiv25\cdot25\cdot121=75625\equiv25\pmod{168} $$ and it's the same as $25+16$ hours passed from “Wed 0:00”: one full day plus one hour (as seen before).