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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.

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    $\begingroup$ I guess there's no daylight saving time in wherever that book was written. Even leap seconds ruin everything here, so those have to be ignored, too. $\endgroup$ – JiK Sep 26 '14 at 9:59
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    $\begingroup$ In $47^{74}$ hours we will be in the grave. Whoever set that problem should have made at least a mild effort to keep the premise from being completely ridiculous. $\endgroup$ – KCd Sep 26 '14 at 12:19
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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).

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  • $\begingroup$ Dear @egreg, so you means two days later? 48 hours? $\endgroup$ – Mouna Mokhiab Sep 26 '14 at 9:58
  • $\begingroup$ I think Thursday? would you please describe it? $\endgroup$ – Mouna Mokhiab Sep 26 '14 at 10:16
  • $\begingroup$ @MounaMokhiab Using that remainder is the same as throwing out whole weeks, which are unimportant. The quotient of the remainder by $24$ will give you the number of days to add. $\endgroup$ – egreg Sep 26 '14 at 10:29
  • $\begingroup$ I know it. but how calculate the remainder by 24? would u please learn me ? $\endgroup$ – Mouna Mokhiab Sep 26 '14 at 10:54
  • $\begingroup$ Dear @Egreg. I confused. please let me know how many days should be add? $\endgroup$ – Mouna Mokhiab Sep 26 '14 at 11:54
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Since $47^{74}$ hours is more than a googol ($10^{100}$) years, by all estimates, the sun will be long gone, and we won't have days anymore.

But if you ignore all that, you should try to reduce $47^{74} \pmod{168}$ (the number of hours in a week).

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