I tried to solve it by Wilson theorem and I got 41 but answer is 1927.


closed as off-topic by Shaun, Claude Leibovici, JMP, user26857, user91500 Aug 28 '17 at 9:53

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    $\begingroup$ Did you notice that if $50! = q \cdot 47^2 + r$, then $r$ has to be divisible by $47$? $\endgroup$ – peter a g May 18 '17 at 19:15
  • $\begingroup$ Didn't get you can you explain in brief.... $\endgroup$ – Sumit Jha May 18 '17 at 19:16
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    $\begingroup$ Please share your full calculation, so we can see where you might have gone wrong $\endgroup$ – Pieter21 May 18 '17 at 19:17
  • $\begingroup$ Well - ignoring the answer below - you want to calculate my $r$. The LHS is divisible by $47$, as is $q\cdot 47^2$, so $r$ must also be divisible by $47$ . Your answer was not... $\endgroup$ – peter a g May 18 '17 at 19:20
  • $\begingroup$ 46!=47k-1 so 50!=(47k-1)*47*48*49*50=>50!/47=(47k-1)*48*49*50=>rem(50!/47^2)=(-1)*1*2*3=-6=41 where I am wrong??? @Peter a g $\endgroup$ – Sumit Jha May 18 '17 at 19:22

You've probably found $$\frac{50!}{47}\equiv 41\pmod{47}$$ but you were looking for $$50!\equiv 41\cdot 47\equiv 1927\pmod{47^2}$$


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