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There is a past question that wasn't solved. It's in the Advanced Maths category.  I'm still curious to see the final answer, if anyone can figure how to do it.

 

Here's that question again: http://web2.0calc.com/questions/wilson-s-theorem

 

Badinage  May 11, 2015

Best Answer 

 #4
avatar+520 
+5

Thanks Mathcad. I get the idea now.

Badinage  May 12, 2015
 #1
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One last attempt at posting this.The software on this system is hopeless.

We can write 36x27!+25 as (36x31!+(31x30x29x28)x25)/31x30x29x28 and can now factor out 31 from the numerator.So the expression is divisible by 31 and we can write

36x27!+25=31n for some n.

then,to use Wilson,we can write 36x27!+25=31!n/30!   and since 31!=30mod 31 ,rearrange to get

n=30!(36x27!+25)/30modulo31  which is divisible by 31.

 

hope this helps.

Mathcad  May 11, 2015
 #2
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Thanks Mathcad, that looks interesting :)

Melody  May 11, 2015
 #3
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+5

Mathcad, I have transfered your answer over onto the original thread.

 

Badinage has asked the question exactly as I have requested but it is better if the answers are all together on the original thread. 

 

Thank you   

Melody  May 11, 2015
 #4
avatar+520 
+5
Best Answer

Thanks Mathcad. I get the idea now.

Badinage  May 12, 2015

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