Remainders

Number Theory Level 3

$101! \pmod{103}$ is equal to :

The answer is 1.

4 solutions

Adarsh Kumar
Aug 20, 2014

By Wilson's theorem we have that 102!=-1(mod103).But 102!(mod103)=101!(mod103) x 102(mod103)=-1.102(mod103)=-1.Now if and only if 101!(mod103)=1 is this possible.

Really very good jump for you that you've gotten to level 5 in NT so soon! Keep it up! Was it only Brilliant that helped you or..anything else? Adarsh Kumar

Krishna Ar - 6 years, 9 months ago

Actually yes.But I have read many books on NT like the 104 NT problems.

Adarsh Kumar - 6 years, 9 months ago

and thank you.

Adarsh Kumar - 6 years, 9 months ago

Which others apart from 104 NT? Like any ones...which helped you to write olympiad level proofs? Adarsh Kumar

Jayakumar Krishnan - 6 years, 9 months ago

@Jayakumar Krishnan – no sorry nothing else than this.104 NT problems didn't give me much info about modular arithmetic,I mainly learned it from Brilliant.

Adarsh Kumar - 6 years, 9 months ago

Indeed its the elegant one

math man - 6 years, 9 months ago

thank you.

Adarsh Kumar - 6 years, 9 months ago

Muhammad Adnan Al Ghafiqi
Sep 23, 2014

asem, ane pake wolfram :v

gapapa bang....... yg penting dpt poin

math man - 6 years, 8 months ago

Sophie Crane
Sep 27, 2014

I just guessed.

Jaiveer Shekhawat
Sep 15, 2014

According to Wilson's theorem: (p-2)! = 1(mod 103) Thus, (103-2)! = 1(mod103) =(101)! = 1(mod 103)

Actually Wilson's theorem states (p-1)! = -1(mod 103), on simplifying we get what u have said..... but,, this was a nice problem

Raushan Sharma - 6 years, 4 months ago

Remainders

The answer is 1.

4 solutions

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