A huge remainder

Number Theory Level 3

Find the remainder when $(270!)^2$ is divided by 541. - Do not use WolframAlpha. - You may want to use the fact that $541$ is prime.

The answer is 540.

2 solutions

Adarsh Kumar
Oct 16, 2014

From Wilson's theorem we have $540!\equiv-1\pmod{541}.$ Now.let us assume that $270!\equiv a\pmod{541}.$ Multiplying both sides by $540*539*538*.........271$ ,we get that $540*539*538*.........271*270!\equiv-1\pmod{541}.$ $\Longrightarrow [540\pmod{541}*539\pmod{541}*538\pmod{541}*......271\pmod{541}*\\270!\pmod{541}]\pmod{541}=(-1)$ $\Longrightarrow[(-1)*(-2)*(-3)*.....*(-270)\\*(270!)]\pmod{541}=-1\\ \Longrightarrow (270!*270!)\pmod{541}=(-1)\\ \Longrightarrow (270!)^{2}\equiv(-1)\pmod{541}\equiv540\pmod{541}.$ And we are done.

@Sean Ty

Adarsh Kumar - 6 years, 8 months ago

A corollary of Wilson's Theorem states that iff a prime $p$ is written in the form $4k+1$ , then $(2k!)^{2} \equiv -1 \pmod p$ .

Sean Ty - 6 years, 8 months ago

nice bit of info.

Adarsh Kumar - 6 years, 7 months ago

@Adarsh Kumar – This problem is pretty straightforward to do with Wilson's Theorem. Mostly if you know the corollary (The one I just gave.) If so, then this problem can be solved mentally!

Sean Ty - 6 years, 7 months ago

@Sean Ty – Well,it is a world of the people who have the largest knowledge bank!!

Adarsh Kumar - 6 years, 7 months ago

@Adarsh Kumar – Ya Same with me .....Schools ARE again starting

sakshi rathore - 5 years, 11 months ago

Really elegant. Shows your level of clarity :)

Krishna Ar - 6 years, 8 months ago

thank you!!

Adarsh Kumar - 6 years, 7 months ago

So cloooose :(

Ghany M - 5 years, 7 months ago

Xiaoying Qin
Oct 24, 2015

You REALLY shouldn't say Don't use WolframAlpha- I bet 1/2 of the solvers did.

A huge remainder

The answer is 540.

2 solutions

0 pending reports