I Was Vey Amaze At The Solution 1

2012 ! mod 2017 = ? \large 2012! \text{ mod } 2017 = \, ?


Notation: ! ! is the factorial notation. For example, 8 ! = 1 × 2 × 3 × × 8 8! = 1\times2\times3\times\cdots\times8 .


For more problems like this, try answering this set .


The answer is 84.

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2 solutions

Kushal Bose
Mar 3, 2017

From Wilson's Theorem 2016 ! 1 2016 ( m o d 2017 ) 2016! \equiv -1 \equiv 2016 \pmod{2017}

2015 ! 1 ( m o d 2017 ) 2012 ! × ( 2013 × 2014 × 2015 ) 1 ( m o d 2017 ) 2015! \equiv 1 \pmod{2017} \\ 2012! \times(2013 \times 2014 \times 2015) \equiv 1 \pmod{2017}

2013 × 2014 × 2015 ( 4 ) ( 3 ) ( 2 ) 24 ( m o d 2017 ) 2013 \times 2014 \times 2015 \equiv (-4)(-3)(-2) \equiv -24 \pmod{2017}

Let, 2012 ! r ( m o d 2017 ) 2012! \equiv r \pmod{2017}

So, combining these two results we get 2015 ! ( 24 ) ( r ) 1 ( m o d 2017 ) 2015! \equiv (-24)(r) \equiv 1 \pmod{2017}

So, ( 24 ) ( r ) 1 ( m o d 2017 ) (-24)(r) \equiv 1 \pmod{2017} .

So, it can be said that 24 2017 k + 1 24 | 2017k+1 for some positive integer k k .Then k k must be odd.Let's assume that k = 2 l + 1 k=2l+1 for some integer l l .Putting this 2017 k + 1 = 2017 ( 2 l + 1 ) + 1 = 4034 l + 2018 2 l + 2 ( m o d 24 ) 24 2 l + 2 2017k+1=2017(2l+1)+1=4034l+2018 \equiv 2l+2 \pmod{24} \implies 24 | 2l+2 .So, the smallest l l is 11 11 .Then k = 23 k=23

( 24 ) ( r ) = 2017.23 + 1 = > 24 ( r ) = 46392 = > ( r ) = 1933 = > r = 2017 1933 = 84 (-24)(r) = 2017.23+1 => 24(-r)=46392 => (-r)=1933 => r=2017-1933=84

I think 2011 ! ? ( m o d 2017 ) 2011! \equiv ? \pmod{2017} can be found this way

Kushal Bose - 4 years, 3 months ago

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I'll try sir. Thanks. :)

Christian Daang - 4 years, 3 months ago

Confirm sir, this way can find the 2011! mod 2017. :)

Christian Daang - 4 years, 3 months ago
Christian Daang
Mar 3, 2017

By Wilson's Theorem,

2016 ! m o d 2017 = 2016 ( ( 2016 2015 2014 2013 ) m o d 2017 ( 2012 ! m o d 2017 ) ) = 2016 ( ( ( 1 ) ( 2 ) ( 3 ) ( 4 ) ) × ( 2012 ! m o d 2017 ) ) = 2016 2012 ! m o d 2017 = 2016 4 ! = 84 . \displaystyle \begin{aligned} 2016! \mod 2017 & = 2016 \\ \implies \big( (2016*2015*2014*2013) \mod 2017 *(2012! \mod 2017) \big ) & = 2016 \\ \implies \big( ((-1)*(-2)*(-3)*(-4)) \times (2012! \mod 2017) \big ) & = 2016 \\ \implies 2012! \mod 2017 & = \cfrac{2016}{4!} = \boxed{84} . \end{aligned}

If 2012 is replaced by 2011 then if this method works ? Surely not

Then what will be 2011 ! ? ( m o d 2017 ) 2011! \equiv ? \pmod{2017}

Kushal Bose - 4 years, 3 months ago

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Ow, I see. XD I have no idea for that. XD

Can you share it?

Christian Daang - 4 years, 3 months ago

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give me some time

Kushal Bose - 4 years, 3 months ago

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@Kushal Bose Ok sir, thanks. :)

Christian Daang - 4 years, 3 months ago

I think it should be 790.

Kushagra Sahni - 4 years, 3 months ago

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yes, it is 790. :)

By Wilson's theorem, 2016 ! m o d 2017 2016 2016! \mod 2017 \equiv 2016

Hence, 2015 ! m o d 2017 1 = 2011 ! × 2012 × 2013 × 2014 × 2015 m o d 2017 1 2015! \mod 2017 \equiv 1 \\ = 2011! \times 2012 \times 2013 \times 2014 \times 2015 \mod 2017 \equiv 1

Let r = 2011 ! m o d 2017 r = 2011! \mod 2017

( r × ( 5 ) × ( 4 ) × ( 3 ) × ( 2 ) ) m o d 2017 1 120 r m o d 2017 1 120 r = 2017 k + 1 \implies \big( r \times (-5) \times (-4) \times (-3) \times (-2) \big) \mod 2017 \equiv 1 \\ 120r \mod 2017 \equiv 1 \\ \implies 120r = 2017k + 1

Since 2017k must end in 9, k must end in 7 so let k = 7 + 10 n k = 7 + 10n

120 r = 2017 ( 7 + 10 n ) + 1 120 r = 14120 + 20170 n 12 r = 1412 + 2017 n \implies 120r = 2017(7 + 10n) + 1 \\ 120r = 14120 + 20170n \\ 12r = 1412 + 2017n

Applying modulo 12,

0 m o d 12 ( 8 + n ) n + 8 = 12 v n = 4 , v = 1 0 \mod 12 \equiv (8 + n) \\ n + 8 = 12v \implies n = 4 \ , \ v = 1

r = 1412 + ( 2017 4 ) 12 = 790 \begin{aligned} \therefore r & = \cfrac{1412 + (2017*4)}{12} \\ & = \boxed{790} \end{aligned}

Christian Daang - 4 years, 3 months ago

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@Christian Daang There is typo in the solution.

It must be ' 2017 k 2017k ends in 9 9 ' and not ' k k ends in 9 9 '.

Ankit Kumar Jain - 4 years, 3 months ago

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@Ankit Kumar Jain oh yah. I will fix it. :)

Christian Daang - 4 years, 3 months ago

@Christian Daang Nicely done

Kushal Bose - 4 years, 3 months ago

Show how you solve this

Kushal Bose - 4 years, 3 months ago

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@Kushal Bose Do you know the answer? Like did you check it on wolfram alpha or any other place.

Kushagra Sahni - 4 years, 3 months ago

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@Kushagra Sahni No,I have not tried yet

Kushal Bose - 4 years, 3 months ago

The same method works for 2011 2011 as well with just a bit of modification.

2016 ! 2011 ! 2012 2013 2014 2015 2016 ( 5 ) ( 4 ) ( 3 ) ( 2 ) ( 1 ) 2016 ( m o d 2017 ) 2016! \equiv{2011!\cdot 2012 \cdot 2013 \cdot 2014 \cdot 2015 \cdot 2016}\equiv{(-5)\cdot (-4) \cdot (-3) \cdot (-2)\cdot (-1)}\equiv{2016}\pmod {2017} ------------------- By Wilson's Theorem

2011 ! ( 5 ) 84 ( m o d 2017 ) 2011!\cdot (-5)\equiv{84}\pmod{2017}

Now , we want to add 2017 2017 some number of times so as to make the unit digit of the RHS of equivalence 0 0 or 5 5 . We see that this can be done by adding 2017 2017 3 3 times.

Hence , 2011 ! ( 5 ) 84 + 6051 6135 ( m o d 2017 ) 2011! \cdot (-5)\equiv{84 + 6051}\equiv{6135}\pmod{2017}

2011 ! ( 1 ) 1227 790 ( m o d 2017 ) \Rightarrow 2011! \cdot (-1)\equiv{1227}\equiv{-790}\pmod{2017}

2011 ! 790 ( m o d 2017 ) \Rightarrow 2011! \equiv{790}\pmod{2017}

Ankit Kumar Jain - 4 years, 3 months ago

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@Kushal Bose @Christian Daang Hope this works too!!

Ankit Kumar Jain - 4 years, 3 months ago

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