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I think 2 0 1 1 ! ≡ ? ( m o d 2 0 1 7 ) can be found this way
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I'll try sir. Thanks. :)
Confirm sir, this way can find the 2011! mod 2017. :)
By Wilson's Theorem,
2 0 1 6 ! m o d 2 0 1 7 ⟹ ( ( 2 0 1 6 ∗ 2 0 1 5 ∗ 2 0 1 4 ∗ 2 0 1 3 ) m o d 2 0 1 7 ∗ ( 2 0 1 2 ! m o d 2 0 1 7 ) ) ⟹ ( ( ( − 1 ) ∗ ( − 2 ) ∗ ( − 3 ) ∗ ( − 4 ) ) × ( 2 0 1 2 ! m o d 2 0 1 7 ) ) ⟹ 2 0 1 2 ! m o d 2 0 1 7 = 2 0 1 6 = 2 0 1 6 = 2 0 1 6 = 4 ! 2 0 1 6 = 8 4 .
If 2012 is replaced by 2011 then if this method works ? Surely not
Then what will be 2 0 1 1 ! ≡ ? ( m o d 2 0 1 7 )
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give me some time
I think it should be 790.
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yes, it is 790. :)
By Wilson's theorem, 2 0 1 6 ! m o d 2 0 1 7 ≡ 2 0 1 6
Hence, 2 0 1 5 ! m o d 2 0 1 7 ≡ 1 = 2 0 1 1 ! × 2 0 1 2 × 2 0 1 3 × 2 0 1 4 × 2 0 1 5 m o d 2 0 1 7 ≡ 1
Let r = 2 0 1 1 ! m o d 2 0 1 7
⟹ ( r × ( − 5 ) × ( − 4 ) × ( − 3 ) × ( − 2 ) ) m o d 2 0 1 7 ≡ 1 1 2 0 r m o d 2 0 1 7 ≡ 1 ⟹ 1 2 0 r = 2 0 1 7 k + 1
Since 2017k must end in 9, k must end in 7 so let k = 7 + 1 0 n
⟹ 1 2 0 r = 2 0 1 7 ( 7 + 1 0 n ) + 1 1 2 0 r = 1 4 1 2 0 + 2 0 1 7 0 n 1 2 r = 1 4 1 2 + 2 0 1 7 n
Applying modulo 12,
0 m o d 1 2 ≡ ( 8 + n ) n + 8 = 1 2 v ⟹ n = 4 , v = 1
∴ r = 1 2 1 4 1 2 + ( 2 0 1 7 ∗ 4 ) = 7 9 0
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@Christian Daang – There is typo in the solution.
It must be ' 2 0 1 7 k ends in 9 ' and not ' k ends in 9 '.
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@Ankit Kumar Jain – oh yah. I will fix it. :)
@Christian Daang – Nicely done
Show how you solve this
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@Kushal Bose – Do you know the answer? Like did you check it on wolfram alpha or any other place.
The same method works for 2 0 1 1 as well with just a bit of modification.
2 0 1 6 ! ≡ 2 0 1 1 ! ⋅ 2 0 1 2 ⋅ 2 0 1 3 ⋅ 2 0 1 4 ⋅ 2 0 1 5 ⋅ 2 0 1 6 ≡ ( − 5 ) ⋅ ( − 4 ) ⋅ ( − 3 ) ⋅ ( − 2 ) ⋅ ( − 1 ) ≡ 2 0 1 6 ( m o d 2 0 1 7 ) ------------------- By Wilson's Theorem
2 0 1 1 ! ⋅ ( − 5 ) ≡ 8 4 ( m o d 2 0 1 7 )
Now , we want to add 2 0 1 7 some number of times so as to make the unit digit of the RHS of equivalence 0 or 5 . We see that this can be done by adding 2 0 1 7 3 times.
Hence , 2 0 1 1 ! ⋅ ( − 5 ) ≡ 8 4 + 6 0 5 1 ≡ 6 1 3 5 ( m o d 2 0 1 7 )
⇒ 2 0 1 1 ! ⋅ ( − 1 ) ≡ 1 2 2 7 ≡ − 7 9 0 ( m o d 2 0 1 7 )
⇒ 2 0 1 1 ! ≡ 7 9 0 ( m o d 2 0 1 7 )
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@Kushal Bose @Christian Daang Hope this works too!!
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From Wilson's Theorem 2 0 1 6 ! ≡ − 1 ≡ 2 0 1 6 ( m o d 2 0 1 7 )
2 0 1 5 ! ≡ 1 ( m o d 2 0 1 7 ) 2 0 1 2 ! × ( 2 0 1 3 × 2 0 1 4 × 2 0 1 5 ) ≡ 1 ( m o d 2 0 1 7 )
2 0 1 3 × 2 0 1 4 × 2 0 1 5 ≡ ( − 4 ) ( − 3 ) ( − 2 ) ≡ − 2 4 ( m o d 2 0 1 7 )
Let, 2 0 1 2 ! ≡ r ( m o d 2 0 1 7 )
So, combining these two results we get 2 0 1 5 ! ≡ ( − 2 4 ) ( r ) ≡ 1 ( m o d 2 0 1 7 )
So, ( − 2 4 ) ( r ) ≡ 1 ( m o d 2 0 1 7 ) .
So, it can be said that 2 4 ∣ 2 0 1 7 k + 1 for some positive integer k .Then k must be odd.Let's assume that k = 2 l + 1 for some integer l .Putting this 2 0 1 7 k + 1 = 2 0 1 7 ( 2 l + 1 ) + 1 = 4 0 3 4 l + 2 0 1 8 ≡ 2 l + 2 ( m o d 2 4 ) ⟹ 2 4 ∣ 2 l + 2 .So, the smallest l is 1 1 .Then k = 2 3
( − 2 4 ) ( r ) = 2 0 1 7 . 2 3 + 1 = > 2 4 ( − r ) = 4 6 3 9 2 = > ( − r ) = 1 9 3 3 = > r = 2 0 1 7 − 1 9 3 3 = 8 4