Sunday, February 4, 2024

2024/010) Prove that $18 ! \equiv -1 \pmod {437} $

 We first factorize 437

$437  = 19 * 23$

Now let compute mod relative to19 and 23

As 19 is prime number as as per Wilson'sTheorem we have

  $18 ! \equiv -1 \pmod {19}\cdots(1) $

As 23 is prime number as as per Wilson'sTheorem we have

  $22 ! \equiv -1 \pmod {23} $

Now  

$22  \equiv -1 \pmod {23}\cdots(2) $

$21  \equiv -2 \pmod {23} \cdots(3)$

$20  \equiv -3 \pmod {23}\cdots(4) $

$19  \equiv -4 \pmod {23}\cdots(5) $

As $22! = 22 * 21 * 20 * 19 * 18! $

So $22 ! \equiv -1 \pmod {23} $

$\implies  22 * 21 * 20 *  19 *18 ! \equiv -1 \pmod {23} $

 $\implies  (-1) *(-2) * (-3) * (-4) *18 ! \equiv -1 \pmod {23} $

 $\implies  24 *18 ! \equiv -1 \pmod {23} $

 $\implies  24 *18 ! \equiv -1 \pmod {23} $

$\implies  1 *18 ! \equiv -1 \pmod {23} $

$\implies  18 ! \equiv -1 \pmod {23}\cdots(6) $

Using (1) and (3) we get 

  $18 ! \equiv -1 \pmod {437} $

Proved

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