Students

Remainder Using Wilson's Theorem (Modular Arithmetic)

From a recent test in Level II session:

Finding the remainder when 1991! is divided by 2003.

2003 is a prime number and by Wilson's Theorem:

Step 1: (p−1)! ≅ (p−1) (mod p) where p is a prime number
    2002! ≅ 2002 (mod 2003)
    2001! ≅ 1 (mod 2003)

Step 2: 1991!.2000.2001 ≅ 1 (mod 2003)
    => 1991!.(−3).(−2) ≅ 1 (mod 2003)
    => 1991! ≅ 1/6 (mod 2003)
    => 1991! ≅ (2004/6) (mod 2003)
    => 1991! ≅ 334 (mod 2003)

The answer is, 334

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