Remainder of 34! divided by 37 using Wilson's Theorem

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SUMMARY

The discussion focuses on finding the remainder of 34! divided by 37 using Wilson's Theorem, which states that for a prime p, (p-1)! ≡ -1 (mod p). Participants clarify that 34! can be expressed as 34! * 35 ≡ 1 (mod 37), leading to the equation 35x ≡ 1 (mod 37). The solution involves finding the modular multiplicative inverse of 35 modulo 37, ultimately determining that the remainder is 19. The extended Euclidean algorithm is recommended for solving similar modular equations.

PREREQUISITES
  • Understanding of Wilson's Theorem
  • Familiarity with modular arithmetic
  • Knowledge of the extended Euclidean algorithm
  • Basic skills in solving linear congruences
NEXT STEPS
  • Study the application of Wilson's Theorem in combinatorial problems
  • Learn about modular multiplicative inverses and their computation
  • Explore the extended Euclidean algorithm in depth
  • Practice solving linear congruences using various methods
USEFUL FOR

Mathematics students, particularly those studying number theory, educators teaching modular arithmetic, and anyone interested in combinatorial mathematics and its applications.

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Homework Statement



Find the remainder when 34! is divided by 37.

Homework Equations



Wilson's Theorem

The Attempt at a Solution



I understand that (p-1)! = (-1)(mod p) and that (p-2)! = (1)(mod p). I don't understand how to apply this to (p-3)! though.
 
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you know that (p-2)! = 1 (mod p). So (p-3)!*(p-2) = 1 (mod p). In this situation, 34!*35 = 1 (mod 37). Call 34! 'x' and then solve 35x = 1 mod 37, which has a unique solution since gcd(35,37) = 1.
 
So do you do..

1 = 35x
1 = (-2)x
1 = (-2)(-18)

R = -18 + 37 = 19 ??
 
35*19 isn't 1 mod 37. Don't you mean 1=(-2)(-19)? It's easy enough to check your answers with a quick calculation.
 
You're right. Thanks for the help.
 
I'm trying to find 33! / 37 now.

I have gotten to (-3)x = 18 (mod 37)... but I can't figure out what x is.
 

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