Math Problem Statement

Find the multiplicative inverse of (97!) modulo (103).

Solution

To find the multiplicative inverse of 97!97! modulo 103103, we need to solve for xx in the equation:

97!x1(mod103)97! \cdot x \equiv 1 \pmod{103}

This means xx is the multiplicative inverse of 97!97! modulo 103103. We can find it using Fermat's Little Theorem, which is applicable because 103103 is a prime number.

Step-by-step process:

  1. Fermat's Little Theorem: For any integer aa and a prime pp, Fermat's Little Theorem states that: ap11(modp)a^{p-1} \equiv 1 \pmod{p} which implies: ap2a1(modp)a^{p-2} \equiv a^{-1} \pmod{p} Thus, the inverse of 97!97! modulo 103103 is: (97!)101(mod103)(97!)^{101} \pmod{103}

  2. Simplifying using Wilson's Theorem: Wilson's theorem states that for a prime pp, (p1)!1(modp)(p-1)! \equiv -1 \pmod{p} Therefore, for p=103p = 103, 102!1(mod103)102! \equiv -1 \pmod{103} Since 102!=102101100999897!102! = 102 \cdot 101 \cdot 100 \cdot 99 \cdot 98 \cdot 97!, we can break it down: 102!=102101100999897!102! = 102 \cdot 101 \cdot 100 \cdot 99 \cdot 98 \cdot 97! We can simplify modulo 103103:

    • 1021(mod103)102 \equiv -1 \pmod{103}
    • 1012(mod103)101 \equiv -2 \pmod{103}
    • 1003(mod103)100 \equiv -3 \pmod{103}
    • 994(mod103)99 \equiv -4 \pmod{103}
    • 985(mod103)98 \equiv -5 \pmod{103}

    So, 102!(1)(2)(3)(4)(5)97!(1)512097!12097!(mod103)102! \equiv (-1)(-2)(-3)(-4)(-5) \cdot 97! \equiv (-1)^5 \cdot 120 \cdot 97! \equiv -120 \cdot 97! \pmod{103} We know that 102!1(mod103)102! \equiv -1 \pmod{103}, thus: 12097!1(mod103)-120 \cdot 97! \equiv -1 \pmod{103} Multiplying both sides by 1-1: 12097!1(mod103)120 \cdot 97! \equiv 1 \pmod{103}

  3. Finding the inverse of 120 modulo 103: To isolate 97!97!, we need the inverse of 120(mod103)120 \pmod{103}. Using the extended Euclidean algorithm:

    • 120=1103+17120 = 1 \cdot 103 + 17
    • 103=617+1103 = 6 \cdot 17 + 1
    • 17=171+017 = 17 \cdot 1 + 0

    Reversing the steps, we get: 1=103617=1036(1201103)=710361201 = 103 - 6 \cdot 17 = 103 - 6 \cdot (120 - 1 \cdot 103) = 7 \cdot 103 - 6 \cdot 120 Thus, 697(mod103)-6 \equiv 97 \pmod{103}, meaning the inverse of 120120 modulo 103103 is 9797.

  4. Conclusion: The multiplicative inverse of 97!97! modulo 103103 is: 9797

Would you like more details or have any questions?

Here are 5 related questions:

  1. How do you use Fermat's Little Theorem to find inverses modulo primes?
  2. What are the applications of Wilson's Theorem in number theory?
  3. How does the extended Euclidean algorithm help in finding modular inverses?
  4. Why is it important to compute factorials modulo primes efficiently?
  5. How would this method change for a non-prime modulus?

Tip: Always check if a modulus is prime before applying Fermat's Little Theorem or Wilson's Theorem.

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Math Problem Analysis

Mathematical Concepts

Number Theory
Modular Arithmetic
Factorials

Formulas

a^(p-2) ≡ a^(-1) (mod p) from Fermat's Little Theorem
(p-1)! ≡ -1 (mod p) from Wilson's Theorem
Extended Euclidean Algorithm for finding modular inverses

Theorems

Fermat's Little Theorem
Wilson's Theorem
Extended Euclidean Algorithm

Suitable Grade Level

Grades 11-12 and College-level