Solve Modular Math Problem: A,B→C

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Discussion Overview

The discussion revolves around solving a modular math problem involving variables A, B, and C, specifically finding C = f(A,B) under the condition that 1 + BC ≡ 0 (mod 2^A), where A is a positive integer and B is a positive odd integer. The scope includes mathematical reasoning and exploration of algorithms related to modular arithmetic.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant suggests using Euclid's Algorithm on B and 2^A to solve the problem.
  • Another participant questions the applicability of the GCD algorithm to the problem and requests an example with specific values (2^A = 64, B = 15).
  • A participant mentions that the extended Euclidean algorithm provides two numbers that relate to the GCD of two numbers.
  • There is a claim that since B is odd and 2^A is a power of two, they are coprime, which may influence the solution.
  • One participant proposes rewriting the equation in a different form to clarify the relationship between the variables.

Areas of Agreement / Disagreement

Participants express differing views on the applicability of the Euclidean algorithm and its relevance to the problem, indicating that the discussion remains unresolved with multiple competing perspectives.

Contextual Notes

There are limitations regarding the assumptions made about the relationship between B and powers of two, as well as the specific mathematical steps required to reach a solution.

Gokul43201
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Given, A>0 and odd B>0, find C = f(A,B), satisfying :

1 + BC == 0 (mod 2^A)
 
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Euclid's Algorithm on B and 2**A will do it. sorry, my six key is buggered so i can't do powers in pseudotex.
 
matt,

I'm not sure I follow. You are talking about the GCD algorithm, right ?

I don't see how this works. Can you show me how, on an example, say 2^A = 64, B = 15 ?
 
Recall that the extended Euclidean algorithm will give you two numbers, u and v, such that u * m + v * n = gcd(m, n)...
 
Yes, I've used it to solve linear diophantine eqns...but don't see how it can be used here.
 
B is odd, 2**A is a power of two, they are coprime.
 
Maybe it would help if I rewrite it as

(-u) * m + gcd(m, n) = v * n

?
 
Thanks,

I never cease to amaze myself !
 

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