Solving Modular Arithmetic: x\equiv2 (mod km)

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Discussion Overview

The discussion revolves around the properties of modular arithmetic, specifically the conditions under which multiple modular congruences can be combined into a single congruence. Participants explore whether the congruences x≡2 (mod k) and x≡2 (mod m) imply x≡2 (mod km) and the necessary conditions for this to hold.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant suggests that if x≡2 (mod k) and x≡2 (mod m), then it follows that x≡2 (mod km), questioning the validity of this implication.
  • Another participant counters this by providing a specific example where k=4, m=8, and x=10, demonstrating that while x satisfies the first two congruences, it does not satisfy the combined congruence x≡2 (mod 32), indicating a need for additional conditions.
  • A third participant proposes that if k and m are coprime (i.e., (k,m) = 1), then the earlier congruences can be combined to yield x≡a (mod km), outlining a reasoning process to support this claim.
  • A subsequent reply expresses agreement with the third participant's reasoning, suggesting that the argument is valid.

Areas of Agreement / Disagreement

Participants express disagreement regarding the initial assumption that the modular congruences can be combined without additional conditions. There is a proposal for a specific condition (coprimality) that may allow for the combination, but no consensus is reached on the broader implications or the necessity of this condition.

Contextual Notes

The discussion highlights the importance of conditions such as coprimality in modular arithmetic, but does not resolve the implications of these conditions or provide a comprehensive framework for all cases.

Who May Find This Useful

Individuals interested in modular arithmetic, number theory, or those studying congruences in mathematics may find this discussion relevant.

theIBnerd
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i might be making it up, but i am confused.

can we say:

[tex]x\equiv[/tex]2 (mod k)
[tex]x\equiv[/tex]2 (mod m)
hence
[tex]x\equiv[/tex]2 (mod km) by km i mean k multiplied by m.

if not, what is the result? or can it be found?

thank you in advance.
 
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No.

k=4
m=8
x=10
x=2(mod 4) and x=2(mod 8) but x=10(mod 32)

In general, if you have something mod m and something mod k, and want to discuss what happens mod mk, then you need a condition on m and k being coprime, or something similar.
 
thank you for your answer.

i think i found sth:

say (k,m) = 1

x=a (mod k)
x=a (mod m)

x=kt+a and x=my+a
kt=my
t=mb
y=kb

then x=kmb+a
x-a=kmb
x-a=0 (mod km)
x=a (mod km)

it is valid, isn't it? any counterexamples?
 
That looks pretty good to me
 
:) then my problem is solved. now i should get back to work.
 
Yeah, I hate when that happens!:-p
 

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