Congruence problem finding smallest modulus.

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Homework Help Overview

The problem involves solving the congruence equation 20x ≡ 16 (mod 92) and finding the smallest possible modulus. The discussion centers around the steps taken to manipulate the equation and the calculations involved in determining the greatest common divisor (gcd).

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the method of expressing the congruence as a linear Diophantine equation and explore the calculation of the gcd. There are attempts to clarify the steps taken and the reasoning behind them, with some questioning the validity of the approach used to arrive at the final modulus.

Discussion Status

The discussion is ongoing, with participants providing guidance on explaining the steps taken in the calculations. There is a recognition of the need for clarity in the reasoning process, and some participants express confusion about the results obtained, particularly regarding the final equivalence statements.

Contextual Notes

There is mention of a potential misunderstanding stemming from a previously learned method that may not apply correctly to this problem. Participants are navigating through the implications of using different coefficients in their calculations.

sg001
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Homework Statement



Solve the congruence 20x \equiv16 (mod 92)
Give your answer to the smallest possible modulus.



Homework Equations





The Attempt at a Solution



so 20x -92y=16

Hence,

92=20(4) +12...

gcd = 4

working back,

I have -16(-9)= z (mod 92)

so z = 52

and if I divide through by my gcd =4

36\equiv 13 (mod 23)

but the answer says -36\equiv 10 (mod 23)

??

Im lost.
 
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Try explaining what you're doing, and do it without so much shorthand.
 
Hurkyl said:
Try explaining what you're doing, and do it without so much shorthand.

Ok, I was being lazy

Here it is step by step.

20x\equiv16(mod92)

so..

20x - 92y =16

20 goes into 92, 4 times with 12 remainder


Hence,

92 = 20(4) + 12

12 = 92 - 20(4)

20= 12(1) + 8

8= 20 - 12(1)

12= 8(1) + 4

4= 12 - 8(1)

8= 4(2) + 0 \Rightarrow gcd = 4


working backwards

4= 12 - (20-12)

= (92-20(4)) - (20-12)

= (92-20(4)) - (20-(92-20(4)))

= 92(2) -9(20)

so..

-16(-9) = (?) (mod92)

144 = 52 (mod92)


Now dividing through by gcd...

36\equiv 13 (mod 23)


But the answer says

-36\equiv10(mod23)

I don't understand?
 
sg001 said:
20x - 92y =16
...
4 = 92(2) -9(20)
I was more interested in having you explain what you're doing and why you're doing it, than detailing the individual steps in actually doing it.

This far, I can determine you intended to find gcd(92,20), along with coefficients u,v satisfying
gcd(92,20) = 92u + 20v​
I won't ask why you went off into this calculation, since it's frequently useful enough that it's probably fine to do habitually and without comment.However, I can't tell what you were intending to do with the rest of your work:

so..

-16(-9) = (?) (mod92)

144 = 52 (mod92)Now dividing through by gcd...

36\equiv 13 (mod 23)But the answer says

-36\equiv10(mod23)

I don't understand?
 
Last edited:
Hurkyl said:
I was more interested in having you explain what you're doing and why you're doing it, than detailing the individual steps in actually doing it.

This far, I can determine you intended to find gcd(92,20), along with coefficients u,v satisfying
gcd(92,20) = 92u + 20v​
I won't ask why you went off into this calculation, since it's frequently useful enough that it's probably fine to do habitually and without comment.


However, I can't tell what you were intending to do with the rest of your work:

I figured it out I just was shown a bad way with a simpler problem and have been using that approach since.(I was only using one coefficient instead of the two).

Thanks for the help thoe.
 

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