Problems with inverses in arithmetic in ring z

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The discussion focuses on calculating the expression 7*11 + 9*11^-1 in the group Z20, with a particular emphasis on finding the multiplicative inverse of 11 modulo 20. The initial attempt at a solution led to confusion regarding the correct answer, which a computer program indicated was 16. Participants emphasize the importance of determining 11^-1 mod 20 first, as this is crucial for solving the problem correctly. The calculation reveals that the multiplicative inverse of 11 mod 20 is 11 itself, which satisfies the equation 11m = 1 (mod 20). Understanding the difference between ring congruence arithmetic and regular congruence arithmetic is also raised as a point of confusion in the discussion.
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Homework Statement


Calculate 7*11 + 9*11^-1 in the group Z20


Homework Equations





The Attempt at a Solution


77+ (9*1/11) in group Z20
77 + 9/11
17 +11x= 20mod+9
My solution was 12, this makes 149 on both sides when you multiply the mod times 7.
I am doing independent study and the computer program I am using says the answer is 16.
What am I doing wrong? Btw, what is the difference between ring congruence arithmetic and regular congruence arithmetic? They seem pretty similar.

 
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I'm not really sure what you are doing. What is 20mod+9 supposed to mean? At some point you need to figure out what 11^(-1) mod 20 is. Why not do it first?
 
morrowcosom said:
bump

You can 'bump' a post if no one has given any response. I don't think that's the case here.
 
The first thing you absolutely need to do is to figure out what 11^-1 mod 20 is.
 
The first thing you absolutely need to do is to figure out what 11^-1 mod 20 is

original problem: calculate 7*11+9*11^-1 with mod 20

So in order to figure out the inverse of 11^-1 mod 20, I would do the typical
11x=mod 20+1, and from here would I go 17+99x=mod 20+1(I get 16 on this one, the supposedly correct solution), or some other route?
 
To find the multiplicative inverse of 11 mod 20, you want a number, m, such that 11m= 1 (mod 20) or 11m= 1+ 20n for some integer n.

That is the same as solving the diophantine equation 11m- 20n= 1.

11 divides into 20 once with remainder 9: 20- 11= 9.

9 divides into 11 once with remainder 2: 11- 9= 2.

2 divides into 9 four times with remainder 1: 9- 4(2)= 1.

Replace the "2" in that equation with 11- 9: 9- 4(11- 9)= 5(9)- 4(11)= 1.

Replace the "9" in that equation with 20- 11: 5(20- 11)- 4(11)= 5(20)- 9(11)= 1.
(Of course: 100- 99= 1.)

A solution to 11m- 20n= 1 is m= -9, n= -5. Thus, the multiplicative inverse of 11, mod 20, is -9= 20- 9= 11 (mod 20). That is, the multiplicative inverse of 11, mod 20, is 11 itself.

Check: 11(11)= 121= 6(20)+ 1= 1 (mod 20).
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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