In need of major assistance with Congurences

[iceman2013]

I am trying to wrap my mind around some practice problems but i am not understanding. Anyone know how to solve something like this?

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Solve each of these congruences using the modular inverses
found in parts (b), (c), and (d) of Exercise 6.
a)34x ≡ 77 (mod 89)

b)144x ≡ 4 (mod 233)

c)200x ≡ 13 (mod 1001)

Here is my solutions from exercise 6

b) inverse of 34 modulo 89 is -34
c) inverse of 144 modulo 233 is 89
d)inverse of 200 modulo 1001 is -5.
Any help would be much appreciated!!!!

 

[Deleted member 4993]

I am trying to wrap my mind around some practice problems but i am not understanding. Anyone know how to solve something like this?
Solve each of these congruences using the modular inverses
found in parts (b), (c), and (d) of Exercise 6.
a)34x ≡ 77 (mod 89)

b)144x ≡ 4 (mod 233)

c)200x ≡ 13 (mod 1001)

Here is my solutions from exercise 6

b) inverse of 34 modulo 89 is -34
c) inverse of 144 modulo 233 is 89
d)inverse of 200 modulo 1001 is -5.
Any help would be much appreciated!!!!

Do you know the "meaning of"

A ≡ B (mod C)

If yes - tell us.

If no - search textbook, google - find out and tell us.

 

[HallsofIvy]

I am trying to wrap my mind around some practice problems but i am not understanding. Anyone know how to solve something like this?
Solve each of these congruences using the modular inverses
found in parts (b), (c), and (d) of Exercise 6.
a)34x ≡ 77 (mod 89)

b)144x ≡ 4 (mod 233)

c)200x ≡ 13 (mod 1001)

Here is my solutions from exercise 6

b) inverse of 34 modulo 89 is -34
c) inverse of 144 modulo 233 is 89
d)inverse of 200 modulo 1001 is -5.
Any help would be much appreciated!!!!

There are different ways of thinking of "modulo"- you can think "1 mod 89" as the set of numbers {1, -88, 90, -177, ...} or as just the numbers from 0 to 83, in this case, 1.

Also, we have both addition and multiplication defined for modulo operations. Actually, I, at first, thought that you were confusing "additive inverse" with "multiplicative inverse" because, of course, 34+ (-34)= 0 which is 0 modulo anything. But then I realised that -34 is also the multiplicative inverse of 34 modulo 89: (-34)(34)= -1156= -13(89)+ 1 , so (-34)(34)= 1(mod 89). It is odd that it happens to also be the additive inverse! Although I would say the multiplicative invers is either "the set of all numbers that differ from -34 by a multiple of 89" or the single number 89- 34= 55, depending upon which of the conventions above you use. It is true that 34(55)= 1870= 21(89)+ 1.

So to solve 34x= 77 (mod 89), multiply both sides by -34 (or, better, 55). That will give x= 55(77) (mod 89).
55(77)= 4235= 47(89)+ 52.