If you've done the aforementioned exercise 6, then you already know
[tex]34^{-1}\equiv55\pmod{89}[/tex]
[tex]144^{-1}\equiv89\pmod{233}[/tex]
[tex]200^{-1}\equiv996\pmod{1001}[/tex]
a. In case you don't already have the above, you can use the Euclidean algorithm:
89 = 2*34 + 21
34 = 1*21 + 13
21 = 1*13 + 8
13 = 1*8 + 5
8 = 1*5 + 3
5 = 1*3 + 2
3 = 1*2 + 1
Then we can write 1 as a linear combination of 89 and 34, namely
[tex]1=13\cdot89+(-34)\cdot34[/tex]
Taken mod 89, we find that
[tex]34^{-1}\equiv-34\equiv55\pmod{89}[/tex]
Then
[tex]55\cdot34x\equiv x\pmod{89}[/tex]
[tex]55\cdot77\equiv4235\equiv52\pmod{89}[/tex]
so that [tex]x=52+89n[/tex] for any integer [tex]n[/tex] is a solution to the congruence.
b. [tex]x=123+233n[/tex]
c. [tex]x=936+1001n[/tex]
