Solve each of these congruences using the modular in-verses 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)

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 + 2134 = 1*21 + 1321 = 1*13 + 813 = 1*8 + 58 = 1*5 + 35 = 1*3 + 23 = 1*2 + 1Then 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]