(A) Convince yourself that 50 is the inverse of 5 modulo 83, and that 12 is the inverse of 7 modulo 83. Find the inverse of 35 modulo 8

(A) Convince yourself that 50 is the inverse of 5 modulo 83, and that 12 is the inverse of 7 modulo 83. Find the inverse of 35 modulo 83. (Give your answer as a nonnegative integer that is less than 83.)

$$\small{\text{ The inverse of $35 \pmod{83} = \mathbf{ 19 }$, because $35 \cdot 19 \equiv 1 \pmod {83}$ }}$$

(b) Find the inverse of 49 modulo 83. (Give your answer as a nonnegative integer that is less than 83.)

$$\small{\text{ The inverse of $49 \pmod{83} = \mathbf{ 61 }$, because $49 \cdot 61 \equiv 1 \pmod {83}$ }}$$

Thanks Heureka, I want to look at this too :)

(A) Convince yourself that 50 is the inverse of 5 modulo 83,

If 50 is the inverse of 5 modulo 83 then 5*50 (mod83) must equal 1

5*50=250 = 83*3+1 = 1 (mod 83)        ok I'm convinced 

and that 12 is the inverse of 7 modulo 83.

12*7=84 = 83+1 = 1  (mod 83)             Yep - I am happy.

Find the inverse of 35 modulo 83. (Give your answer as a nonnegative integer that is less than 83.)

Well since 5 is  the inverse of 50 and 7 is the inverse of 12 THEN  5*7 is the inverse of 50*12 (mod83)

Check

35*(50*12)= 35*600 = 21000 = 253*83 + 1 = 1 mod83

So            The inverse of 35 mod 83 = 600

However

It seems to me that there must be an infinite number of inverses if 35 lets see if I can find some more.

600= 7*83+ 19

So any number of the form 83N+19 where N is an integer will be the inverse of 35 (mod83)

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I'm still having problems with part B   

It seems to me that there must be an infinite number of inverses of 35 ...

True, but the question specifies that the answer to be supplied must be a nonnegative integer that is less than 83.

. 

Thanks Alan I didn't notice that.

I wasn't trying to be smart.  I was just answering the question in an attempt to learn from it. I am new to modulo arithmetic.