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Find 4^{-1}*9^{-1} (mod 35), as a residue modulo 35. (Give an answer between 0 and 34, inclusive.)

 May 16, 2022
 #1
avatar+1 
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\(4^{-1} = \frac{1}{4}\\ 9^{-1} = \frac{1}{9}\\ \frac{1}{4} \cdot \frac{1}{9} = \frac{1}{36} \)

Therefore, the answer is 1/36 (mod 35).

 May 16, 2022
 #2
avatar+117494 
+1

A reasonable guess Pureant but unfortunately it is not correct.   

* I will give you thumbs up for your attempt ;)

 

Think about normal numbers

The multiplication inverse of  4 is 1/4  .

This is becasue  4 * 1/4 = 1

It is the same with modular arithemetic.

The number that is the inverse of 4 is the one that can be multipled by 4 to get 1.

So the inverse of 4 mod 35   is  B such that    4B=1  mod35

 

 

4*9=36 which is 1 mod 35

therefore the inverse of 4 is 9 and the inverse of 9 is 4  mod35

 

so the question becomes  9*4 = 36  = 1 mod35

 May 16, 2022
edited by Melody  May 16, 2022
edited by Melody  May 16, 2022

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