Find 4^{-1}*9^{-1} (mod 35), as a residue modulo 35. (Give an answer between 0 and 34, inclusive.)
+1 \(4^{-1} = \frac{1}{4}\\ 9^{-1} = \frac{1}{9}\\ \frac{1}{4} \cdot \frac{1}{9} = \frac{1}{36} \)
Therefore, the answer is 1/36 (mod 35).
+118608 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
