1. ## inverse modulo

Find the inverse modulo m of each integer n below:

a) n=8 and m=35

b) n=51 and m=99

2. The answer to a is 22.

3. Originally Posted by icemanfan
The answer to a is 22.

could u give more details to that??

4. Hello,
Originally Posted by rmpatel5
Find the inverse modulo m of each integer n below:

a) n=8 and m=35
Find p such that $8p=1 \bmod 35$

Euclidian division :

$35=8 \times 4+3 \implies 3=35-8 \times 4$

$8=3 \times 3-1 \implies 1=3 \times 3-8=3 \times (35-8 \times 4)-8=3 \times 35-12 \times 8-8=3 \times 35-13 \times 8$

Therefore $(-13) \times 8=1-3 \times 35 =1 \bmod 35$

So the inverse is $-13 \bmod 35$, that is to say $22 \bmod 35$

b) n=51 and m=99
Try to do it...

5. Originally Posted by rmpatel5
b) n=51 and m=99
Okay, there was a trap here...

You're looking for q such that $51q=1 \bmod 99$
But this is not possible since 51 and 99 are not coprime.

$51q=1 \bmod 99 \Longleftrightarrow \exists k \in \mathbb{Z} ~,~ 51q-1=99k \implies 1=51q-99k=3(17q-33k)$

So 3 must divide 1, which is not possible.