# Examples

```Find the multiplicative inverse of 23 in Z26
First of all, 23 has an inverse in Z/26Z because gcd(26,23)=1. So use the Euclidean
algorithm to show that gcd is indeed 1. Going backward on the Euclidean algorithm, you
will able to write 1=26s+23t for some s and t. Thus 23t≡1 mod 26. So t is an inverse of
23 in Z/26Z. To find this t:
26 = (1)(23) + 3
23 = (7)(3) + 2
3 = (1)(2) + 1
2 = (2)(1) + 0
Hence the gcd is 1. Now going backward on the Euclidean algorithm, you get
1 = 3 - (1)(2)
Using the second line, you get
1 = 3 - (1)(23 - (7)(3))
1 = 3 - 23 + (7)(3)
1 = (-1)(23) + 8(3)
using the first line
1 = (-1)(23) + 8(26 - (1)(23))
1 = -1(23) + 8(26) + (-8)(23)
1 = -9(23) + 8(26)
Hence the inverse of 23 is -9. Moreover −9≡17 mod 26. So −9 or equivalently 17 is the
inverse of 23.
Alternative solution: In this case the numbers are small enough that the easiest
approach is trial and error: find a multiple of 23 that leaves a remainder of 1 when
divided by 26. But we can be a bit cleverer than that: notice that 23 is −3 in Z26. Now
(−3)(9)=−27=−1 in Z26, so (−3)(−9)=1 in Z26: −9 is the multiplicative inverse of −3.
Finally, −9=26−9=17 in Z26, and we’ve got the answer: the multiplicative inverse of 23
in Z26 is 17. As a check, 17⋅23=391=1+15⋅26.
```