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4.3. MULTIPLICATIVE INVERSES 61
(2) Compute gcd(12, 45) and find a solution to 12x ≡ gcd (mod 45).
From 45 ÷ 12,
From 12 ÷ 9,
From 9 ÷ 3,
{
q = 3, r = 9, and
0 − (3 × 1) = −3 :
{
q = 1, r = 3, and
1 − (1 × −3) = 4 :
{
q = 3, r = 0, and
−3 − (3 × 4) = −15 :
Setup:
q
r
c
q
r
c
q
r
c
q
r
c
45
0
45
0
45
0
45
0
12
1
3
12
1
3
12
1
3
12
1
9
-3
1
9
−3
1
9
−3
3
4
3
3
4
0
So gcd(12, 45) = 3 and x = 4 is a solution to 12x ≡ gcd (mod 45), which
we can check by seeing that (12 × 4)%45 = 3%45. (The entry in the
coefficient row under the final remainder of 0 is ignored, so we didn’t
bother to enter it.)
(3) Compute gcd(27, 50) and find a solution to 27x ≡ gcd (mod 50).
q
r
c
50
0
1
27
1
1
23
−1
5
4
2
1
3
−11
3
1
13
0
So gcd(27, 50) = 1. Since −11 ≡ 39 (mod 50), we have 27 × 39 ≡ 1
(mod 50).
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄
As we’ve seen, gcd(a, n) always divides (ab)%n, for any choice of integer b.
That is, (ab)%n is always a multiple of gcd(a, n). So for there to be a solution
x to ax%n = 1 we must have gcd(a, n) = 1. And if so, then the solution x is
found in the coefficient line.
These concepts are important enough that they have names. If two numbers
have a gcd of 1, they are said to be relatively prime. And when two numbers
a and n are relatively prime then they each have a multiplicative inverse
with respect to the other – there are solutions x and y to (ax)%n = 1 and
(ny)%a = 1.