gcd(a, b)

Theorem.   If b = a*q + r, then gcd(a,b) = gcd(a,r)

proof

suppose d|a and d|b. Since r = b - aq, d|r.  Every divisor of a and b
is a divisor of a and r.

suppose d|r and d|a. Then since b = a*q + r, d|b.  Every divisor of a and r
is a divisor of a and b.

gcd(a,b) = gcd(a,r)

gcd(102, 119)

119 = 102(1) + 17

gcd(102, 119) = gcd(102, 17)

102 = 17(6) + 0

gcd(102, 17) = gcd(17, 0) = 17

Euclid's Algorithm for GCDs.

b = (b//a)*a + b%a