Find the largest number that divides both inputs evenly.
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First integer
Second integer
GCF(a, b) = GCF(b, a mod b) until b = 0, then GCF = aRepeatedly replace a with b and b with remainder until remainder is 0 — last non-zero remainder is GCF.
Efficient even for large numbers
GCF is largest common divisor; LCM is smallest common multiple. Relationship: GCF(a,b) × LCM(a,b) = a × b.
No — GCF is always ≤ the smaller of the two numbers.