$\gcd(105, 84) = 21$

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• $\gcd(105, 945) = 105$
• $\gcd(105, 3465) = 105$ (since $3465 \div 105 = 33$)
• $\gcd(105, 9009)$: $9009 \div 105 = 85.8$, $105 \cdot 85 = 8925$, remainder $84$
• Now $\gcd(21, 19305)$: $19305 \div 21 = 919.285...$, $21 \cdot 919 = 19299$, remainder $6$
• $\gcd(21, 6) = 3$
• So GCD so far is $3$ — but we must check if $3$ is always a divisor.

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