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📅 May 3, 2026 👤 scraface

May 03, 2026

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$7 \cdot 9 \cdot 11 \cdot 13 = 9009$
$9 \cdot 11 \cdot 13 \cdot 15 = 19305$
Now compute GCD of these values:
$\gcd(105, 945) = 105$
$\gcd(105, 3465) = 105$ (since $3465 \div 105 = 33$)
$\gcd(105, 9009)$: $9009 \div 105 = 85.8$, $105 \cdot 85 = 8925$, rema...
$\gcd(105, 84) = 21$
Now $\gcd(21, 19305)$: $19305 \div 21 = 919.285...$, $21 \cdot 919 = 1...
$\gcd(21, 6) = 3$
So GCD so far is $3$ — but we must check if $3$ is always a divisor.

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