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Home » Additionally, since both $a$ and $b$ are even, $d$ must be even (because any common divisor of two even numbers is even). - AMAZONAWS

Additionally, since both $a$ and $b$ are even, $d$ must be even (because any common divisor of two even numbers is even). - AMAZONAWS

📅 March 11, 2026 👤 scraface
Mar 11, 2026
Additionally, since both $a$ and $b$ are even, $d$ must be even (because any common divisor of two even numbers is even).

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