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Home » Play Pokémon Gold Like a Pro: Ultimate Guide to the Classic Game’s Hidden Gems - Appcentric

Play Pokémon Gold Like a Pro: Ultimate Guide to the Classic Game’s Hidden Gems - Appcentric

📅 February 27, 2026 👤 scraface

Feb 27, 2026

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📌 $\gcd(105, 84) = 21$

📌 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.

📌 But earlier logic: in any four consecutive odd integers, one is divisible by 3 — yes, because the step is 2, and modulo 3, the odd residues are 1 and 2, and over four terms: positions mod 3 cycle every 3 steps, but span 8 steps total? The sequence of odd integers mod 3: starts at $n \mod 3$, then $n+2$, $n+4 \equiv n+1$, $n+6 \equiv n$, so cycle: $n, n+2 \equiv n+2, n+1, n$. So values: $a, a+2, a+1, a \mod 3$. So set: $a, a+1, a+2$ — all residues mod 3. So one is divisible by 3.

📌 Thus, always divisible by 3.

📌 But not by 9, as shown.

📌 Is it divisible by any higher number? Try divisibility by 15? No — $1 \cdot 3 \cdot 5 \cdot 7 = 105$, divisible by 15, but $3 \cdot 5 \cdot 7 \cdot 9 = 945$, also divisible by 15. But $5 \cdot 7 \cdot 9 \cdot 11 = 3465$, divisible by 5 and 3, but 5 and 7, but 3465 ÷ 15 = 231 — yes. But is it always divisible by $5$? No — try $7,9,11,13$: $7\cdot9\cdot11\cdot13 = 9009$, ends with 9 — not divisible by 5. So not always divisible by 5.

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