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. - Abu Waleed Tea