But what about $3 \cdot 5 = 15$? We saw $120$ in previous problem, but here product $1 \cdot 3 \cdot 5 \cdot 7 = 105$, divisible by $15$, but $3 \cdot 5 = 15$ — yes. But $3 \cdot 3 = 9$: $105$ not divisible by $9$. So $9$ does not always divide $P$.

Wed, 06 May 2026 07:05:26 +0000

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