But is that multiple divisible by $9$? Not always — e.g., $n = 3$: 3 (div by 3 but not 9), $n = 9$: 9 (div by 9), $n = 15$: 15 (not div by 9). So only guaranteed $3^1$.

February 22, 2026

But is that multiple divisible by $9$? Not always — e.g., $n = 3$: 3 (div by 3 but not 9), $n = 9$: 9 (div by 9), $n = 15$: 15 (not div by 9). So only guaranteed $3^1$.

Related Articles

Now check if $3^3 = 27$ always divides $P$:
Among four consecutive odd integers, one is divisible by 3. If one is divisible by 9, we get $3^2$, and if another is divisible by 3 (which happens in most cases), but since the step is 2, two of them can be divisible by 3 only if spaced by 6 — but only one in every three odd numbers is divisible by 3. So only one multiple of 3. So $3^1$ is guaranteed, $3^2$ is possible, but not guaranteed.
Wait: recurring pattern — in any four consecutive odd numbers, the residues mod 3 cycle. Since step is 2, and $2 \cdot 3 \equiv 0$, but over four steps, one must hit 0 mod 3. So exactly one multiple of 3 per such quartet.
Similarly, for $5$, not always.
Now check if divisible by $9$: yes, as shown above.
Check for $3 \cdot 9 = 9$: yes.

Trending Articles