@Labib Rashid
–
True. But you could save yourself from the trouble by just putting \[\displaystyle \sum_{i=1}^\infty \lfloor\frac{n}{p^i}\rfloor\].
–
Mursalin Habib
·
2 years, 12 months ago

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@Mursalin Habib
–
He asked for an algorithm.
A code with your algo would run for an infinite time.
–
Labib Rashid
·
2 years, 12 months ago

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@Labib Rashid
–
Oh! I think I missed the word 'algorithm'. But I don't think the OP meant algorithm literally [It's tagged with number theory]. Despite that, you're right. That algorithm will go on forever in the natural sense. I just posted that because I think that looks cool :)
–
Mursalin Habib
·
2 years, 12 months ago

## Comments

Sort by:

TopNewestThe highest power of p that divides n! is

\(\displaystyle\sum_{i=1}^{j}\left \lfloor \frac{n}{p^i} \right \rfloor\)

,where j is the biggest integer, for which \(p^j<n\) – Bogdan Simeonov · 3 years ago

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– Labib Rashid · 3 years ago

I think it should be \(p^j \leq n\)Log in to reply

– Mursalin Habib · 2 years, 12 months ago

True. But you could save yourself from the trouble by just putting \[\displaystyle \sum_{i=1}^\infty \lfloor\frac{n}{p^i}\rfloor\].Log in to reply

– Labib Rashid · 2 years, 12 months ago

He asked for an algorithm. A code with your algo would run for an infinite time.Log in to reply

– Mursalin Habib · 2 years, 12 months ago

Oh! I think I missed the word 'algorithm'. But I don't think the OP meant algorithm literally [It's tagged with number theory]. Despite that, you're right. That algorithm will go on forever in the natural sense. I just posted that because I think that looks cool :)Log in to reply