# A different way to find the highest power

For a prime $p>n$ for $n\in\mathbb{N}$, you've probably seen this formula somewhere for the highest power of $p$ dividing $n!$:

$\displaystyle \sum _{ i\ge 1 }^{ \quad }{ \left\lfloor \frac { n }{ { p }^{ i } } \right\rfloor }$

In fact, there's a much more surprising formula (let's face it, this one isn't that hard to derive): suppose $n={ ({ b }_{ 0 }{ b }_{ 1 }{ b }_{ 2 }...{ b }_{ r }) }_{ p }$, or

$n={ b }_{ 0 }{ p }^{ r }+{ b }_{ 1 }{ p }^{ r-1 }...+{ b }_{ r }$

and let $\delta (n)$ be the digit sum of $n$ base $p$. Then the highest power of $p$ dividing $n!$ is

$\displaystyle \frac { n-\delta (n) }{ p-1 }$

Wow. That's a heck of a lot shorter, isn't it? But why does this even work? Let's prove it below...

$\frac { n-\delta (n) }{ p-1 } = \frac { n-({ b }_{ 0 }+{ b }_{ 1 }+...{ b }_{ r }) }{ p-1 } =\frac { \displaystyle \sum _{ k=0 }^{ r }{ { b }_{ k }({ p }^{ r-k }-1) } }{ p-1 }$

Then factor ${p}^{r-k}-1$ (by difference of nth powers) and cancel the $p-1$ in the denominator. This results in

$\displaystyle \sum _{ k=0 }^{ r }{ { b }_{ k }\left( 1+p+{ p }^{ 2 }...{ p }^{ r-k-1 } \right) } = b_{r-1}+(b_{r-2}p+b_{r-2}) \dots$

Now write this sum out - term by term - in columns (not a typo, I mean columns - there's a trick here):

$\displaystyle \sum _{ k=0 }^{ r }{ { b }_{ k }\left( \sum _{ j=0 }^{ r-k-1 }{ { p }^{ j } } \right) } =({ b }_{ r-1 }+{ b }_{ r-2 }p+\dots+ { b }_{ 0 }{ p }^{ r-1 })$ $\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad+(b_{r-2}+ \dots + b_0 p^{r-2})$ $\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\vdots$ $\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad +b_0$

Okay, while each term was written by column before, let's look at the rows now. In fact, each row is equal to a term in this familiar expression:

$\displaystyle \sum _{ k=1}^{ r }{ \left\lfloor \frac { n }{ { p }^{ k } } \right\rfloor }$

because each time you divide by $p$ again one term in the base $p$ expansion vanishes (that associated with $p^0$) and the others go down an exponent of $p$. So, we have

$\displaystyle \sum _{ k=1}^{ r }{ \left\lfloor \frac { n }{ { p }^{ k } } \right\rfloor } = \displaystyle \frac { n-\delta (n) }{ p-1 }$ Note by Dylan Pentland
4 years, 5 months ago

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Wow! Nice, let me see whether I can apply to this question.

- 4 years, 5 months ago