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If $N$ is expressed as a polynomial in a prime $p$, with each of the the coefficients less than $p$, and $s$ is the sum of these coefficients, prove that the power of $p$ contained in $N!$ is $\frac{(N-s)}{(p-1)}$.

Here's my approach:—

As $N$ be expressed as a polynomial in a prime $p$,

$$N=a_np^x+a_{n-1}p^{x-1}+\dots +a_1p+a_0$$

We know that the highest power of prime $p$ contained in $N!$ is given by:

$$w_p(N!) = \left \lfloor \frac{N}{p} \right \rfloor + \left \lfloor \frac{N}{p^2} \right \rfloor + \left \lfloor \frac{N}{p^3} \right \rfloor + \cdots \cdots$$

As each of the coefficients is less than $p$ so $\left \lfloor \frac{a_0}{p} \right \rfloor\left \lfloor \frac{a_1}{p^2} \right \rfloor\dots\dots$ can be neglected

So,

$$\left \lfloor \frac{N}{p}\right \rfloor=a_np^{x-1}+a_{n-1}p^{x-2}+\cdots +a_2p+a_1$$

$$w_p(N!)=a_np^{n-1}+(a_n+a_{n-1})p^{n-2}+ \cdots +(a_n+a_{n-1}+ \cdots +a_1)$$

Now, I don't know how to get $\frac{(N-s)}{(p-1)}$ from here. If there are any better ways please let me know.

Any help would be highly appreciated

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Crocogator
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1 Answers1

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The laat sum may be rewritten as $$w_p(N!)=a_n(p^{n-1}+p^{n-2}+\dots+1)+a_{n-1}(p^{n-2}+p^{n-3}+\dots+1)+\dots+a_1$$ $$=a_n\frac{p^n-1}{p-1}+a_{n-1}\frac{p^{n-1}-1}{p-1}+\dots+a_1\frac{p-1}{p-1}$$ $$=\frac{a_np^n+a_{n-1}p^{n-1}+\dots+a_1p}{p-1}-\frac{a_n+a_{n-1}+\dots+a_1}{p-1}$$ $$=\frac{a_np^n+a_{n-1}p^{n-1}+\dots+a_1p+\color{blue}{a_0}}{p-1}-\frac{a_n+a_{n-1}+\dots+a_1+\color{blue}{a_0}}{p-1}=\frac{N-s}{p-1}$$

Parcly Taxel
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