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What is the highest power of n in $(n^r-1)!$ where $n$, $r$ are positive integers? The answer supplied is $\frac{n^r-nr+r-1}{n-1}$

  1. The highest power of $n$ in $n^r! = r +$ the highest power of $n$ in $(n^r-1)!$ as $n^r$ contributes $r$ powers.

    So, the highest power of $n$ in $n^r!$ needs to be $\frac{n^r-nr+r-1}{n-1} + r =\sum_{1≤t<r}n^t $.

    This is clearly true if $n$ were prime according to this.

  2. Again, if $r=1$, $\frac{n^r-nr+r-1}{n-1}$ becomes $0$.

    But according to this, the highest power of composite $n$ in $(n-1)! ≥ 1 $.

    So, the answer does not seem to hold for $r=1$ if $n$ is composite.

  3. If $n=\prod p_i^{a_i}$, where $p_i$ are distinct primes, $n^r=\prod p_i^{ra_i}$

    So, the highest power $(q_i)$ of $p_i$ in $n^r!$ is $$\sum_{1≤t<∞}\left\lfloor\frac{n^r}{p_i^t} \right\rfloor =\sum_{1≤t≤ra_i}\frac{n^r}{p_i^t} +\sum_{ra_i+1≤t<∞}\left\lfloor\frac{n^r}{p_i^t} \right\rfloor$$ as $p^t|n^r$ for $t≤ra_i$.

    So, the highest power of $p_i^{a_i}$ in $n^r!$ is $\left\lfloor \frac{q_i}{a_i} \right\rfloor=P_i$(say).

So, the highest power of n=$\prod p_i^{a_i}$ in $n^r!$ is $min(P_i)$

I could not proceed anymore.

Community
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lab bhattacharjee
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1 Answers1

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If $p$ is prime, the number of factors of $p$ in $n!$ is $\frac{n-\sigma_p(n)}{p-1}$, where $\sigma_p(n)$ is the sum of the digits of $n$ in base $p$. $\sigma_p(p^r-1)=(p-1)r$. Thus, the number of factors of $p$ in $(p^r-1)!$ is $$ \frac{p^r-1-(p-1)r}{p-1} $$ This is the formula for $p$ a prime. For a composite number, the count will be greater because there will be combinations of the factors of $n$ that give extra factors of $n$ in $(n^r-1)!$.

For example, consider the number of factors of $4$ in $(4^2-1)!$ $$ 1\cdot\color{#C00000}{2}\cdot3\cdot\color{#00A000}{4}\cdot5\cdot\color{#C00000}{6}\cdot7\cdot\color{#00A000}{8}\cdot9\cdot\color{#C00000}{10}\cdot11\cdot\color{#00A000}{12}\cdot13\cdot\color{#C00000}{14}\cdot15 $$ The number counted by $\frac{n^r-1-(n-1)r}{n-1}=\frac{16-1-3\cdot2}{4-1}=3$ is the count of the powers of $4$ in the individual factors of $15!$ (shown in $\color{#00A000}{\text{green}}$). However, this does not count the factors of $2$ contributed by the powers of $2$ in other individual factors of $15!$ (shown in $\color{#C00000}{\text{red}}$). There are $4$ powers of $2$ giving an extra $2$ powers of $4$ in $15!$.

A Strategy for Composites

For each prime $p$ that divides $n$, count the number of factors of $p$ that divide $(n^r-1)!$, and from that, compute the greatest power of $n$ that divides $(n^r-1)!$.

In the previous example, count the number of factors of $2$ that divide $15!$: $\frac{15-4}{2-1}=11$. Thus, we get $\lfloor11/2\rfloor=5$ as the power of $4$ that divides $15!$

Let's try another example. The highest power of $12$ that divides $12^3-1$. There are two primes that divide $12$: $2$ and $3$.

For $p=2$: $12^3-1=11010111111_{\text{two}}$ so $\sigma_2(12^3-1)=9$. Therefore, the number of factors of $2$ is $\frac{12^3-1-9}{2-1}=1718$.

For $p=3$: $12^3-1=2100222_{\text{three}}$ so $\sigma_3(12^3-1)=9$. Therefore, the number of factors of $3$ is $\frac{12^3-1-9}{3-1}=859$.

Thus, the highest power of $12$ that divides $(12^3-1)!$ would be $\min\left(\frac{1718}{2},859\right)=859$.

robjohn
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