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How does one find the highest power of a prime $p$ that divides $N!$ and other related products?

Related question: How many zeros are there at the end of $N!$?


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4 Answers 4

up vote 19 down vote accepted

Largest power of a prime dividing $N!$

In general, the highest power of a prime $p$ dividing $N!$ is given by

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

The first term appears since you want to count the number of terms less than $N$ and are multiples of $p$ and each of these contribute one $p$ to $N!$. But then when you have multiples of $p^2$ you are not multiplying just one $p$ but you are multiplying two of these primes $p$ to the product. So you now count the number of multiple of $p^2$ less than $N$ and add them. This is captured by the second term $\displaystyle \left \lfloor \frac{N}{p^2} \right \rfloor$. Repeat this to account for higher powers of $p$ less than $N$.

Number of zeros at the end of $N!$

The number of zeros at the end of $N!$ is given by $$\left \lfloor \frac{N}{5} \right \rfloor + \left \lfloor \frac{N}{5^2} \right \rfloor + \left \lfloor \frac{N}{5^3} \right \rfloor + \cdots$$ where $\left \lfloor \frac{x}{y} \right \rfloor$ is the greatest integer $\leq \frac{x}{y}$.

To make it clear, write $N!$ as a product of primes $N! = 2^{\alpha_2} 3^{\alpha_2} 5^{\alpha_5} 7^{\alpha_7} 11^{\alpha_{11}} \ldots$ where $\alpha_i \in \mathbb{N}$.

Note that $\alpha_5 < \alpha_2$, $\forall N$. (Why?)

The number of zeros at the end of $N!$ is the highest power of $10$ dividing $N!$

If $10^{\alpha}$ divides $N!$ and since $10 = 2 \times 5$, $2^{\alpha} | N!$ and $5^{\alpha} | N!$. Further since $\alpha_5 < \alpha_2$, the highest power of $10$ dividing $N!$ is the highest power of $5$ dividing $N!$ which is $\alpha_5$.

Note that there will be 

 1. A jump of $1$ zero going from $(N-1)!$ to $N!$ if $5 || N$

 2. A jump of $2$ zero going from $(N-1)!$ to $N!$ if $5^2 || N$

 3. A jump of $3$ zero going from $(N-1)!$ to $N!$ if $5^3 || N$ and in general

 4. A jump of $k$ zero going from $(N-1)!$ to $N!$ if $5^k || N$

where $a || b$ means $a$ divides $b$ and $\gcd\left(a,\dfrac{b}{a} \right)$ = 1.

Largest power of a prime dividing other related products

In general, if we want to find the highest power of a prime $p$ dividing numbers like $\displaystyle 1 \times 3 \times 5 \times \cdots (2N-1)$, $\displaystyle P(N,r)$, $\displaystyle \binom{N}{r}$, the key is to write them in terms of factorials.

For instance, $$\displaystyle 1 \times 3 \times 5 \times \cdots (2N-1) = \frac{(2N)!}{2^N N!}.$$ Hence, the largest power of a prime, $p>2$, dividing $\displaystyle 1 \times 3 \times 5 \times \cdots (2N-1)$ is given by $s_p((2N)!) - s_p(N!)$, where $s_p(N!)$ is defined above. If $p = 2$, then the answer is $s_p((2N)!) - s_p(N!) - N$.

Similarly, $$\displaystyle P(N,r) = \frac{N!}{(N-r)!}.$$ Hence, the largest power of a prime, dividing $\displaystyle P(N,r)$ is given by $s_p((N)!) - s_p((N-r)!)$, where $s_p(N!)$ is defined above.

Similarly, $$\displaystyle C(N,r) = \binom{N}{r} = \frac{N!}{r!(N-r)!}.$$ Hence, the largest power of a prime, dividing $\displaystyle C(N,r)$ is given by $$s_p((N)!) - s_p(r!) - s_p((N-r)!)$$where $s_p(N!)$ is defined above.

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Yes you are right. I have deleted the corresponding part in the post. –  user17762 May 5 '12 at 1:19
    
In my example, I meant highest power of $12$ that divides $3!$. If $m$ is square-free, then indeed the largest prime that divides $m$ is the obstruction. –  André Nicolas May 5 '12 at 1:32
    
@AndréNicolas Yes I recognized it. Thanks. For instance, in the highest power of $24$ dividing $10!$ the prime $2$ acts as the bottleneck since $2^3$ divides $24$ whereas only $3^1$ divides $24$ and since $10$ is square-free it works. Is there a generic formula known for composite $m$? –  user17762 May 5 '12 at 1:36

de Polignac's formula named after Alphonse de Polignac, gives the prime decomposition of the factorial $n!$.

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2  
Sometimes the formula is also attributed to Legendre. –  J. M. May 5 '12 at 4:46

Define $ e_p(n) = s_p(n!), $ as Marvis is using $s_p$ for the highest exponent function...

We get $$ e_p(n) = s_p(n!) = \sum_{i \geq 1} \left\lfloor \frac{n}{p^i} \right\rfloor $$

I thought I would prove this by mathematical induction. We need, for any positive integers $m,n,$

LEMMA:

(A) If $n + 1 \equiv 0 \pmod m,$ then $$ \left\lfloor \frac{n + 1}{m} \right\rfloor = 1 + \left\lfloor \frac{n}{m} \right\rfloor $$ (B) If $n + 1 \neq 0 \pmod m,$ then $$ \left\lfloor \frac{n + 1}{m} \right\rfloor = \left\lfloor \frac{n}{m} \right\rfloor $$

For $n < p,$ we know that $p$ does not divide $n!$ so that $e_p(n) = s_p(n!)$ is $0.$ But all the $\left\lfloor \frac{n}{p^i} \right\rfloor$ are $0$ as well. So the base cases of the induction is true.

Now for induction, increasing $n$ by $1.$ If $n+1$ is not divisible by $p,$ then $e_p(n+1) = e_p(n),$ while part (A) of the Lemma says that the sum does not change.

If $n+1$ is divisible by $p,$ let $s_p(n+1) = k.$ That is, there is some number $c \neq 0 \pmod p$ such that $n+1 = c p^k.$ From the Lemma, part (B), all the $\left\lfloor \frac{n}{p^i} \right\rfloor$ increase by $1$ for $i \leq k,$ but stay the same for $i > k.$ So the sum increases by exactly $k.$ But, of course, $e_p(n+1) = s_p((n+1)!) = s_p(n!) + s_p(n+1) = e_p(n) + k.$ So both sides of the middle equation increase by the same $s_p(n+1) = k,$ completing the proof by induction.

Note that it is not necessary to have $n$ divisible by $p$ to get nonzero $e_p(n).$ All that is necessary is that $n \geq p,$ because we are not factoring $n,$ we are factoring $n!$

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For something completely different, see page 114 of Concrete Mathematics by Graham, Knuth, Patashnik. The highest power of a prime $p$ dividing $n!$ is $n$ minus the sum of the integer digits of $n$ in base $p$, all divided by $p-1$. In Mathematica you would write:

(n-Total[IntegerDigits[n,p]])/(p-1)

Hence, the number of trailing zeros of $n!$ is

(n-Total[IntegerDigits[n,5]])/4

I've found this method to be much faster than the sum of Floor functions...

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