# janmr blog

## Prime Factors of Factorial Numbers October 30, 2010

Factorial numbers, $n! = 1 \cdot 2 \cdots n$, grow very fast with $n$. In fact, $n! \sim \sqrt{2 \pi n} (n/e)^n$ according to Stirling's approximation. The prime factors of a factorial number, however, are all relatively small, and the complete factorization of $n!$ is quite easy to obtain.

We will make use of the following fundamental theorem:

$p \mid a b$ for a prime $p$, then $p \mid a$ or $p \mid b$.

(Here, $p \mid a$ means that $p$ divides $a$.) This is called Euclid's First Theorem or Euclid's Lemma. For most, it is intuitively clear, but a proof can be found in, e.g., Hardy and Wright: An Introduction to the Theory of Numbers.

An application of this theorem to factorial numbers is that if a prime $p$ is a divisor of $n!$ then $p$ must be a divisor of at least one of the numbers $1, 2, \ldots, n$. This immediately implies

Every prime factor of $n!$ is less than or equal to $n$.

Conversely, every prime number between 2 and $n$ must be a prime factor of $n!$.

Let us introduce the notation $d_a(b)$ as the number of times $a$ divides into $b$. Put more precisely, $d_a(b) = k$ if and only if $b/a^k$ is an integer while $b/a^{k+1}$ is not.

We now seek to determine $d_p(n!)$ for all primes $p \leq n$. From Euclid's First Theorem and the Fundamental Theorem of Arithmetic follows:

$d_p(n!) = d_p(1) + d_p(2) + \cdots + d_p(n)$

The trick here is not to consider the right-hand side term by term, but rather as a whole. Let us take

$42! = 1405006117752879898543142606244511569936384000000000$

and $p=3$ as an example. How many of the numbers 1, 2, …, 42 are divisible by 3? Exactly $\lfloor 42/3 \rfloor = 14$ of them. But this is not the total count, because some of them are divisible by 3 multiple times. So how many are divisible by $3^2$? $\lfloor 42/3^2 \rfloor = 4$ of them. Similarly, $\lfloor 42/3^3 \rfloor = 1$. And $\lfloor 42/3^4 \rfloor = \lfloor 42/3^5 \rfloor = \ldots = 0$. So we have

$d_3(42!) = 14+4+1 = 19.$

This procedure is easily generalized and we have

(1)

$d_p(n!) = \sum_{k=1}^\infty \left\lfloor \frac{n}{p^k} \right\rfloor = \sum_{k=1}^{\lfloor \log_p(n) \rfloor} \left\lfloor \frac{n}{p^k} \right\rfloor.$

This identity was found by the french mathematician Adrien-Marie Legendre (see also Proofs From The Book, page 8, where it is called Legendre's Theorem).

Doing this for all primes in our example, we get

$42! = 2^{39} \cdot 3^{19} \cdot 5^9 \cdot 7^6 \cdot 11^3 \cdot 13^3 \cdot 17^2 \cdot 19^2 \cdot 23 \cdot 29 \cdot 31 \cdot 37 \cdot 41.$

Notice how the exponents do not increase as the prime numbers increase. This is true in general. Assume that $p$ and $q$ are both primes and $p < q$. Then $\log_p(n) \geq \log_q(n)$ and $n/p^k \geq n/q^k$ for all positive integers $k$. Using this in equation (1) we get

(2)

$d_p(n!) \geq d_q(n!) \; \text{for primes p, q with p < q}$

and thus

$d_2(n!) \geq d_3(n!) \geq d_5(n!) \geq d_7(n!) \geq d_{11}(n!) \geq \ldots$

What about $d_k(n!)$ for composite numbers $k$? Given the factorization of both $n!$ and $k$, this is easy to compute. But if, e.g., the multiplicity of all prime factors of $k$ are the same, then the relation (2) can be used. Consider $d_{10}(m)$ for a positive integer $m$. Since $10 = 2 \cdot 5$ then

$d_{10}(m) = \min\{ d_2(m), d_5(m) \}.$

But if $m=n!$ then we can use (2) and we have

$d_{10}(n!) = d_5(n!).$

For instance,

$d_{10}(42!) = d_5(42!) = \lfloor 42/5 \rfloor + \lfloor 42/5^2 \rfloor = 8 + 1 = 9,$

so there are 9 trailing zeros in the decimal representation of 42!.