This is a proof from a book on number theory I'm reading. I'm having a hard time following. I think there's a variable here that means two different things at two different times...


If n is a positive integer and p is a prime, then $p^e \vert\vert n!$, where $e=\lfloor\frac{n}{p}\rfloor + \lfloor\frac{n}{p^2}\rfloor + \dots + \lfloor\frac{n}{p^r}\rfloor$ and r is determined by n by the inequality $p^r \le n <p^{r+1}$


For a given integer k, the multiples of $p^k$ that do not exceed n are $p^k, 2p^k, \cdots, qp^k$ , where q is the largest integer such that $qp^k \le n$. But this says that q is the largest integer not exceeding $n/p^k$, so that $q= \lfloor n/p^k \rfloor$. Thus, $\lfloor n/p^k\rfloor$ gives the number of positive multiples of $p^k$ that do not exceed n.


I understand the proof so far. We've simply chosen some power of p and shown that $\lfloor n/p^k \rfloor$ is the largest integer that we can multiply by $p^k$ and stay under n. Here's where I start to get confused.

Now, if $1 \le m \le n$, then $m=qp^k$ with $(q,p) = 1, 0 \le k \le r$, and m contributes precisely k to the total exponent e with which p appears in the canonical representation of n!

Interlude 2:

How exactly does $m=qp^k$ follow from $1 \le m \le n$? Also, this q seems to mean something other than $\lfloor n/p^k\rfloor$ since we seem to be considering the set of all numbers less than or equal to n and greater than or equal to 1 that have the form $m=qp^k$?

Moreover, m is counted precisely k times by the sum

$\lfloor\frac{n}{p}\rfloor + \lfloor\frac{n}{p^2}\rfloor + \dots + \lfloor\frac{n}{p^r}\rfloor$,

once as a multiple of p, once as a multiple of $p^2$, ... , once as a multiple of $p^k$, and no more. Of course, if $k=0$, then m is not counted in the sum. Therefore, the sum above accounts exactly for the contribution of each m between 1 and n to the exponent e as claimed.

Interlude 3:

What exactly is meant here by counted? I really don't understand the last part of this proof at all...

  • $\begingroup$ Re: Interlude 2, I don't think it's the same $q$. We're just writing $m$ as a power of $p$ (possibly $p^0$) times a number prime to $p$. We can always do this. $\endgroup$ – Dylan Moreland Feb 18 '12 at 5:02
  • $\begingroup$ I thought so too. However I'm still confused at the last part of the proof. I always use different variables to avoid confusion, this really is annoying. $\endgroup$ – user17137 Feb 18 '12 at 5:03
  • $\begingroup$ Re: Interlude 3, Think back to the beginning of the proof. $\lfloor n/p^s \rfloor$ counts the multiples of $p^s$ that are $\leq n$, and for $s = 1, \ldots, k$, your $m$ is such a multiple. So $m$ contributes $k$ to the exponent $e$ and $k$ to this sum on the RHS. There might be a better way of organizing this -- if no one else has done this when I get back from walking the dog, I'll give it a shot. $\endgroup$ – Dylan Moreland Feb 18 '12 at 5:07

Interlude 2. You are considering the numbers of the form $qp^k$ that are between $1$ and $n$, because you are trying to figure out how many multiples of $p^k$ you have in $n!$. $m=qp^k$ does not follow from $1\leq m\leq n$; it is an extra condition. That is, "Let's now look at the numbers of the form $qp^k$ with $\gcd(p,q)=1$ that are between $1$ and $n$."

Interlude 3. In order to figure out the largest power of $p$ that divides $n!$, we need to count how many times $p$ occurs in the prime factorization of $n!$; this is equivalent to counting how many times it occurs in the prime factorizations of each number $m$, $1\leq m\leq n$, and then adding. It will appear once for every multiple of $p$ that is not a multipe of $p^2$; twice for every multiple of $p^2$ that is not a multiple of $p^3$; three times for each multiple of $p^3$ that is not a multiple of $p^4$; etc. But that is hard to count.

Alternatively: once for each multiple of $p$; then one more for each multiple of $p^2$; then one more for each multiple of $p^3$; then one more for each multiple of $p^4$; etc. That's easy to count, because we just figured out exactly how many multiples of $p$, how many multiples of $p^2$, etc. there are between $1$ and $n$.

There are $\lfloor\frac{n}{p}\rfloor$ multiiples of $p$; $\lfloor\frac{n}{p^2}\rfloor$ multiples of $p^2$; $\lfloor \frac{n}{p^3}\rfloor$ multiples of $p^3$; etc. Add them up, we get the number of times $p$ shows up in the factorization.

What they are saying is, instead: focus on a single $m = qp^k$, $\gcd(p,q)=1$. How many times do we count $m$? Once for $p$, once for $p^2$, once for $p^3$, etc. up until we get to $p^k$.

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  • $\begingroup$ Thanks a bunch. I get it now. This is actually a really clever way to count the number of times p goes into n! I think I would have been helped if the book had labeled the content before the first interlude a lemma, and then not used q twice in the proof. But it makes perfect sense now. $\endgroup$ – user17137 Feb 18 '12 at 6:26
  • $\begingroup$ Arturo, I really can't thank you enough. Understanding this proof absolutely made my night. The formula itself is quite beautiful. $\endgroup$ – user17137 Feb 18 '12 at 6:41
  • $\begingroup$ @Josh: Glad to help. $\endgroup$ – Arturo Magidin Feb 18 '12 at 7:21

Maybe the easiest way to follow the proof is to work through a numerical example. What power of 3 divides 100-factorial? Up to 100, the number of multiples of 3 is $[100/3]=33$, and each of these contributes a factor of 3. The number of multiples of $3^2$ is $[100/9]=11$, and each of these contributes an additional factor of 3. The number of multiples of $3^3$ is $[100/27]=3$, and each of these numbers contributes yet one more 3 to the product. Finally, the number of multiples of $3^4$ is $[100/81]=1$, giving yet one more factor of 3. All told, $$[100/3]+[100/9]+[100/27]+[100/81]$$

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This is referred to as Legendre's formula (1830) and is very popular in contest mathematics books. A complete treatment is given in pages 122-128 in Number Theory: Structures, Examples, and Problems by Titu Andreescu and Dorin Andrica.

As in $p$-adic Numbers by Fernando Q. Gouvea, I will refer to the highest exponent with which a prime $p$ divides a nonzero integer $n$ as $v_p(n).$

That is, if $n \equiv 0 \pmod {p^k}$ and $n \neq 0 \pmod {p^{k+1}},$ then we say $v_p(n) = k.$ In particular, A and A define $e_p(n) = v_p(n!)$ and call it Legendre's function. The notation $v_p(n)$ makes a good deal of sense in English language writing, as the function $v_p$ is the "$p$-adic valuation," see page 25 of Gouvea. Ireland and Rosen

denote this by $ \mbox{ord}_p \; n = v_p(n),$ see pages 3-15.

Quoted without proof in How to find maximum $x$ that $k^x$ divides $n!$

So, Theorem 6.5.1 on page 123 of Andreescu and Andrica is $$ e_p(n) = v_p(n!) = \sum_{i \geq 1} \left\lfloor \frac{n}{p^i} \right\rfloor = \frac{n - S_p(n)}{p-1} $$ where $S_p(n)$ is the sum of the digits of $n$ when written in base $p.$

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


(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 $e_p(n) = v_p(n!)$ 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 (B) of the Lemma says that the sum does not change.

If $n+1$ is divisible by $p,$ let $v_p(n+1) = k.$ That is, there is some number $c \neq 0 \pmod p$ such that $n+1 = c p^k.$ From the Lemma, part (A), all the $\left\lfloor \frac{n}{p^i} \right\rfloor$ increase by $1$ for $1 \leq i \leq k,$ but stay the same for $i > k.$ So the sum increases by exactly $k.$ But, of course, $e_p(n+1) = v_p((n+1)!) = v_p(n!) + v_p(n+1) = e_p(n) + k.$ So both sides of the middle equation increase by the same $v_p(n+1) = k,$ 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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  • $\begingroup$ Your encyclopedic seeming knowledge of problem solving literature is frightening. $\endgroup$ – user17137 Feb 18 '12 at 6:27
  • $\begingroup$ @Josh, dogs like me. $\endgroup$ – Will Jagy Feb 18 '12 at 7:06

One can give a very nice proof using the Von Mangoldt function. We define $\Lambda(n)$, as follows. It equals $\log p$ is $n=p^k$ for some prime $p$; and $0$ else. Observe then that if $n=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ then $$\log n=\sum_{i=1}^k \alpha_i \log p_i= \sum_{d\mid n}\Lambda (d)$$

The last equality holds by observing we have $\log p$ accounted $\alpha_i$ times for $p^j\mid n,j=1,\ldots,\alpha_i$, and any composite divisor dies. Now, $$\log m!=\sum_{b=1}^m\log b=\sum_{b=1}^m\sum_{d\mid b}\Lambda(d)$$

It suffices we note that $$\sum_{b=1}^m\sum_{d\mid b}\Lambda(d)=\sum_{b=1}^m \sum_{d\geqslant 1}[d\mid b]\Lambda(d)=\sum_{d\geqslant 1}\Lambda(d)\sum_{b=1}^m[d\mid b]$$

and of course $\displaystyle\sum_{b=1}^m[d\mid b]=\left\lfloor \dfrac m d\right\rfloor$, since it counts the number of multiples of $d$ less than $m$. Thus

$$\log m!=\sum_{b=1}^m\log b=\sum_{d\geqslant 1} \Lambda(d)\left\lfloor\frac md \right\rfloor$$

Since the Von Mangoldt function is nonzero only at prime powers, $$\log m!=\sum_{i\geqslant 1}\sum_{p\;\rm prime}\Lambda(p^i)\left\lfloor\frac{m}{p^i}\right\rfloor=\sum_{i\geqslant 1}\sum_{p\;\rm prime}\log p\left\lfloor\frac{m}{p^i}\right\rfloor$$

Hence the result: if we define $v_p(n)=\displaystyle\sum_{i\geqslant 1} \left\lfloor\frac{n}{p^i}\right\rfloor$ $$n!=\prod_{p\rm\; prime}p^{v_p(n)}$$

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Let $n$ be a positive integer, and let $p$ be prime. We want to find the largest $e$ such that $p^e$ divides $n!$. We will understand what's going on as soon as we have thoroughly absorbed one numerical example. So let $p=5$ and $n=888$.

By the Fundamental Theorem of Arithmetic, the $5$'s come from the $5$'s that divide the various numbers from $1$ to $888$. For example, the number $5$ contributes one $5$. So do $10$, $15$, and $20$. But $25$ is divisible by $5^2$, so contributes two $5$'s. But $30$ contributes only one, as do $35$, $40$, and $45$. However, $50$ is more generous and contributes two. And some numbers are more generous still. For instance, $125$ contributes three, while $625$ contributes four.

Imagine that the taxperson wants to collect $1$ dollar fpr every $5$ that a number "has." So the number $10$ will have to pay a tax of $1$ dollar, while $75$ will have to pay $2$ dollars, and $675$ will have to pay $4$ dollars.

The taxperson collects the tax as follows. First she gets $1$ dollar from every number that is divisible by $5$. These numbers are $1\cdot 5$, $2\cdot 5$, $3\cdot 5$, $4\cdot 5$, $5\cdot 5$, and so on up to $117\cdot 5$. There are $117$ such numbers. To put it another way, the number of such numbers is $$\left\lfloor \frac{888}{5}\right\rfloor.$$

Note that $25$, $50$, $75$, and so on still owe money. So the taxperson collects $1$ dollar from every number that still owes money. These are the multiples of $25$, and there are $$\left\lfloor \frac{888}{5^2}\right\rfloor$$ of them. But the multiples of $125$ still owe money. So the taxperson collects $1$ dollar from each multiple of $5^3$. There are $$\left\lfloor \frac{888}{5^3}\right\rfloor$$ of them. Finally, $1$ dollar is collected from the lone number which still owes money. The number of dollars collected in this final stage is $$\left\lfloor \frac{888}{5^4}\right\rfloor.$$ Now everybody has paid up. The total amount collected is $$ \left\lfloor \frac{888}{5^1}\right\rfloor+\left\lfloor \frac{888}{5^2}\right\rfloor +\left\lfloor \frac{888}{5^2}\right\rfloor+\left\lfloor \frac{888}{5^4}\right\rfloor.$$

Precisely the same idea works for every positive integer $n$, and every prime $p$.

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  • $\begingroup$ This explanation also made my night. How entertaining. $\endgroup$ – user17137 Feb 18 '12 at 7:12
  • $\begingroup$ @Josh: I came up with this version when dealing with "contest kids." In North American society at least, even relatively young people have a sophisticated understanding of money. $\endgroup$ – André Nicolas Feb 18 '12 at 7:22

Here is a concise way of expressing the argument. In the equalities below, $j$ and $k$ are positive integers, and $k$ does not exceed $n$. Denoting by $e$ the integer we want to compute, we have $$ e =\sum_{j\ge1,\,p^j|n!}1 =\sum_{k=1}^n\ \sum_{j\ge1,\,p^j|k}1 =\sum_{j=1}^\infty\ \sum_{1\le k\le n,\,p^j|k}1 =\sum_{j=1}^\infty\ \left\lfloor\frac{n}{p^j}\right\rfloor. $$ There are four equalities. The justifications of the first and the last ones are the same. The other equalities are obvious. Thus, it suffices to convince yourself that the first equality holds. This is very clearly explained in the other answers. (I just wanted to break the reasoning into a small number of elementary steps.)

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  • $\begingroup$ The lack of summation indices might be confusing to some. $\endgroup$ – Pedro Tamaroff Mar 14 '14 at 19:06
  • $\begingroup$ @PedroTamaroff - Dear Pedro: Thanks for your comment. I edited. $\endgroup$ – Pierre-Yves Gaillard Mar 15 '14 at 7:35

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