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I need to solve $3! \cdot 5! \cdot 7! = n!$ for $n$.

I have tried simplifying as follows:

$$\begin{array}{} 3! \cdot 5 \cdot 4 \cdot 3! \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3! &= n! \\ (3!)^3 \cdot 5^2 \cdot 4^2 \cdot 7 \cdot 6 &= n! \\ 6^3 \cdot 5^2 \cdot 4^2 \cdot 7 \cdot 6 &= n! \\ 6^4 \cdot 5^2 \cdot 4^2 \cdot 7 &= n! \\ \end{array}$$

I really didn't see this helping me.

I then tried $6 \cdot 6 \cdot 6 \cdot 6 \cdot 25 \cdot 16 \cdot 7$, but $25$ only has $5$ as a double factor.

Any ideas?

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8 Answers 8

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We have

$$3!\cdot 5!\cdot 7!=(1\cdot 2\cdot 3)\cdot (1 \cdot 2\cdot 3\cdot 4\cdot 5)\cdot 1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7,$$

and combining some of those gives

$$1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot \underbrace{(2\cdot 4)}_8\cdot \underbrace{(3\cdot 3)}_9\cdot \underbrace{(2\cdot 5)}_{10}=10!$$

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If the formula is true it can only possibly be $8!$, $9!$, or $10!$ because $11!$ and larger have a factor of $11$. $8!$ doesn't have a large enough power of $3$, and $9!$ doesn't have a large enough power of $5$, so if the formula holds it must be $10!$.

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    $\begingroup$ I like this argument -- it avoids actual multiplication completely. But it ends up "necessary" and not "sufficient" (don't know if I'm using those words correctly) - you still have to multiply to verify the equality. $\endgroup$ Sep 18, 2015 at 5:59
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    $\begingroup$ @RossPresser Once you know $n$, the verification here ends up being quite simple: $3!5!7! = 10!$ iff $3!5! = 8 \cdot 9 \cdot 10$ iff $6 \cdot 120 = 72 \cdot 10$, which is the case. (In fact, this "verification" approach is reasonable even without knowing $n$ in advance; see A. Blass' response.) $\endgroup$ Sep 18, 2015 at 7:58
  • $\begingroup$ @Ross I never actually did the computation. I figured if we were willing to assume the equation is correct this is sufficient. $\endgroup$ Sep 18, 2015 at 11:52
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    $\begingroup$ If the formula is true it can only be $10!$ because $5!$ has a factor of $5$. $\endgroup$
    – N. S.
    Sep 18, 2015 at 15:49
  • $\begingroup$ @N.S: that's a good tweak. Figuring out that "$8!$ doesn't have a large enough power of $3$" as Matt says, requires being able to count past 2. Effort. So I'm much more comfortable observing that neither $8!$ nor $9!$ has a large enough power of $5$ $\endgroup$ Sep 18, 2015 at 22:51
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You already have some clever answers; here's less clever approach. You want $$ 3!\cdot5!\cdot7!=n!=7!\cdot8\cdot9\cdots\cdot n, $$ so, cancelling 7! and computing that $3!\cdot 5!=6\cdot120=720$, you want $$ 720=8\cdot9\cdots\cdot n. $$ Now start dividing both sides by 8, then 9, etc. until you get the answer. Dividing by 8 gives you $90=9\cdots n$, and then you either see the answer already or divide both sides by 9 to get the result.

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    $\begingroup$ This is quite possibly what Samir Khan actually did, except that he expressed $3!5!$ as a product rather than multiplying it out to $720$. Then pulling out a $2$ and a $4$ to the front is like dividing by $8$. $\endgroup$ Sep 18, 2015 at 12:06
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    $\begingroup$ There are at least two other answers (aside from Samir's and this) that factor out an 8, then a 9. These are all useful answers in their own ways, because each shows a different level of sophistication you can bring to the problem. And this one in particular is very straightforward after the initial insight. $\endgroup$
    – David K
    Sep 18, 2015 at 13:38
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$$\begin{align} 3! \cdot 5! \cdot 7! &= 6 \cdot 120 \cdot 7! \\ &= 6 \cdot 15 \cdot 8! \\ &= 2 \cdot 5 \cdot 9! \\ &= 10 \cdot 9! \\ &= 10! \end{align}$$

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  • $\begingroup$ Are there other examples of a product of 3 or more factorials (each greater than 1) equalling a factorial? $\endgroup$ Sep 19, 2015 at 6:17
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    $\begingroup$ many: for example take the product of two factorials: n = p! q!, then (n-1) ! p ! q ! = n ! $\endgroup$
    – aka.nice
    Sep 20, 2015 at 22:08
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It should work if we start factoring consecutive integers out of the expression. You have it boiled down to this $$ 6 \cdot 6^3\cdot5^2 \cdot 4^2 \cdot 7 =$$ $$2\cdot 3 \cdot (6^3 \cdot 5^2 \cdot 4^2 \cdot 7)=2 \cdot 3 \cdot 4\cdot 5\cdot 6\cdot 7\cdot (6^2 \cdot 5\cdot 4)$$ This is a good start, but now we need an 8. we get an 8 by using our 4 and a factor of 2 from one of our 6s. This yields: $$2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot(3\cdot6\cdot5)$$ We can get a 9 by using our 3 and a factor of 3 from our remaining 6. What remains:$$2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot(2\cdot5)=10!$$ How about that?

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Notice, $\color{red}{3}$, $\color{red}{5}$, $\color{red}{7}$ are prime numbers (can't be factorized), then the factorials can be successively reduced as follows $$3!5!7!=(3\cdot 2!)\cdot (5\cdot 4!)\cdot (7\cdot 6!)$$ $$=\color{red}{3}\cdot \color{red}{5}\cdot \color{red}{7}\cdot (2!)\cdot (4!)\cdot (6!)$$ $$=\color{red}{3}\cdot \color{red}{5}\cdot \color{red}{7}\cdot (2\cdot 1)\cdot (4\cdot 3!)\cdot (6\cdot 5!)$$ $$=1\cdot 2\cdot \color{red}{3}\cdot4\cdot \color{red}{5}\cdot 6\cdot \color{red}{7}\cdot (3! )\cdot (5!)$$

$$=1\cdot 2\cdot 3\cdot4\cdot 5\cdot 6\cdot 7\cdot (3\cdot 2\cdot1 )\cdot (5\cdot 4\cdot 3\cdot 2\cdot 1)$$ $$=1\cdot 2\cdot 3\cdot4\cdot 5\cdot 6\cdot 7\cdot ((2\cdot 4)\cdot(3\cdot 3) \cdot(2\cdot 5) )$$ $$=1\cdot 2\cdot 3\cdot4\cdot 5\cdot 6\cdot 7 \cdot8\cdot9\cdot 10$$$$=10!=n!$$ $$\color{red}{n=10}$$

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One can (and possibly should) be even less clever than Andreas proposes. Just compute 3! 5! 7! (getting 3628800) and then use brute force to find a number whose factorial that is. E.g., obviously the number is bigger than 7!, so start with 8! and work upwards. 8! = 40320, too small. 9! = 362880, too small. 10! = 3628800, just right -- and we're done.

If the number were much larger, requiring a search of a larger range of possible n, you could search in a marginally less-brainless way. E.g., find one n that's too small and one that's too large, then repeatedly try one in the middle and halve the range size.

Note that this requires neither intelligence nor good luck, and doesn't really have anything to do with factorials!

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  • $\begingroup$ The only problem I have with this answer is "(and possibly should)." Your second paragraph can be turned into a fast algorithm (it's essentially binary search) and is possibly the best way to do it on a computer but then, at least in my opinion, it no longer qualifies as "less clever." Good answer otherwise though. $\endgroup$ Sep 18, 2015 at 23:21
  • $\begingroup$ I said "and possibly should" because, as Einstein is alleged to have said, "chalk is cheaper than grey matter" and if there's a stupid way to do something effectively it's often a good choice. Only "possibly", of course, because Andreas's solution is pretty parsimonious with the grey matter too. I agree that once you go from sequential to binary search it's no longer "less clever" -- but it's still less mathematically clever. $\endgroup$ Sep 21, 2015 at 11:11
  • $\begingroup$ (+1) Because factorials grow so fast, $n$ can't be much more than 7, which is helpful in terms of reducing things to check. This answer inspired my own one which also works based on whether things look clearly too big or too small, just with some very rough bounds put on! $\endgroup$
    – Silverfish
    Aug 2, 2023 at 18:02
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Here's an approach by thinking about orders of magnitude. Factorials can be approximated well using Stirling's approximation but even the crude bounds in this answer deal with the problem, and a tougher generalisation, quite quickly.

We seek $n! = 7! \times 5! \times 3!$ and, given how quickly factorials grow, it's reasonable to think $n$ cannot be much more than $7$. In fact there are no solutions known for $a! = b! \times c! \times \cdots$ with $a > b + 3$ (where without loss of generality $b \ge c \ge \cdots$).

With $n = 7 + k$, and clearly $k \ge 1$, consider solving $(7+k)! = 7! \times 5! \times 3!$. We need $$ (7+1)(7+2)\cdots (7+k) = 5! \times 3! $$ so an obvious inequality is $$ (7+1)^k \le 5! \times 3! \le (7+k)^k $$

and if $k>1$ these bounds become strict. Indeed it's clear that $k \ne 1$ since $(7+1)^1 = 8 \ll 5! \times 3!$. We could also make a better estimate using the arithmetic mean of $7+1, 7+2, \dots, 7+k$ which is $7 + (k+1)/2$, so that $(7 + (k+1)/2)^k \approx 5! \times 3!$.

The AM-GM inequality then gives that $(7 + (k+1)/2)^k > (7+1)\cdots (7+k)$ for $k>1$. So we can use the tighter inequality

$$ 7+1 \lt \sqrt[k]{5! \times 3!} \lt 7+ \frac{k+1}{2} $$

and since $5! \times 3! = 120 \times 6 = 720$ is quite small, we will only need to consider the first few roots. If a $k$-th root lies within plausible bounds then we will check if $(7+1)\cdots (7+k)=720$.

For $k=2$, $\sqrt{5! \times 3!} = 26.83\dots > 7 + 3/2$ is no good.

For $k=3$, $\sqrt[3]{5! \times 3!} = 8.96\dots \in (7 + 1, 7 + 4/2)$ is plausible. In other words, we can see $5! \times 3! \approx 9^3$ but need to check if it exactly equals $8 \times 9 \times 10$. Both are $720$ so we have found a solution $10! = 7! \times 5! \times 3!$.

We could stop there as if this question has a solution it is clearly unique, but the next step is illustrative had the question been slightly different and we had not found a solution at $k=3$. For $k=4$, $\sqrt[4]{5! \times 3!} = 5.18\dots < 7 + 1$ is no good, and it's clear higher roots will take us even further below the lower bound. So if we hadn't found a solution already for lower $k$, there wouldn't be one at all.


This method is "overkill" in some ways, but the calculations aren't terribly difficult — tables of logarithms have been around for over four centuries. What's nice is that it easily generalises to the more general question of finding integer $a$ and $b$ such that

$$a! = b! \times 5! \times 3!$$

for which we obtain the inequality

$$ b+1 \le \sqrt[k]{5! \times 3!} \le b + \frac{k+1}{2} $$

with $a = b + k$, $k \ge 1$. Again the inequality becomes strict if $k > 1$.

For $k = 1$ we have $b + 1 = 720$ so $b = 719$ and $a = b + k = 720$. So we find the solution $720! = 719! \times 5! \times 3!$.

For $k=2$, $\sqrt{5! \times 3!} = 26.83\dots \notin (b + 1, b + 3/2)$ as that would require a fractional part between $0$ and $0.5$.

For $k=3$, $\sqrt[3]{5! \times 3!} = 8.96\dots \in (b + 1, b + 4/2)$ is plausible. It would require $6.96\dots < b < 7.96\dots$ and hence $b = 7$. We have already confirmed $8 \times 9 \times 10 = 720$, so have recovered the solution $10! = 7! \times 5! \times 3!$.

For $k=4$, $\sqrt[4]{5! \times 3!} = 5.18\dots \in (b + 1, b + 5/2)$ would require $2.68\dots < b < 4.18\dots$ so $b=3$ or $b=4$. But neither work, since $4 \times 5 \times 6 \times 7 > 720$ and so clearly $5 \times 6 \times 7 \times 8 > 720$ also.

For $k=5$, $\sqrt[5]{5! \times 3!} = 3.72\dots \in (b + 1, b + 6/2)$ would require $0.72\dots < b < 2.72\dots$ so $b=1$ or $b=2$. For $b = 1$, we need to check $2 \times 3 \times 4 \times 5 \times 6 = 720$, which it does. So we have found the solution $6! = 1! \times 5! \times 3!$. For $b = 2$, clearly the product $3 \times 4 \times 5 \times 6 \times 7 > 720$.

For $k=6$, $\sqrt[6]{5! \times 3!} = 2.99\dots \in (b + 1, b + 7/2)$ would require $-0.51 \dots < b < 1.99 \dots$ so $b=0$ or $b=1$. For $b=0$, we need to check $1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720$, which it does. So we have found the solution $6! = 0! \times 5! \times 3!$. For $b = 1$, clearly the product $2 \times 3 \times 4 \times 5 \times 6 \times 7 > 720$.

For $k>6$, we know $\sqrt[k]{5! \times 3!} < 3$ and so it will only lie in $(b + 1, b + (k+1)/2)$ for non-negative $b$ if $b=0$ or $b=1$. Either way the products $(b+1)(b+2) \cdots (b+k)$ will be too large, as they will be bigger than in the case for $k=6$.

Overall we obtain the exact solutions $720! = 719! \times 5! \times 3!$, $10! = 7! \times 5! \times 3!$, $6! = 1! \times 5! \times 3!$ and $6! = 0! \times 5! \times 3!$. Examining our results for $k=2$ and $k=4$ more closely, our method has also found the "near(ish) misses":

$$\frac{27!}{25!5!3!} = 0.975; \frac{28!}{26!5!3!} = 1.05; \frac{6!}{5!3!2!} = 0.5; \frac{7!}{5!3!3!} = 1.1\dot{6} $$

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