Find $n$, where its factorial is a product of factorials

Question

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?

Answer 1

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

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

Answer 2

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 $10!$.

Answer 3

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.

Answer 4

$$\begin{align} 3! \cdot 5! \cdot 7! &= 6 \cdot 120 \cdot 7! \\ &= 6 \cdot 15 \cdot 8! \\ &= 2 \cdot 5 \cdot 9! \\ &= 10 \cdot 9! \\ &= 10! \end{align}$$

Answer 5

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?

Answer 6

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 $$3!5!7!=(3\cdot 2!)\cdot (5\cdot 4!)\cdot (7\cdot 6!)$$ $$=\color{red}{3}\cdot \color{red}{5}\cdot \color{red}{7}\cdot (2!)\cdot (4!)\cdot (6!)$$ $$=\color{red}{3}\cdot \color{red}{5}\cdot \color{red}{7}\cdot (2\cdot 1)\cdot (4\cdot 3!)\cdot (6\cdot 5!)$$ $$=1\cdot 2\cdot \color{red}{3}\cdot4\cdot \color{red}{5}\cdot 6\cdot \color{red}{7}\cdot (3! )\cdot (5!)$$

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

Answer 7

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!

Answer 8

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":

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