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The question is defined as follows:

$$\frac{(x!)^3}{x}-1=3455$$

I first did the basics which was getting rid of the 1 onto the left and getting rid of the $x$, then I factorised both sides to get an expression as seen below:

$$x!=3456^{1/3}\cdot x^{1/3}$$

$$x(x-1)!=x\cdot 3456^{1/3}\cdot x^{-2/3}$$

$$3456^{1/3}/(x-1)!=x^{2/3}$$

I then got rid of the power of x and tried using an iterative process, however it got messy because negative integers and square roots don’t go well together.

So I wanted to know: can you solve this question without trial and error? And if so, please can you exemplify your justification in mathematical notation?

NickD
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Benny Copeland
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    I'd get the prime factorization of 3456 and then try matching powers of the prime factors. – NickD Jul 24 '18 at 17:18
  • Please use MathJax the next time you are asking an question. I just fixed that for you. – mrtaurho Jul 24 '18 at 17:18
  • Maybe you were only supposed to find the small "nice" positive root, which can be guessed fairly easily. – dxiv Jul 24 '18 at 17:29

3 Answers3

1

We have $3456=2^7\cdot 3^3$. So, you have

$$(x!)^3=3456x=2^7\cdot 3^3\cdot x.$$

In particular, no prime $>3$ can divide $x!$, and so $x\leq 4$. You also need $2^7 \cdot 3^3 \cdot x$ to be a cube, so $2x$ should be a cube, which means $x=4$. Now you can just test it and it works! so you're done.

Carl Schildkraut
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"Without trial and error" seems excessive, as the function is very quickly growing.

Grossly, you can use

$$x!^2\le \frac{x!^3}{x}\le x!^3$$ which gives you the range

$$x!\in(15,59)$$

and only $4!$ can do.

But my first reaction would be to try $3$ to $5$ without any preliminary effort.

It is worth to notice that Stirling is not helpful here.

For larger values, I would consider the bracketing between $\frac{\log(m+1)}{3\log\log(m+1)}$ and $\frac{\log(m+1)}{2\log\log(m+1)}$ and work by dichotomy.

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Let $\,a_n=\dfrac{\left(n!\right)^3}{n}\,$ then $\,a_{n+1}=n(n+1)^2a_n\,$. It's quite obvious that $\,a_n\,$ grows very fast, and the first few terms are $\,a_1=1, a_2=4, a_3 = 72, a_4=3456\,$. The latter gives the solution $\,x=4\,$.

dxiv
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  • @YvesDaoust It starts as a (computationally cheaper) way to narrow down $,n = \lfloor x \rfloor,$, which would be useful in any approach to solve the problem. Then, yes, it so happens that it does in fact find the solution. FWIW all of the answers posted thus far are one form or another of trial and error. – dxiv Jul 24 '18 at 18:25
  • @YvesDaoust Fine, I can restate the conclusion in my answer as $,x \in [4, 5),$ instead ;-) It's still finding the range with less calculations than factoring or extracting roots. – dxiv Jul 24 '18 at 18:46
  • @YvesDaoust I'll just note that the recursive calculation scales better, and leave it at that. As an example, the other two methods don't work well if the RHS were $6220799$ instead of $3455$. – dxiv Jul 24 '18 at 19:05
  • You probably mean $62207999$. –  Jul 25 '18 at 06:22
  • @YvesDaoust Right, of course. And thank you for the downvote, though it wouldn't be appropriate to return the favor. – dxiv Jul 25 '18 at 06:34
  • Your answer is nothing than pure trial and error. No attempt to restrict the range or eliminate solutions a priori. –  Jul 25 '18 at 06:40
  • @YvesDaoust I am curious how you decide that x! ∈ (15,59) means only 4! can do other than by pure trial and error. – dxiv Jul 25 '18 at 06:45
  • $\log n!/\log\log n!$ is a good approximation of $n$. –  Jul 25 '18 at 07:04
  • @YvesDaoust You'll (obviously) win for (very) large values. My answer still works better for smaller values, like OP's - integer multiplication is a lot cheaper than square/cube roots, factorials and logs. – dxiv Jul 25 '18 at 07:22
  • You didn't read the second part of my answer. –  Jul 25 '18 at 07:32