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I'm stuck with a problem that apparently is simple.

I need to find $x$ in following equation:

$$ x!=\left(x-1\right)! + 96 $$

How can I solve it?

After some passage I've found:

$$ \left(x-1\right)\left(x-1\right)! = 96 $$

But then I don't know how to continue...

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  • $\begingroup$ hint: the function $(x-1)\cdot(x-1)!$ is increasing. Try out some values for $x$. $\endgroup$
    – Giulio
    Jun 19 at 14:48
  • $\begingroup$ $(x-1)!$ is a product of subsequent integers. If we multiply the product by the largest of them, we obtain $96.$ It suffices to check few values of $x.$ $\endgroup$
    – user376343
    Jun 19 at 14:52
  • $\begingroup$ By trying out I've found that the solution is 5, but I'd like to know if there's a more efficient way that trial and error... $\endgroup$
    – Jepessen
    Jun 19 at 14:53
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If you factorize $96=(2^5)(3^1)=(1.2.3.4)\times 4$ you can notice it is already in the form $n!\times n$ for $n=4$ and there is no other way to arrange $2^5\times 3$ to make a product of consecutive numbers appear.

Therefore using your equation we get $x-1=n\iff x=5$


Edit:

  • For larger $n$ in $x!=(x-1)!+n$ but still reasonably low, the factorization is still a quick method, because $n$ being divisible by a factorial it will factorize easily (i.e, not a product of big primes).

For instance let $n=334764638208000=(2^{19})(3^6)(5^3)(7^2)(11)(13)$

We can start by examinining $\dfrac{n}{13!}=(2^9)(3)(5)(7)$

From there, if I try to continue the sequence $\, 1,2,\cdots,13\ $ it is not difficult to figure out that $(14\times 15\times 16)\times 16$ comes next, and the solution is $x=17$.


  • For very large $n$ this can become tedious,but we can switch to asymptotic approximation of the inverse factorial.

See this post for instance Inverse of a factorial

$$ n\sim e\exp\left(\operatorname{W}\left(\frac1{e}\log\left(\frac{n!}{\sqrt{2\pi}}\right)\right)\right)-\frac12\tag{1} $$

Notice also that $(x-1)!\le (x-1)(x-1)!\le x!$ therefore $n$ is squeezed between two consecutive factorial.

E.g. for $n=96$ then $\overbrace{4!}^{24} \le 96\le \overbrace{5!}^{120}$ and just need to confirm that $x=5$ is effectively solution.

For large $n$, it suffice then to calculate the above approximation and test just a few cases to find the suitable $x$ (provided $n$ fits the equation).

E.g. $(334764638208000)(!^{-1})\approx 16.978$ and you just need to check which if $x=17$ solves the equation, if not check also $16$ and $18$.

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  • $\begingroup$ Its a wonderful answer but don't you think we have more of used hit and trial. This answer can help in questions having small values of x but what about numbers having large values of x. Even the OP asked to avoid trial and error in the comment. $\endgroup$ Jun 19 at 15:42
  • $\begingroup$ @JitendraSingh What about now ? $\endgroup$
    – zwim
    Jun 19 at 16:30
  • $\begingroup$ Yes now the answer is perfect not perfect but a brilliant answer that definitely deserved a upvote from me $\endgroup$ Jun 19 at 16:40
  • $\begingroup$ Thanks for the answer. I didn't think that for solving this type of equation at the end trial and error or approximations were the right way to do it. $\endgroup$
    – Jepessen
    Jun 20 at 7:21
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Write $(x-1)(x-1)!$ as $(x-1)^2(x-2)!$
There are only two ways to write $96$ in this form: $2^2 \cdot 4!$ and $4^2 \cdot 3!$ (There is also $1^2 \cdot 96$ but $96$ isn't a factorial).
Inspection shows that $96=4^2\cdot 3!$ gives the solution $x=5$. Reducing the problem to two possibilities is hardly trial and error.

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  • $\begingroup$ Thanks, but inspections and trial and error can be fine for little values of the constant. I don't know if there's a more general way to solve it, for example by solving $x! = (x-1)! + 5901025694562095573881198946226948593732812800000000$ (which solution is 43)... $\endgroup$
    – Jepessen
    Jun 20 at 7:23
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First, we can move all the terms containing $x$ to the left side: $x!-(x-1)!=96$

Now, we can write them in expanded form(this is sometimes helpful with problems involving multiple factorials): $(x*(x-1)*(x-2)*(x-3)*...)-((x-1)*(x-2)*(x-3)*(x-4)*...)=96$

Aha! We can factor out a $(x-1)*(x-2)*...$, which is equivalent to a $(x-1)!$ We get: $x(x-1)!-1(x-1)!=96$

So, $(x-1)(x-1)!=96$

From here, we notice two things:

a) $(x-1)!$ is bigger than $(x-1)$ b) $(x-1)!$ has to be less than $96$

This greatly limits our possible choices.

From here, we can either prime factorize $96$, or notice that $96$ is a multiple of $24$. How does that help? Well, $24=4!$, and $96=4*24$. So that means $96=4*4!$

Now, we know $96=(x-1)*(x-1)!$, so $(x-1)*(x-1)!=4*4!$, so $x-1=4$, so

$\boxed{x=5}$

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  • $\begingroup$ Its a wonderful answer but don't you think we have more of used hit and trial. This answer can help in questions having small values of x but what about numbers having large values of x. Even the OP asked to avoid trial and error in the comment. $\endgroup$ Jun 19 at 15:42
  • $\begingroup$ @JitendraSingh I wanted to choose a different route to go, because Peter Phipps answer already covered prime factorization, so I just assumed it. The method for higher numbers would be to prime factorize whatever the integer is on the left side, and then, if there are integer solutions for $x$, be of the form $x!*x$. And then you just simplify $\endgroup$ Jun 20 at 15:10

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