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

Question

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...

Answer 1

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 $x-1=n\iff x=5$

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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 $18$.

Answer 2

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

$\boxed{x=5}$

Answer 3

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.