10
$\begingroup$

Solve the following equation : $$4 (x+1)!= x! (2 x-6)!$$ My turn :$$24(x+1)!=6 x! (2x-6)!$$ $$4!(x+1)!=3!x!(2x-6)!$$ $$\frac{(x+1)!}{3!x!}=\frac{(2x-6)!}{4!}$$ I tried to get a formula of a permutation in both sides but i could not do it

$\endgroup$
1
  • 2
    $\begingroup$ $4(x+1)!$ on LHS means the $4[(x+1)!]$ or $[4(x+1)]!$? $\endgroup$
    – Yuta
    Sep 7, 2018 at 3:25

2 Answers 2

13
$\begingroup$

Bringing $x!$ to the left hand side gives $\frac{4(x+1)!}{x!} = (2x - 6)! \implies 4x + 4 = (2x-6)!$. Note that $(2x-6)!$ is a multiple of $4$, so this forces $2x - 6 \geq 4$, as no other smaller factorial is a multiple of $4$.

Finally, note that $2x - 6 = 4 \implies x = 5$ works, so we have one solution.There is no other solution, because if $x > 5$ and $4x + 4 < (2x - 6)!$, then $$4(x+1) + 4 = 4x + 8 < (4x+4)(2x-5)(2x-4) < (2(x+1) - 6)!$$

where the middle inequality is obvious from the fact that $\frac{4x+8}{4x+4} = 1+\frac{4}{4x+4}$ which is smaller than $2$ if $x >1$ and the product of $2x - 5$ and $2x - 4$ is of course at least $2$ if $x > 3$. Consequently, by induction there is no solution for $x > 5$ (base case easily verified) and clearly , for $2x - 6 \geq 0$ we must have $x \geq 3$ and one sees that $x = 3,4$ do not work.

$\endgroup$
11
$\begingroup$

This $$\frac{(x+1)!}{3!x!}=\frac{(2x-6)!}{4!}$$ is rewritten as $$\frac{(x+1)x!}{3!x!}=\frac{(2x-6)!}{4!}$$ or even more $$\frac{(x+1)x!}{3!x!}=\frac{(2x-6)!}{4 \cdot 3!}$$ will give you \begin{equation} x+1 = \frac{1}{4} (2x - 6)! \end{equation} which is valid for $x = 5$.

$\endgroup$
2
  • 4
    $\begingroup$ You would need to add that for any greater $x$ the right hand side grows faster, so the solution is unique. $\endgroup$
    – Andrei
    Sep 7, 2018 at 3:27
  • $\begingroup$ could you please edit it @Andrei ? $\endgroup$ Sep 7, 2018 at 3:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .