How do I solve $\frac{(n+2)!-n!}{5!} = 330$?

Question

I know the answer is $n=6$, but can't figure out how to solve. I tried dividing by $n!$, but didn't work because there isn't one in RHS to simplify... also tried using Gamma function properties, but didn't work either...

Any help would be appreciated.

Thanks.

Answer 1

Multiply both sides by 5!. That gives you:

$n!((n+1)(n+2)-1) = 330\cdot 5!$

So, you now have the general format of a solution. We know that $n!$ divides $330\cdot 5!$, so $n\le 6$. Trial and error will get you there quickly.

$1!(2\cdot 3-1) = 5\cdot 1! \neq 330\cdot 5!$

$2!(3\cdot 4-1) = 11\cdot 2! \neq 330\cdot 5!$

$3!(4\cdot 5-1) = 19\cdot 3! \neq 330\cdot 5!$

$4!(5\cdot 6-1) = 29\cdot 4! \neq 330\cdot 5!$

$5!(6\cdot 7-1) = 41\cdot 5! \neq 330\cdot 5!$

$6!(7\cdot 8-1) = 55\cdot 6! = 55\cdot 6\cdot 5! = 330\cdot 5!$

Answer 2

I don't see a way to solve for n but I do see a way to see the only solution is 6

Multiply both sides by 5! And factor out $n!$ to get the equivalent equation

$$ (n^2+3n+1)n!=55(6!) $$

The left hand side is a strictly increasing function for $n>0$. So for any value of the right hand side, if there is a solution it is unique.

A quick check off $n=6$

$$ 36+18+1=55 $$

So $n=6$ is the only solution

Answer 3

Write $330 = 2\times3\times5\times11 $ & write $(n+2)!-n! = n!\times{(n^2+3n+1)}$

Clearly the RHS doesn't have 7 as its factor, so you must have $n\le6$

The other thing which we can see is that after simplification, we will be getting a quadratic equation of form $n^2 +3n-m =0$ where $m =\frac{330\times (5!)}{n!} -1$. Whatever integer $m$ might be, we need to have discriminant non-negative as well as a pure square, to have legitimate answer. As we already know, $n\le6$ and by quadratic equation, we know that $n = \frac{-3 + \sqrt D}2$, so $\sqrt D \le 15 $ means $ D\le 225$

Thus, $9+4m \le 225 $ or $m\le54$

or $\frac{330 \times5!}{n!} -1 \le 54 $

or $\frac{330 \times5!}{55}\le n!$

thus $6! \le n!$

or $n\ge6$

Thus, you have only one solution n=6

Answer 4

Because of $a_n = (5!)^{-1}((n+2)! - n!)$ in $\mathbb{N}$, so $n \geq 5$, and $a_n \equiv 0\pmod{n}$ for $n \geq 6$. However, $330 = 6 \cdot 55 \not \equiv 0 \pmod{7}$ therefore $a_n = 330$ for some of $n \leq 6$. Therefore $a_n = 330$ at $n = 6$.