Interesting Property Of Seven!

My first approach was to do some brute force hand calcs to get a feel for the question and try to pick up any patterns. After a while I realised $\frac{n!}{3!} < n!$ and therefore $\frac{n!}{3!} < \frac{(n+k)!}{(3+k)!} < n!$ for certain values of k up to some limit.

So if we want to find a solution we can set the second two parts of the inequality to be equal: $\frac{(n+k)!}{(3+k)!} = n!$ I figured since both sides have n! in some forms I could probably cancel it out... And with some more algebra I arrived at this product series $\prod_{i=1}^k (n+i) = 3!\cdot \prod_{i=1}^k (3+k)$

And here the RHS does not change as you increase n, and I also thought (since the question suggested there was a maximum number) k would be 1 for the largest n, and so we have $n+1=3!*(3+1)$ and so $n=\boxed{23}$

I think I was half trying to find a simple iterative equation I could plug into python but it was much more satisfying arriving at the solution on paper. Great question!

Just a general formula for problems such as this. Say we want the max integer n such that n! can be written as the product of (n-k) consecutive integers. $N=(k+1)!-1$

Consider $n=23$ . $5×6×7×...×23×24=\frac {24!}{4!}=23!$ hence it is possible.

Suppose $n> 23$ .

Then $4×5×6×...×n=\frac {n!}{3!}<n!$ are $n-3$ consecutive numbers with product less than $n!$ . This means the smallest among the $n-3$ numbers is at least 5.

But $5×6×7×...×(n+1)=\frac {n!}{4!}×(n+1)>\frac {n!}{4!}×25> n!$ , so there exists no such $n-3$ consecutive numbers.

In conclusion, the answer is 23.

I will start by saying my solution is not as concise or clever as the others. But it led to some interesting things I've never seen before. I set it up with an equation using the Pochhammer function, $m(m+1)_{n-4} = n!$ considering only the greater part of the first quadrant. Due to the asymptotic nature of this relation, there is not a solution at $m =4,$ but there is at $m=5.$ Sure enough, the solution is at $\boxed{(5,23)}.$

Now here's the interesting part. Unless my calculator is mistaken, the graph makes a very very very nice pattern in the other quadrants (of course, substituting the gamma function for factorial). In fact, I might even put it on a quilt!

Let $k:=3$ and write the $n-k$ consecutive integers as $\begin{aligned} (n+i);\:\ldots;\: (i+k+1),&&&&&&&& p_i&:=(n+i)\cdot\ldots\cdot (i+k+1)=\frac{(n+i)!}{(k+i)!},&&&p_0&=\frac{n!}{k!}<n! \end{aligned}$

Clearly $p_i$ increases as $i$ increases, so if $p_1>n!$ , we will not find a solution: $\begin{aligned} p_1&\overset{!}{\leq}n! & \Rightarrow&& \frac{(n+1)!}{(k+1)!}&\overset{!}{\leq}n!&\Rightarrow&&n&\overset{!}{\leq}(k+1)!-1 \end{aligned}$

Luckily the factorial of the maximum value $n=(k+1)!-1\quad (*)$ can be represented by $n!=n\cdot\ldots\cdot (k+2)\cdot \red{(k+1)!}\underset{(*)}{=}\red{(n+1)}\cdot n\cdot\ldots\cdot (k+2)\qquad|(n-k)\text{ consecutive integers!}$

and the answer is $n=(3+1)!-1=\boxed{23}$

We have to find a number $p$ such that satisfies the equality $n!=(p+n)(p-1+n)(p-2+n)\cdots(p-(n-4)+n)$ , where $p\geq1$ . However, by checking we find, for $n=4$ , $p=20$ , for $n=6$ , $p=4$ , for $n=7$ , $p=3$ , for $n=23$ , $p=1$ , so we must stop here, and conclude, the required result is $n=23$ .