A number theory problem by Daniel Chiu

Since $\frac {x!}{x - k} = x$ , this means $x! = x^2 - kx$ . By observation, you can tell that all factorials greater than 3 (excluding 0) are greater the number's square. This leaves the value of $x$ to $3$ , $2$ and $1$ . Now we need to use a bit of trial and error. If $x$ were $3$ or $2$ , $k$ would have to be $1$ which is not allowed. Therefore, $x$ has a value of $1$ .

It can be understood from the question that we must find the value of $x > 0$ satisfying

$\frac{x!}{x-k} = x$

We will rewrite this as $x! = x(x-1)\dots 2\cdot 1 = x(x-k)$ .

Dividing both sides by $x(x-k)$ , we have on the LHS a product of all positive integers less than $x$ and not equal to $x - k$ , and on the RHS we have $1$ . This implies that all of the positive integers less than $x$ and not equal to $x - k$ must equal $1$ , if any such exist.

With that established, by speculation it can be found that the only possibilities are

$\frac{3!}{3 - 1} = 3, \:\textrm{or}\: \frac{2!}{2 - 1} = 2, \:\textrm{or}\: \frac{1!}{1 - 0} = 1$

It is given that $k \neq 1$ , and so finally we arrive at $k = 0\:$ and $\:x = \boxed{1}$

As $x$ is the answer to the problem, we must have that $\frac{x!}{x-k} = x$ , Solving for $k$ ,

$\frac{x!}{x-k} = x$ $x! = x^2 - k x$ $x! - x^2 = -k x$ $\frac{x!-x^2}{-x} = k$ $x - (x-1)! = k$

Trying out a few values for $x$ , it can be seen that for any $x > 3$ the value of $k$ is negative as $(x-1)!$ grows much faster than $x$ . For $x=3$ and $x=2$ , the value of $k$ is 1, which is not allowed by the restriction in the problem. For $x=1$ , however, $k=0$ , and as $x$ must be a positive integer (due to the factorial of $x-1$ ), we must have $x=1$ and $k=0$ . This produces a final answer of $\boxed{1}$ .