Let n = 2003. The least positive integer ‘k’ for which  $k.n^2\left(n^2-1^2\right)\left(n^2-2^2\right)\left(n^2-3^2\right)...\left\{n^2-{\left(n-1\right)}^2\right\}=r!$ for some positive integer ‘r’ is

$$\begin{gathered} k.n^2(n-1^2)(n^2-2^2)(n^2-3^2)\mathrm{.......}\{n^2-{(n-1)}^2\}=r! \\ \Rightarrow k.n^2(n-1)(n+1)(n-2)(n+2)(n-3)(n+3)\mathrm{.....}(n+n-)(n-n+1)=r! \\ \Rightarrow r!=k.n\mathrm{.1.2.3......}(n-1).n.(n+1)(n+2)\mathrm{......}(2n-1) \end{gathered}$$

To convert R.H.S as a factorial we must have k = 2 which will convert 2n!
$\therefore$ The least value of ‘k’ is 2.