Question

The least positive integer $k$ for which $k\left(n^2\right)\left(n^2-1^2\right)\left(n^2-2^2\right)\left(n^2-3^2\right)\dots \left[n^2-{\left(n-1\right)}^2\right]=r!$ for some positive integers $r$ is?

• A. $2002$
• B. $2004$
• C. $1$
• D. $2$

Answer 1

The correct option is D

$2$

Explanation for the correct option:

Given, $k\left(n^2\right)\left(n^2-1^2\right)\left(n^2-2^2\right)\left(n^2-3^2\right)\dots \left[n^2-{\left(n-1\right)}^2\right]=r!$ and $r$ is a positive integer

From the identity $a^2-b^2=\left(a+b\right)\left(a-b\right)$,
$\begin{array}{rcl}k\left(n^2\right)\left(n^2-1^2\right)\left(n^2-2^2\right)\left(n^2-3^2\right)\dots \left[n^2-{\left(n-1\right)}^2\right]& =& k{n}^2\left(n+1\right)\left(n-1\right)\left(n+2\right)\left(n-2\right)\left(n+3\right)\left(n-3\right)\dots \left(n+n-1\right)\left(n-n+1\right)\\ & =& k{n}^2\left(n+1\right)\left(n-1\right)\left(n+2\right)\left(n-2\right)\left(n+3\right)\left(n-3\right)\dots \left(2n-1\right)\left(1\right)\\ & =& kn·1·2·3·\dots ·\left(n-2\right)\left(n-1\right)n\left(n+1\right)\left(n+2\right)\dots \left(2n-1\right)\\ & =& kn\left(2n-1\right)!\end{array}$

$\therefore kn\left(2n-1\right)!=r!$

We observe that, if $k=2$,
$\begin{array}{cc}& 2n\left(2n-1\right)!=r!\\ \Rightarrow & \left(2n\right)!=r!\end{array}$

Which is true.

Hence, option(D) is correct.