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For a fixed positive integer n, if

$D = \left|\begin{array}{ccc} n! & (n + 1)! & (n + 2)! \\ (n + 1)! & (n + 2)! & (n + 3)! \\ (n + 2)! & (n + 3)! & (n + 4)! \end{array} \right|$

show that $\left(\dfrac{D}{(n!)^{3}} - 4 \right)$ is divisible by $n$.

Any ideas on how to go about solving this??

Thank You in advance.

  • 2
    Direct calculation leads to $D/(n!)^3 - 4= 2 n [5+n(4+n)]$. So it is not only divisible by $n$ but also by $2$ and by $5+n(4+n)$.2011-03-04
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    First you can get rid of $n!^3$ by taking it out of every row. Then you are left with a simple 3x3 matrix. Calculate the determinant subtract 4 and there you go.2011-03-04
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    can u show some steps on how to do it. Thanks2011-03-04
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    I'm just curious, why does something like this get a downvote?2011-03-04

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