If $(n + 1)! = 90 \times (n - 1)!,$ find n.

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Solution

We have

$(n + 1)! = 90 \times (n - 1)!$

$\Rightarrow (n + 1) \times n \times (n - 1)! = 90 \times (n - 1)!$

$\Rightarrow (n + 1) n = 90$

$\Rightarrow (n + 1) = 10 \times 9$

[ writing 90 as product of two consecutive integers]

$\Rightarrow n = 9$

Hence, $n = 9$

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