Indian Institute of Foreign Trade 2012 Solved Paper

It is given that Z = 31!
$\mathit{X}=31!+1$
$\mathit{X}+1=31!+2$ this is divisible by 2
$\mathit{X}+2=31!+3$ this is divisible by 3
$\mathit{X}+3=31!+4$ this is divisible by 4
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$\mathit{X}+29=31!+30$ this is divisible by 30
$\mathit{X}+30=31!+31$ this is divisible by 31
Hence, none of $\mathit{X}+1,\mathit{X}+2,...,\mathit{X}+29,\mathit{X}+30$ is a prime number
Hence, option D is the correct answer.