For each positive integer n , consider the highest common factor h_...

$\mathrm{n}!+1$ is not divisible by $1,2, \ldots, \mathrm{n}$ $(\mathrm{n}+1)$ ! is divisible by $1,2, \ldots, \mathrm{n}$ so $\mathrm{HCF} \geq \mathrm{n}+1$ also ( $\mathrm{n}+1$ )! is not divisible by $\mathrm{n}+2, \mathrm{n}+3 \ldots$. so HCF can be $\mathrm{n}+1$ only Let us start by taking $\mathrm{n}=99$ $\Rightarrow 99!+1$ and 100 ! $\mathrm{HCF}=100$ is not possible as 100 divides 99 ! composite number will not be able to make it so let us take prime i.e. $\mathrm{n}=97$ now 96 ! +1 and 97 ! are both divisible by 97 so $\mathrm{HCF}=97$ (by Wilson's theorem $(p-1)!+1$ ) is divisible by $p$ )