Consider the following two statements:I. If n | vedclass

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(D) Statement $I$: If $n$ is a composite number,then $n$ divides $(n-1)!$. For $n=4$,$(n-1)! = 3! = 6$. Since $4$ does not divide $6$,Statement $I$ is

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$I$ is false and $II$ is true

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(D) Statement $I$: If $n$ is a composite number,then $n$ divides $(n-1)!$. For $n=4$,$(n-1)! = 3! = 6$. Since $4$ does not divide $6$,Statement $I$ is false. Statement $II$: We need to check if $n^3+2n^2+n = n(n+1)^2$ divides $n!$. This is equivalent to checking if $(n+1)^2$ divides $(n-1)!$. For $n=3k-1$ where $k > 3$,we have $n+1 = 3k$. Then $(n+1)^2 = 9k^2$. For $(n-1)! = (3k-2)!$,if $k$ is large enough,the product $(3k-2)!$ contains factors $3k$ and $3k-3$ (or similar multiples of $3$),ensuring the presence of $3^2$ and $k^2$ in the prime factorization. Thus,$(n+1)^2$ divides $(n-1)!$ for infinitely many $n$. Statement $II$ is true.

Consider the following two statements:
$I.$ If $n$ is a composite number,then $n$ divides $(n-1)!$.
$II.$ There are infinitely many natural numbers $n$ such that $n^3+2n^2+n$ divides $n!$.

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Consider the following two statements:$I.$ If $n$ is a composite number,then $n$ divides $(n-1)!$.$II.$ There are infinitely many natural numbers $n$ such that $n^3+2n^2+n$ divides $n!$.