Let N = n(n+1)(n+2)(n+3) where n is a natural | vedclass

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(D) The expression $N = n(n+1)(n+2)(n+3)$ is the product of four consecutive integers.Among any four consecutive integers,one is divisible by $4$,anot

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$d$ is even

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(D) The expression $N = n(n+1)(n+2)(n+3)$ is the product of four consecutive integers.Among any four consecutive integers,one is divisible by $4$,another is divisible by $2$,and at least one is divisible by $3$.Thus,$N$ is always divisible by $4 	imes 2 	imes 3 = 24$.We can rewrite $N$ as:$N = [n(n+3)] 	imes [(n+1)(n+2)]$$N = (n^2 + 3n)(n^2 + 3n + 2)$Let $x = n^2 + 3n$. Then $N = x(x+2) = x^2 + 2x = (x+1)^2 - 1$.Since $N = (n^2 + 3n + 1)^2 - 1$,$N$ is one less than a perfect square.For any natural number $n \ge 1$,$N > 0$. Since $N$ is not a perfect square,the number of divisors $d$ of $N$ must be even (because divisors of a non-perfect square always come in pairs $(a, b)$ such that $ab = N$ and $a 
eq b$).

Let $N = n(n+1)(n+2)(n+3)$ where $n$ is a natural number,and $d$ is the number of divisors of $N$. Which of the following is true?

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