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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$d$ is even

Vedclass · Answer

(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?

Similar Questions

The greatest value of $x$ satisfying $21 \equiv 385 \pmod{x}$ and $587 \equiv 167 \pmod{x}$ is

The largest value of $n$, for which $40^n$ divides $60!$, is

Among the positive divisors of the number $12600$,if $n_1$ is the number of divisors which are multiples of $3$ and $n_2$ is the number of divisors which are multiples of $14$,then $n_1 + n_2 =$

The coefficient of $x^3 y^4 z^5$ in the expansion of $(x y+y z+x z)^6$ is

The number of natural number solutions to the equation $xyz = 2^5 \times 3^2 \times 5^2$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module