The number of positive integers n such that n | vedclass

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

Question

(C) We are given that $(n+3)$ divides $(n^3-3)$.Using polynomial division,we can write:$\frac{n^3-3}{n+3} = \frac{n^3+27-30}{n+3} = \frac{(n+3)(n^2-3n

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) We are given that $(n+3)$ divides $(n^3-3)$.Using polynomial division,we can write:$\frac{n^3-3}{n+3} = \frac{n^3+27-30}{n+3} = \frac{(n+3)(n^2-3n+9)-30}{n+3} = (n^2-3n+9) - \frac{30}{n+3}$.For the expression to be an integer,$(n+3)$ must be a divisor of $30$.Since $n$ is a positive integer,$n \ge 1$,so $n+3 \ge 4$.The divisors of $30$ are $1, 2, 3, 5, 6, 10, 15, 30$.Considering the condition $n+3 \ge 4$,the possible values for $(n+3)$ are $5, 6, 10, 15, 30$.Calculating $n$ for each case:$n+3 = 5 \Rightarrow n = 2$$n+3 = 6 \Rightarrow n = 3$$n+3 = 10 \Rightarrow n = 7$$n+3 = 15 \Rightarrow n = 12$$n+3 = 30 \Rightarrow n = 27$Thus,there are $5$ such positive integers $n$.

The number of positive integers $n$ such that $n+3$ divides $n^3-3$ is

Similar Questions

If $n$ is a factor of $72$,such that $xy = n$,then the number of ordered pairs $(x, y)$ is: (where $x, y \in N$)

The number $111...1$ ($91$ times) is

If $(1-x^3)^{10} = \sum_{r=0}^{10} a_r x^r (1-x)^{30-2r}$,then $\frac{9a_9}{a_{10}}$ is equal to . . . . . . .

Let $N$ be the least positive integer such that whenever a non-zero digit $c \in \{1, 2, \dots, 9\}$ is written after the last digit of $N$,the resulting number is divisible by $c$. The sum of the digits of $N$ is:

The arithmetic mean of five natural numbers is $40$. The largest exceeds the smallest number by $10$. If $\alpha$ is the maximum possible value for the largest of these $5$ numbers,then the number of positive integral divisors of $\alpha$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module