The number of positive integers n such that n | vedclass

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(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

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$5$

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(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

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