Question

Number of values of natural numbers $n\ge 1$ such that $\frac{n^3+3}{n^2+7}$ is also an integer is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B 2

We may write
$\frac{n^3+3}{n^2+7}=7-\frac{7n-3}{n^2+7}$
So, $\frac{n^3+3}{n^2+7}$ is an integer if and only if $\frac{7n-3}{n^2+7}$ is an integer.
Now $\frac{7n-3}{n^2+7}\le \frac{7n}{n^2}\le \frac{7}{n}\le 1$ if $n\ge 7$.
Tabulating the values of $\frac{7n-3}{n^2+7}$ for $n=1,2,\cdots ,6$ we get that for $n=2$ and $n=5$ only $\frac{7n-3}{n^2+7}$ is an integer being equal to $1.$
Thus $\frac{n^2+3}{n^3+7}$ is the integer $2-1=1$ or $5-1=4$ when $n=2$ or $n=5$ respectively.

Hence two valus of $n$ satisfy the reqiored condition.