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The number of natural numbers for which $(15n^2+8n+6)/n$ is a natural number is?

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Answer 1 · 2018-01-07T04:17:25

$4$

$(15n^2+8n+6)/n in N$

$15n+8+6/n in N$

We have

$15n+8 in N$ for all $n in N$

So we just need to find all $n$ for $6/n in N$

In other words,

$n | 6$ , or $n$ is a factor of $6$ , and $n in {1,2,3,6}$ (because $6/n > 0$ )

There are $4$ values for $n$ that satisfy the equation $(15n^2+8n+6)/n in N$ .

Checking all solutions for $n$ :

$n=1, (15n^2+8n+6)/n=29$

$n=2, (15n^2+8n+6)/n=41$

$n=3, (15n^2+8n+6)/n=55$

$n=6, (15n^2+8n+6)/n=99$