I face problem while attempting questions like this one.

\[\huge \frac{n^{2}-9}{n-1}\]

**Find sum of all values of \(n\) as an integer for which \(\frac{n^{2}-9}{n-1}\) is also a integer.**

**Options are :**

\(\huge 1)\)**0**

\(\huge 2)\)**7**

\(\huge 3)\)**8**

\(\huge 4)\)**9**

Can anyone tell me how to do these type of questions with efficiency?

**Thanks in advance.**

## Comments

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TopNewestNote first that \(\dfrac{n^{2} - 9}{n - 1} = n + 1 - \dfrac{8}{n - 1}.\)

Thus the given expression will be an integer for all \(n\) such that \(n - 1\) divides \(8.\)

Since there are \(8\) integer divisors of \(8,\) namely \(\pm 1, \pm 2, \pm 4, \pm 8,\) there will be \(8\) values of \(n\) for which the given expression is also an integer.

(The \(8\) values of \(n\) are \(-7, -3, -1, 0, 2, 3, 5, 9.\)) – Brian Charlesworth · 1 year, 2 months ago

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– Akshat Sharda · 1 year, 2 months ago

Thank you again.:-)Log in to reply