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# Need help

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?

Note by Akshat Sharda
2 years, 5 months ago

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Note 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 $$-7, -3, -1, 0, 2, 3, 5, 9.$$)

- 2 years, 5 months ago

Thank you again.:-)

- 2 years, 5 months ago