- #1

subhradeep mahata

- 120

- 13

## Homework Statement

Find the number of natural numbers n such that (n

^{2}-900)/ (n-100) is an integer.

## Homework Equations

## The Attempt at a Solution

I have done the following:

(n

^{2}-900-9100+9100)/ (n-100)

or,{ (n

^{2}- (100)

^{2}) + 9100 }/ (n-100)

or, (n+100) + (9100)/(n-100)

I hope you can understand this. Also, please let me know if there is a way to insert fractions here, without using "/".

Now, coming back to the question. We clearly have to check the number of factors of 9100, which is 36.

According to my book, the correct answer is 36.

Now, let's consider a factor of 9100, say 91.

for 91, n=191. But, can't the denominator of (9100/n-100) be negative?

I mean, what if (n-100) = -91, and n=9? We have considered only the positive factors of 9100, not the negative factors. So, i think the answer should be well over 36. Please help me out.