Finding number of natural numbers N such that............

[subhradeep mahata]

Homework Statement

Find the number of natural numbers n such that (n²-900)/ (n-100) is an integer.

Homework Equations

The Attempt at a Solution

I have done the following:
(n²-900-9100+9100)/ (n-100)
or,{ (n² - (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.

 

[jedishrfu]

The problem says natural numbers for n right so only integers greater than zero are allowed meaning n can’t be negative. That is your restriction. The expression can evaluate to positive or negative integer values.

 

[subhradeep mahata]

Yes, that is what I am saying.
9 is not a factor of 9100, but if we put 9 into the original equation, we get the result as 9, which is an integer.
But, we haven't counted that as it is not a factor of 9100.

 

[Mark44]

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

Yes. For example $\frac 3 5$, using our implementation of LaTeX. I wrote this as # #\frac 3 5 # # (without the spaces between the # characters). See our tutorial here: https://www.physicsforums.com/help/latexhelp/

Yes, that is what I am saying.
9 is not a factor of 9100, but if we put 9 into the original equation, we get the result as 9, which is an integer.
But, we haven't counted that as it is not a factor of 9100.

If n = 9, $\frac{n^2 - 900}{n - 100}$ is an integer because the expression simplifies to $\frac{81 - 900}{9 - 100} = \frac{-819}{-91} = 9$.

If you divide $n^2 - 900$ by n - 100 using polynomial long division, you get $n + 100 + \frac{9100}{n - 100}$. For the rational expression to be an integer, $\frac{9100}{n - 100}$ must be an integer. Start by factoring 9100 and counting the number of ways that 9100 is evenly divisible by some integer in the range 0 through 100. Should be a relatively short list.

It's not relevant that 9 is not a factor of 9100. What you are looking for are numbers n for which $100 - 9$ divides $n^2 - 900$. Or equivalently, n - 100 divides 9100 (from the work in the previous paragraph). n = 9 is one such number, since we have $\frac {9100}(-91}$ which is an integer, and n = 9 is a natural number..

 

[subhradeep mahata]

Start by factoring 9100 and counting the number of ways that 9100 is evenly divisible by some integer in the range 0 through 100. Should be a relatively short list.

Can you please show me how to do it?

 

[PeroK]

Can you please show me how to do it?

I thought you nearly had the solution in your original post. Note that you started with:

$n$ is a natural number ($n > 0$ or $n \ge 0$?) Depending how "natural" number is defined.

and you got to $\frac{9100}{n-100}$ is an integer.

I would let $k = n -100$, so you have: $k> -100$ and $k|9100$.

It should be easy to count the factors from there. And, you are correct that there are more than $36$.