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

In summary, the problem is asking to find the number of natural numbers n for which the expression (n^2 - 900)/(n - 100) is an integer. After simplifying the expression, it can be seen that for it to be an integer, 9100 must be divisible by n - 100. By setting k = n - 100, we can see that k must be a factor of 9100, and we must count the number of factors in the range of 0 to 100. However, since n is a natural number, k must also be greater than -100. Therefore, the correct answer is greater than 36.

## Homework Statement

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

## 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.

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.

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.

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..

Last edited:
Mark44 said:
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?

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##.

## 1. What is the purpose of finding the number of natural numbers N in a given scenario?

The purpose of this task is to determine the number of natural numbers that satisfy a particular condition or property. This can help in solving various mathematical problems and understanding patterns in numbers.

## 2. How do you find the number of natural numbers N in a given scenario?

The most common method is to use mathematical equations or algorithms to systematically count the numbers that fulfill the given criteria. This may involve using concepts such as prime numbers, factorials, or combinations.

## 3. Can you provide an example of finding the number of natural numbers N in a real-world application?

Sure, one example could be finding the number of ways to arrange a set of objects in a specific order. This can be solved by finding the factorial of the total number of objects.

## 4. Is there a limit to the number of natural numbers in a given scenario?

No, there is no limit to the number of natural numbers in a given scenario. However, the larger the numbers involved, the more complex the calculations may become.

## 5. What are some challenges faced when finding the number of natural numbers N?

One challenge is determining the most efficient method to count the numbers in a given scenario. Another challenge is dealing with large numbers, which may require advanced mathematical techniques or computer programming.

• Precalculus Mathematics Homework Help
Replies
20
Views
2K
• Precalculus Mathematics Homework Help
Replies
12
Views
1K
• Precalculus Mathematics Homework Help
Replies
2
Views
1K
• Precalculus Mathematics Homework Help
Replies
6
Views
796
• General Math
Replies
3
Views
1K
• Precalculus Mathematics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
223
• General Math
Replies
5
Views
983
• Precalculus Mathematics Homework Help
Replies
18
Views
3K
• Introductory Physics Homework Help
Replies
4
Views
1K