Q14 - For how many natural numbers n between 1 and 2014 (both inclusive) is

For how many natural numbers n between 1 and 2014 (both inclusive) is \(\frac{8n}{9999-n}\) an integer?
Correct Answer : 1.
Solution:
We are given that \(\frac{8n}{9999-n}\) is an integer. We can rewrite the expression as follows: $$ \frac{8n}{9999-n} = \frac{8}{(9999/n - 1)} $$ Solution:
\(\frac{8}{9999-n} = \frac{8}{\frac{9999}{n}-1} = integer\)
For integer value \((\frac{9999}{n}-1)\) is a factor of 8.
So possible values of \((\frac{9999}{n}-1) = 1, 2, 4, 8\)
\(\frac{9999}{n} = 2, 3, 5, 9\)
\(n = \frac{9999}{2}, \frac{9999}{3}, \frac{9999}{5}, \frac{9999}{9}\)
Integer \(n = 3333, 1111\)
But n < 2014

So only possible value of n = 1111.
So 1 is correct answer.

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