For how many natural numbers n between 1 and 2014 (both inclusive) is `(8n)/(9999-n)` an integer ?

Answer

To solve the problem of finding how many natural numbers \( n \) between 1 and 2014 (inclusive) make the expression \( \frac{8n}{9999 - n} \) an integer, we can follow these steps:

### Step 1: Set up the equation
We start with the condition that \( \frac{8n}{9999 - n} \) is an integer. Let’s denote this integer as \( k \):
\[ \frac{8n}{9999 - n} = k \]

### Step 2: Rearranging the equation
From the equation above, we can rearrange it to express \( n \):
\[ 8n = k(9999 - n) \]
Expanding this gives:
\[ 8n = 9999k - nk \]
Rearranging terms leads to:
\[ 8n + nk = 9999k \]
Factoring out \( n \) from the left side:
\[ n(8 + k) = 9999k \]
Thus, we can express \( n \) as:
\[ n = \frac{9999k}{8 + k} \]

### Step 3: Determine the bounds for \( n \)
We know that \( n \) must be a natural number between 1 and 2014. Therefore, we need to impose the condition:
\[ 1 \leq n \leq 2014 \]
This leads us to two inequalities:
1. \( n \geq 1 \)
2. \( n \leq 2014 \)

### Step 4: Solve the first inequality
From \( n \geq 1 \):
\[ \frac{9999k}{8 + k} \geq 1 \]
This simplifies to:
\[ 9999k \geq 8 + k \]
Rearranging gives:
\[ 9998k \geq 8 \quad \Rightarrow \quad k \geq \frac{8}{9998} \]
Since \( k \) is a natural number, the smallest possible value for \( k \) is 1.

### Step 5: Solve the second inequality
Now, for the second inequality \( n \leq 2014 \):
\[ \frac{9999k}{8 + k} \leq 2014 \]
This simplifies to:
\[ 9999k \leq 2014(8 + k) \]
Expanding the right side gives:
\[ 9999k \leq 16112 + 2014k \]
Rearranging leads to:
\[ 9999k - 2014k \leq 16112 \quad \Rightarrow \quad 7985k \leq 16112 \]
Thus:
\[ k \leq \frac{16112}{7985} \approx 2.017 \]
Since \( k \) is a natural number, the maximum value for \( k \) is 2.

### Step 6: Check possible values of \( k \)
Now we check the possible values of \( k \):
1. For \( k = 1 \):
\[ n = \frac{9999 \times 1}{8 + 1} = \frac{9999}{9} = 1111 \]
This is a valid natural number within the range.

2. For \( k = 2 \):
\[ n = \frac{9999 \times 2}{8 + 2} = \frac{19998}{10} = 1999.8 \]
This is not a natural number.

### Conclusion
The only valid value of \( n \) is 1111. Therefore, there is only **1 natural number \( n \)** between 1 and 2014 for which \( \frac{8n}{9999 - n} \) is an integer.