Let (S) denotes the number of ordered pairs (x,y) satisfying `(1)/(x)+(1)/(y)=(1)/(n),x,y,n in N`.
Q. `sum_(r=1)^(10)S(r)` equals

Options

A. 47
B. 48
C. 49
D. 50

Explanation

To solve the problem, we need to find the number of ordered pairs \( (x, y) \) that satisfy the equation:

\[ \frac{1}{x} + \frac{1}{y} = \frac{1}{n} \]

where \( x, y, n \) are natural numbers. We will denote the number of such ordered pairs for a specific \( n \) as \( S(n) \). We are tasked with calculating:

\[ \sum_{r=1}^{10} S(r) \]

### Step-by-Step Solution

1. **Rearranging the Equation**:
Start from the equation:
\[ \frac{1}{x} + \frac{1}{y} = \frac{1}{n} \]
We can rewrite this as:
\[ \frac{x + y}{xy} = \frac{1}{n} \]
Multiplying both sides by \( nxy \) gives:
\[ n(x + y) = xy \]

2. **Rearranging Further**:
Rearranging the equation leads to:
\[ xy - nx - ny = 0 \]
This can be rewritten as:
\[ xy - nx - ny + n^2 = n^2 \]
or:
\[ (x - n)(y - n) = n^2 \]

3. **Finding Ordered Pairs**:
Let \( a = x - n \) and \( b = y - n \). Then the equation becomes:
\[ ab = n^2 \]
The number of ordered pairs \( (a, b) \) is equal to the number of divisors of \( n^2 \).

4. **Counting Divisors**:
If \( n \) has the prime factorization:
\[ n = p_1^{e_1} p_2^{e_2} \ldots p_k^{e_k} \]
then the number of divisors \( d(n) \) is given by:
\[ d(n) = (e_1 + 1)(e_2 + 1) \ldots (e_k + 1) \]
Therefore, the number of divisors of \( n^2 \) is:
\[ d(n^2) = (2e_1 + 1)(2e_2 + 1) \ldots (2e_k + 1) \]

5. **Calculating \( S(n) \)**:
Since each divisor \( (a, b) \) corresponds to an ordered pair \( (x, y) \) where \( x = a + n \) and \( y = b + n \), we have:
\[ S(n) = d(n^2) \]

6. **Summing Up**:
Now we need to calculate \( \sum_{r=1}^{10} S(r) \):
- For \( n = 1 \): \( S(1) = d(1^2) = d(1) = 1 \)
- For \( n = 2 \): \( S(2) = d(2^2) = d(4) = 3 \)
- For \( n = 3 \): \( S(3) = d(3^2) = d(9) = 3 \)
- For \( n = 4 \): \( S(4) = d(4^2) = d(16) = 5 \)
- For \( n = 5 \): \( S(5) = d(5^2) = d(25) = 3 \)
- For \( n = 6 \): \( S(6) = d(6^2) = d(36) = 9 \)
- For \( n = 7 \): \( S(7) = d(7^2) = d(49) = 3 \)
- For \( n = 8 \): \( S(8) = d(8^2) = d(64) = 7 \)
- For \( n = 9 \): \( S(9) = d(9^2) = d(81) = 5 \)
- For \( n = 10 \): \( S(10) = d(10^2) = d(100) = 9 \)

Now, summing these values:
\[ \sum_{r=1}^{10} S(r) = 1 + 3 + 3 + 5 + 3 + 9 + 3 + 7 + 5 + 9 = 48 \]

### Final Answer
Thus, the final answer is:
\[ \sum_{r=1}^{10} S(r) = 48 \]