Let `S(n)` denotes the number of ordered pairs `(x,y)` satisfying `1/x+1/y=1/n,AA ,n in N` , then find `S(10)` .

Options

A. 3
B. 6
C. 9
D. 12

Explanation

To find \( S(10) \), which denotes the number of ordered pairs \( (x, y) \) satisfying the equation

\[ \frac{1}{x} + \frac{1}{y} = \frac{1}{10}, \]

we will follow these steps:

### Step 1: Rewrite the equation
Start by rewriting the equation in a more manageable form. We can multiply through by \( xy \) to eliminate the fractions:

\[ y + x = \frac{xy}{10}. \]

### Step 2: Rearrange the equation
Rearranging gives us:

\[ xy - 10x - 10y = 0. \]

### Step 3: Add 100 to both sides
To factor this equation, we can add 100 to both sides:

\[ xy - 10x - 10y + 100 = 100. \]

### Step 4: Factor the left-hand side
Now, we can factor the left-hand side:

\[ (x - 10)(y - 10) = 100. \]

### Step 5: Set \( a = x - 10 \) and \( b = y - 10 \)
Let \( a = x - 10 \) and \( b = y - 10 \). Then we have:

\[ ab = 100. \]

### Step 6: Find the pairs \( (a, b) \)
Now, we need to find all pairs of positive integers \( (a, b) \) such that their product is 100. The pairs are:

1. \( (1, 100) \)
2. \( (2, 50) \)
3. \( (4, 25) \)
4. \( (5, 20) \)
5. \( (10, 10) \)
6. \( (20, 5) \)
7. \( (25, 4) \)
8. \( (50, 2) \)
9. \( (100, 1) \)

### Step 7: Count the pairs
Counting these pairs, we find that there are 9 pairs of \( (a, b) \).

### Step 8: Relate back to \( (x, y) \)
Since \( x = a + 10 \) and \( y = b + 10 \), each pair \( (a, b) \) corresponds directly to a unique pair \( (x, y) \). Therefore, the number of ordered pairs \( (x, y) \) is also 9.

### Final Answer
Thus, we conclude that

\[ S(10) = 9. \]

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