The number of natural number pairs (x, y) in which `x gt y` and `5/x+6/y =1` is :

Answer

To solve the problem of finding the number of natural number pairs \((x, y)\) such that \(x > y\) and \(\frac{5}{x} + \frac{6}{y} = 1\), we can follow these steps:

### Step 1: Rewrite the equation
Start with the given equation:
\[ \frac{5}{x} + \frac{6}{y} = 1 \]
Multiply through by \(xy\) to eliminate the denominators:
\[ 5y + 6x = xy \]

### Step 2: Rearrange the equation
Rearranging gives:
\[ xy - 5y - 6x = 0 \]

### Step 3: Add 30 to both sides
To factor the equation, we can add 30 to both sides:
\[ xy - 5y - 6x + 30 = 30 \]

### Step 4: Factor the left-hand side
Now we can factor the left-hand side:
\[ (x - 5)(y - 6) = 30 \]

### Step 5: Identify factor pairs of 30
Next, we need to find the pairs of natural numbers \((a, b)\) such that \(ab = 30\):
- \(1 \times 30\)
- \(2 \times 15\)
- \(3 \times 10\)
- \(5 \times 6\)

### Step 6: Relate factor pairs back to \(x\) and \(y\)
For each factor pair \((a, b)\), we can set:
- \(x - 5 = a\)
- \(y - 6 = b\)

Thus:
- \(x = a + 5\)
- \(y = b + 6\)

### Step 7: Calculate pairs and check \(x > y\)
Now we calculate \(x\) and \(y\) for each factor pair:
1. For \(1 \times 30\):
- \(x = 1 + 5 = 6\)
- \(y = 30 + 6 = 36\) (not valid since \(x\) is not greater than \(y\))

2. For \(2 \times 15\):
- \(x = 2 + 5 = 7\)
- \(y = 15 + 6 = 21\) (not valid since \(x\) is not greater than \(y\))

3. For \(3 \times 10\):
- \(x = 3 + 5 = 8\)
- \(y = 10 + 6 = 16\) (not valid since \(x\) is not greater than \(y\))

4. For \(5 \times 6\):
- \(x = 5 + 5 = 10\)
- \(y = 6 + 6 = 12\) (not valid since \(x\) is not greater than \(y\))

5. For \(6 \times 5\):
- \(x = 6 + 5 = 11\)
- \(y = 5 + 6 = 11\) (not valid since \(x\) is not greater than \(y\))

6. For \(10 \times 3\):
- \(x = 10 + 5 = 15\)
- \(y = 3 + 6 = 9\) (valid since \(x > y\))

7. For \(15 \times 2\):
- \(x = 15 + 5 = 20\)
- \(y = 2 + 6 = 8\) (valid since \(x > y\))

8. For \(30 \times 1\):
- \(x = 30 + 5 = 35\)
- \(y = 1 + 6 = 7\) (valid since \(x > y\))

### Step 8: Count valid pairs
The valid pairs are:
1. \((15, 9)\)
2. \((20, 8)\)
3. \((35, 7)\)

Thus, the total number of valid pairs \((x, y)\) such that \(x > y\) is **3**.

### Final Answer
The number of natural number pairs \((x, y)\) such that \(x > y\) and \(\frac{5}{x} + \frac{6}{y} = 1\) is **3**.