To find the number of lattice points inside the triangle formed by the lines \(x = 0\), \(y = 0\), and \(9x + 223y = 2007\), we will follow these steps:
### Step 1: Find the vertices of the triangle
The triangle is bounded by the axes and the line \(9x + 223y = 2007\).
1. **Intersection with the y-axis** (where \(x = 0\)):
\[
9(0) + 223y = 2007 \implies 223y = 2007 \implies y = \frac{2007}{223} = 9
\]
So, the point is \((0, 9)\).
2. **Intersection with the x-axis** (where \(y = 0\)):
\[
9x + 223(0) = 2007 \implies 9x = 2007 \implies x = \frac{2007}{9} = 223
\]
So, the point is \((223, 0)\).
3. **The third vertex** is the origin \((0, 0)\).
Thus, the vertices of the triangle are \((0, 0)\), \((0, 9)\), and \((223, 0)\).
### Step 2: Determine the range for \(y\)
The \(y\)-coordinate of the lattice points can take integer values from \(1\) to \(8\) (since it must be less than \(9\)).
### Step 3: Calculate the corresponding \(x\)-values for each \(y\)
For each integer \(y\) from \(1\) to \(8\), we will find the maximum integer \(x\) such that the point \((x, y)\) lies inside the triangle.
The equation of the line can be rearranged to find \(x\):
\[
9x + 223y = 2007 \implies 9x = 2007 - 223y \implies x = \frac{2007 - 223y}{9}
\]
For each \(y\), we calculate \(x\) and find the number of valid integer \(x\) values.
### Step 4: Calculate the number of integer \(x\) values for each \(y\)
1. **For \(y = 1\)**:
\[
x = \frac{2007 - 223 \cdot 1}{9} = \frac{1784}{9} \approx 198.22 \implies x \text{ can be } 1 \text{ to } 198 \quad (198 \text{ values})
\]
2. **For \(y = 2\)**:
\[
x = \frac{2007 - 223 \cdot 2}{9} = \frac{1738}{9} \approx 193.11 \implies x \text{ can be } 1 \text{ to } 193 \quad (193 \text{ values})
\]
3. **For \(y = 3\)**:
\[
x = \frac{2007 - 223 \cdot 3}{9} = \frac{1495}{9} \approx 165.55 \implies x \text{ can be } 1 \text{ to } 165 \quad (165 \text{ values})
\]
4. **For \(y = 4\)**:
\[
x = \frac{2007 - 223 \cdot 4}{9} = \frac{1234}{9} \approx 137.11 \implies x \text{ can be } 1 \text{ to } 137 \quad (137 \text{ values})
\]
5. **For \(y = 5\)**:
\[
x = \frac{2007 - 223 \cdot 5}{9} = \frac{999}{9} = 111 \implies x \text{ can be } 1 \text{ to } 111 \quad (111 \text{ values})
\]
6. **For \(y = 6\)**:
\[
x = \frac{2007 - 223 \cdot 6}{9} = \frac{778}{9} \approx 86.44 \implies x \text{ can be } 1 \text{ to } 86 \quad (86 \text{ values})
\]
7. **For \(y = 7\)**:
\[
x = \frac{2007 - 223 \cdot 7}{9} = \frac{564}{9} \approx 62.67 \implies x \text{ can be } 1 \text{ to } 62 \quad (62 \text{ values})
\]
8. **For \(y = 8\)**:
\[
x = \frac{2007 - 223 \cdot 8}{9} = \frac{356}{9} \approx 39.56 \implies x \text{ can be } 1 \text{ to } 39 \quad (39 \text{ values})
\]
### Step 5: Sum the number of integer \(x\) values
Now, we sum the number of valid \(x\) values for each \(y\):
\[
198 + 193 + 165 + 137 + 111 + 86 + 62 + 39 = 1,011
\]
### Step 6: Find the tens place digit of \(n\)
The total number of lattice points \(n = 1011\). The tens place digit is \(1\).
Thus, the final answer is:
\[
\text{The tens place digit in } n \text{ is } 1.
\]
