If the size of the ground area is $6 \, \text{km} \times 3 \, \text{km}$ and the corresponding photo size in the aerial photograph is $30 \, \text{cm} \times 15 \, \text{cm}$, then the scale of the photograph is $1$
Step 1: Convert ground distances to cm.
Ground size: $6 \,\text{km} \times 3 \,\text{km}$.
\[
6 \,\text{km} = 6000 \,\text{m} = 6000 \times 100 \,\text{cm} = 600000 \,\text{cm}
\]
\[
3 \,\text{km} = 3000 \,\text{m} = 3000 \times 100 \,\text{cm} = 300000 \,\text{cm}
\]
Step 2: Compare photo size with ground size.
On the photo: $30 \,\text{cm} \times 15 \,\text{cm}$.
On the ground: $600000 \,\text{cm} \times 300000 \,\text{cm}$.
Scale ratio:
\[
\frac{\text{Photo length}}{\text{Ground length}} = \frac{30}{600000} = \frac{1}{20000}
\]
\[
\frac{\text{Photo width}}{\text{Ground width}} = \frac{15}{300000} = \frac{1}{20000}
\]
Step 3: Final scale.
Thus the photograph scale is:
\[
1 : 20000
\]
\[
\boxed{1 : 20000}
\]
Consider a five-digit number PQRST that has distinct digits P, Q, R, S, and T, and satisfies the following conditions:
1. \( P<Q \)
2. \( S>P>T \)
3. \( R<T \)
If integers 1 through 5 are used to construct such a number, the value of P is:



