If `p^q-q^r=(p+q)^(r-q),` `pgtrgtqin` Prime numbers less than 11 then p+q is equal to:

Options

A. r(r-q)
B. r(q-p)
C. r(p+q)
D. pq

Answer

To solve the equation \( p^q - q^r = (p + q)^{(r - q)} \) where \( p, q, r \) are prime numbers less than 11 and \( p > r > q \), we can follow these steps:

### Step 1: Identify the prime numbers less than 11
The prime numbers less than 11 are:
\[ 2, 3, 5, 7 \]

### Step 2: Set up the conditions
Given the conditions \( p > r > q \), we can try different combinations of these prime numbers.

### Step 3: Test combinations of \( p, q, r \)
1. **First combination**: Let \( p = 5, q = 2, r = 3 \)
- Calculate \( p^q - q^r \):
\[ 5^2 - 2^3 = 25 - 8 = 17 \]
- Calculate \( (p + q)^{(r - q)} \):
\[ (5 + 2)^{(3 - 2)} = 7^1 = 7 \]
- Check if \( 17 = 7 \): **Not satisfied**.

2. **Second combination**: Let \( p = 7, q = 3, r = 5 \)
- Calculate \( p^q - q^r \):
\[ 7^3 - 3^5 = 343 - 243 = 100 \]
- Calculate \( (p + q)^{(r - q)} \):
\[ (7 + 3)^{(5 - 3)} = 10^2 = 100 \]
- Check if \( 100 = 100 \): **Satisfied**.

### Step 4: Calculate \( p + q \)
From the successful combination \( p = 7 \) and \( q = 3 \):
\[ p + q = 7 + 3 = 10 \]

### Conclusion
Thus, the value of \( p + q \) is:
\[ \boxed{10} \]