Given that the divisors of `n=3^(p)*5^(q)*7^(r)` are of of the form `4lamda+1,lamdage0`. Then,

Options

A. p+r is always even
B. p+q+r is even orr odd
C. q can be any integer
D. if p is even, then r is odd

Explanation

To solve the problem, we need to analyze the divisors of the number \( n = 3^p \cdot 5^q \cdot 7^r \) and determine the conditions under which these divisors can be expressed in the form \( 4\lambda + 1 \) where \( \lambda \geq 0 \).

### Step-by-Step Solution:

1. **Understanding the Divisors**:
The divisors of \( n \) can be expressed in the form \( 3^a \cdot 5^b \cdot 7^c \), where \( 0 \leq a \leq p \), \( 0 \leq b \leq q \), and \( 0 \leq c \leq r \).

2. **Analyzing Each Base**:
- For \( 3^a \): The powers of 3 can be written as \( 3^a \equiv (-1)^a \mod 4 \). This means:
- If \( a \) is even, \( 3^a \equiv 1 \mod 4 \)
- If \( a \) is odd, \( 3^a \equiv 3 \mod 4 \)
- For \( 5^b \): The powers of 5 can be expressed as \( 5^b \equiv 1 \mod 4 \) for all \( b \) since \( 5 \equiv 1 \mod 4 \).
- For \( 7^c \): The powers of 7 can be written as \( 7^c \equiv (-1)^c \mod 4 \). This means:
- If \( c \) is even, \( 7^c \equiv 1 \mod 4 \)
- If \( c \) is odd, \( 7^c \equiv 3 \mod 4 \)

3. **Combining the Results**:
A divisor \( d = 3^a \cdot 5^b \cdot 7^c \) will be of the form \( 4\lambda + 1 \) if:
- \( 3^a \equiv 1 \mod 4 \) (i.e., \( a \) is even)
- \( 5^b \equiv 1 \mod 4 \) (always true)
- \( 7^c \equiv 1 \mod 4 \) (i.e., \( c \) is even)

Therefore, for \( d \equiv 1 \mod 4 \), both \( a \) and \( c \) must be even.

4. **Conditions on \( p \), \( q \), and \( r \)**:
- Since \( a \) can take values from \( 0 \) to \( p \), \( p \) must be even for \( a \) to be even.
- Since \( c \) can take values from \( 0 \) to \( r \), \( r \) must also be even for \( c \) to be even.
- The value of \( b \) (which corresponds to \( q \)) does not affect the modulo condition since \( 5^b \equiv 1 \mod 4 \).

5. **Conclusion**:
- Therefore, \( p \) and \( r \) must be even.
- The sum \( p + q + r \) can be even or odd depending on the value of \( q \) (which can be either even or odd).

### Final Result:
The correct conclusion is that \( p + q + r \) can be either even or odd depending on the value of \( q \).