Define a function `phi:NtoN` as follows `phi(1)=1,phi(P^(n))=P^(n-1)(P-1)` is prime and `n epsilonN` and `phi(mn)=phi(m)phi(n)` if m & n are relatively prime natural numbers.
The number of natural numbers n such that `phi(n)` is odd is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem of finding the number of natural numbers \( n \) such that \( \phi(n) \) is odd, we will analyze the properties of the Euler's totient function \( \phi \) as defined in the question.

### Step-by-Step Solution:

1. **Understanding the function \( \phi \)**:
- We know that \( \phi(1) = 1 \).
- For a prime \( p \) and \( n \in \mathbb{N} \), \( \phi(p^n) = p^{n-1}(p-1) \).
- If \( m \) and \( n \) are relatively prime, then \( \phi(mn) = \phi(m) \phi(n) \).

2. **Calculating \( \phi(2) \)**:
- Since \( 2 \) is the only even prime, we calculate \( \phi(2) \):
\[ \phi(2) = 2^{1-1}(2-1) = 1 \cdot 1 = 1 \]
- Thus, \( \phi(2) \) is odd.

3. **Calculating \( \phi(1) \)**:
- We already know \( \phi(1) = 1 \), which is also odd.

4. **Analyzing \( \phi(n) \) for odd primes**:
- For any odd prime \( p \) (where \( p > 2 \)), we have:
\[ \phi(p^n) = p^{n-1}(p-1) \]
- Here, \( p-1 \) is even (since \( p \) is odd), making \( \phi(p^n) \) even for \( n \geq 1 \).

5. **Analyzing \( \phi(n) \) for composite numbers**:
- For any composite number \( n \) that includes any odd prime factor, \( \phi(n) \) will also be even because it will include factors of the form \( p-1 \) (which is even).
- If \( n \) contains the prime \( 2 \) and any odd prime, \( \phi(n) \) will still be even.

6. **Conclusion**:
- The only natural numbers \( n \) for which \( \phi(n) \) is odd are \( n = 1 \) and \( n = 2 \).
- Therefore, the total number of natural numbers \( n \) such that \( \phi(n) \) is odd is **2**.

### Final Answer:
The number of natural numbers \( n \) such that \( \phi(n) \) is odd is **2**.