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.
If `phi(7^(n))=2058` where `n epsilon N` then the valule of n is

Options

A. 3
B. 4
C. 5
D. 6

Answer

To solve the problem, we need to find the value of \( n \) such that \( \phi(7^n) = 2058 \).

### Step-by-Step Solution:

1. **Understanding the function \( \phi \)**:
The function \( \phi \) is defined as follows:
- \( \phi(1) = 1 \)
- If \( p \) is a prime number and \( n \in \mathbb{N} \), then \( \phi(p^n) = p^{n-1}(p-1) \).
- If \( m \) and \( n \) are relatively prime, then \( \phi(mn) = \phi(m) \phi(n) \).

2. **Applying the function to our case**:
Here, we have \( p = 7 \) (which is prime) and we need to calculate \( \phi(7^n) \):
\[ \phi(7^n) = 7^{n-1}(7 - 1) = 7^{n-1} \cdot 6 \]

3. **Setting up the equation**:
We know from the problem statement that:
\[ \phi(7^n) = 2058 \]
Therefore, we can set up the equation:
\[ 7^{n-1} \cdot 6 = 2058 \]

4. **Solving for \( 7^{n-1} \)**:
To isolate \( 7^{n-1} \), we divide both sides by 6:
\[ 7^{n-1} = \frac{2058}{6} \]
Calculating the right-hand side:
\[ 2058 \div 6 = 343 \]

5. **Expressing 343 as a power of 7**:
Now, we need to express 343 in terms of powers of 7:
\[ 343 = 7^3 \]

6. **Setting the exponents equal**:
From the equation \( 7^{n-1} = 7^3 \), we can equate the exponents:
\[ n - 1 = 3 \]

7. **Solving for \( n \)**:
Adding 1 to both sides gives:
\[ n = 4 \]

### Final Answer:
The value of \( n \) is \( 4 \).

---