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 step by step, we will use the properties of the function \(\phi\) defined in the question.

### Step 1: Write down the equation
We start with the given equation:
\[ \phi(7^n) = 2058 \]

### Step 2: Apply the definition of \(\phi\)
Since \(7\) is a prime number, we can use the formula for \(\phi\) for a prime raised to a power:
\[ \phi(7^n) = 7^{n-1} \cdot (7 - 1) = 7^{n-1} \cdot 6 \]
Thus, we can rewrite the equation as:
\[ 7^{n-1} \cdot 6 = 2058 \]

### Step 3: Simplify the equation
Now, we can divide both sides by \(6\) to isolate \(7^{n-1}\):
\[ 7^{n-1} = \frac{2058}{6} \]

### Step 4: Calculate \(\frac{2058}{6}\)
Now, we perform the division:
\[ \frac{2058}{6} = 343 \]

### Step 5: Rewrite the equation
Now we have:
\[ 7^{n-1} = 343 \]

### Step 6: Express \(343\) as a power of \(7\)
Next, we recognize that:
\[ 343 = 7^3 \]
Thus, we can equate the exponents:
\[ n - 1 = 3 \]

### Step 7: Solve for \(n\)
Now, we solve for \(n\):
\[ n = 3 + 1 = 4 \]

### Conclusion
The value of \(n\) is:
\[ \boxed{4} \]