`3^(6n) + 4^(6n)` is necessarily divisible by 25 when :

Options

A. n is an even integer
B. n is an odd integer
C. only n is a prime
D. `n ge 6, n in I^(+)`

Answer

To determine when \( 3^{6n} + 4^{6n} \) is necessarily divisible by 25, we can analyze the expression step by step.

### Step 1: Analyze the expression
We start with the expression:
\[ 3^{6n} + 4^{6n} \]
We want to find conditions under which this expression is divisible by 25.

### Step 2: Substitute a value for \( n \)
To explore the divisibility, we can substitute a specific value for \( n \). Let's choose \( n = 1 \):
\[ 3^{6 \cdot 1} + 4^{6 \cdot 1} = 3^6 + 4^6 \]

### Step 3: Calculate \( 3^6 \) and \( 4^6 \)
Now we calculate \( 3^6 \) and \( 4^6 \):
\[ 3^6 = 729 \]
\[ 4^6 = 4096 \]

### Step 4: Add the results
Next, we add these two results together:
\[ 729 + 4096 = 4825 \]

### Step 5: Check divisibility by 25
Now we check if 4825 is divisible by 25:
\[ 4825 \div 25 = 193 \]
Since 193 is an integer, \( 4825 \) is divisible by 25.

### Step 6: Generalize for \( n \)
Now, we need to determine if this holds for other values of \( n \). We can analyze the expression modulo 25:
- \( 3^{6n} \mod 25 \)
- \( 4^{6n} \mod 25 \)

Using Euler's theorem, since \( \phi(25) = 20 \):
- \( 3^{20} \equiv 1 \mod 25 \)
- \( 4^{20} \equiv 1 \mod 25 \)

Thus, we can reduce \( 6n \) modulo 20:
- If \( n \) is odd, \( 6n \) is even, and both \( 3^{6n} \) and \( 4^{6n} \) will yield specific residues.

### Step 7: Conclusion
Through our calculations and analysis, we find that \( 3^{6n} + 4^{6n} \) is necessarily divisible by 25 when \( n \) is an odd integer.