IF `n` is a natural number, then `(9^(2n)-4^(2n))` is always divisible by:

Options

A. Only 5
B. Only 13
C. Both 5 and 13
D. None of these

Answer

To solve the problem of whether \( 9^{2n} - 4^{2n} \) is always divisible by a certain number when \( n \) is a natural number, we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression \( 9^{2n} - 4^{2n} \). We can recognize that this is a difference of squares.

\[ 9^{2n} - 4^{2n} = (3^{2n})^2 - (2^{2n})^2 \]

### Step 2: Apply the difference of squares formula
Using the difference of squares formula, \( a^2 - b^2 = (a - b)(a + b) \), we can factor the expression:

\[ (3^{2n} - 2^{2n})(3^{2n} + 2^{2n}) \]

### Step 3: Analyze the factors
Now we need to analyze the two factors \( (3^{2n} - 2^{2n}) \) and \( (3^{2n} + 2^{2n}) \).

1. **For \( 3^{2n} - 2^{2n} \)**:
- This expression is always even because \( 3^{2n} \) is odd and \( 2^{2n} \) is even. The difference of an odd number and an even number is always odd.

2. **For \( 3^{2n} + 2^{2n} \)**:
- This expression is always odd since both \( 3^{2n} \) and \( 2^{2n} \) are even and odd respectively.

### Step 4: Check divisibility by 5
To check if the entire expression \( 9^{2n} - 4^{2n} \) is divisible by 5, we can evaluate the expression modulo 5.

- **Calculating \( 9^{2n} \mod 5 \)**:
- \( 9 \equiv 4 \mod 5 \)
- Therefore, \( 9^{2n} \equiv 4^{2n} \mod 5 \)

- **Calculating \( 4^{2n} \mod 5 \)**:
- \( 4 \equiv -1 \mod 5 \)
- Hence, \( 4^{2n} \equiv (-1)^{2n} \equiv 1 \mod 5 \)

Putting these together:

\[ 9^{2n} - 4^{2n} \equiv 4^{2n} - 4^{2n} \equiv 0 \mod 5 \]

### Conclusion
Thus, \( 9^{2n} - 4^{2n} \) is divisible by 5 for all natural numbers \( n \).

### Final Answer
The expression \( 9^{2n} - 4^{2n} \) is always divisible by **5**.

---