Find the greatest number by which the expression `7^(2n)-3^(2n)` is always exactly divisible.

Options

A. 4
B. 10
C. 20
D. 40

Answer

To find the greatest number by which the expression \( 7^{2n} - 3^{2n} \) is always exactly divisible, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Expression**:
The expression we need to analyze is \( 7^{2n} - 3^{2n} \). This can be rewritten using the difference of squares formula:
\[ 7^{2n} - 3^{2n} = (7^n - 3^n)(7^n + 3^n) \]

2. **Substituting Values for n**:
To find a pattern, let's substitute some small integer values for \( n \).

- For \( n = 1 \):
\[ 7^{2 \cdot 1} - 3^{2 \cdot 1} = 7^2 - 3^2 = 49 - 9 = 40 \]
- For \( n = 2 \):
\[ 7^{2 \cdot 2} - 3^{2 \cdot 2} = 7^4 - 3^4 = 2401 - 81 = 2320 \]

3. **Finding the GCD**:
Now we need to find the greatest common divisor (GCD) of the results obtained:
- The result for \( n = 1 \) is \( 40 \).
- The result for \( n = 2 \) is \( 2320 \).

We can find the GCD of \( 40 \) and \( 2320 \):
- The prime factorization of \( 40 \) is \( 2^3 \times 5 \).
- The prime factorization of \( 2320 \) is \( 2^4 \times 5 \times 29 \).

The GCD is determined by taking the lowest power of each prime:
- For \( 2 \): minimum power is \( 2^3 \).
- For \( 5 \): minimum power is \( 5^1 \).

Thus, the GCD is:
\[ 2^3 \times 5 = 8 \times 5 = 40 \]

4. **Conclusion**:
Therefore, the greatest number by which the expression \( 7^{2n} - 3^{2n} \) is always exactly divisible is \( 40 \).

### Final Answer:
The greatest number by which \( 7^{2n} - 3^{2n} \) is always exactly divisible is **40**.