`(49)^(15)-1` is exactly divisible by:

Options

A. 50
B. 51
C. 29
D. 8

Answer

To determine whether \( (49)^{15} - 1 \) is exactly divisible by a certain number, we can use a property of exponents.

### Step-by-Step Solution:

1. **Identify the expression**: We start with the expression \( (49)^{15} - 1 \).

2. **Rewrite the base**: Notice that \( 49 \) can be expressed as \( 7^2 \). Therefore, we can rewrite the expression as:
\[ (7^2)^{15} - 1 \]

3. **Apply the power of a power property**: Using the property of exponents \( (a^m)^n = a^{m \cdot n} \), we can simplify:
\[ 7^{30} - 1 \]

4. **Recognize the form**: Now we have \( 7^{30} - 1 \). We can use the property that states \( a^n - 1 \) is divisible by \( a - 1 \) for any integer \( a \) and positive integer \( n \).

5. **Check divisibility**: Since \( n = 30 \) is even, we can also use the property that \( a^n - 1 \) is divisible by \( a + 1 \) when \( n \) is even. Here, \( a = 7 \) and \( n = 30 \), thus:
\[ 7^{30} - 1 \text{ is divisible by } 7 + 1 = 8 \]

6. **Conclusion**: Therefore, \( (49)^{15} - 1 \) is exactly divisible by \( 8 \).

### Final Answer:
\( (49)^{15} - 1 \) is exactly divisible by \( 8 \).