`((49)^15 - 1)` is completely divisible by ?

Options

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

Answer

To solve the problem of whether \( (49^{15} - 1) \) is completely divisible by any of the given options, we can follow these steps:

### Step 1: Rewrite the expression
We start by rewriting \( 49^{15} \) in a more manageable form:
\[ 49^{15} = (7^2)^{15} = 7^{30} \]
Thus, we can rewrite the original expression:
\[ 49^{15} - 1 = 7^{30} - 1 \]

### Step 2: Apply the difference of powers formula
We can use the difference of powers formula, which states that \( a^n - b^n \) can be factored as:
\[ a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \ldots + b^{n-1}) \]
In our case, \( a = 7 \), \( b = 1 \), and \( n = 30 \):
\[ 7^{30} - 1 = (7 - 1)(7^{29} + 7^{28} + \ldots + 1) = 6(7^{29} + 7^{28} + \ldots + 1) \]

### Step 3: Check divisibility by the options
Now we need to check if \( 6(7^{29} + 7^{28} + \ldots + 1) \) is divisible by the given options: 50, 51, 29, and 8.

1. **Divisibility by 50**:
- \( 50 = 2 \times 5^2 \)
- The expression is divisible by 2 (since \( 6 \) is even), but not by \( 25 \) (as \( 7^{29} + 7^{28} + \ldots + 1 \) is not divisible by \( 25 \)).
- **Conclusion**: Not divisible by 50.

2. **Divisibility by 51**:
- \( 51 = 3 \times 17 \)
- The expression is divisible by \( 3 \) (since \( 6 \) is divisible by \( 3 \)), but we need to check \( 7^{29} + 7^{28} + \ldots + 1 \) for divisibility by \( 17 \).
- Using Fermat's Little Theorem, \( 7^{16} \equiv 1 \mod 17 \). Thus, \( 7^{29} \equiv 7^{13} \mod 17 \) and we can compute \( 7^{13} \) modulo \( 17 \). After calculating, we find it is not divisible by \( 17 \).
- **Conclusion**: Not divisible by 51.

3. **Divisibility by 29**:
- \( 29 \) is a prime number. Using Fermat's theorem again, \( 7^{28} \equiv 1 \mod 29 \). Therefore, \( 7^{30} \equiv 7^2 \mod 29 \).
- Thus, \( 7^{30} - 1 \equiv 7^2 - 1 \equiv 49 - 1 \equiv 48 \mod 29 \) which is not divisible by \( 29 \).
- **Conclusion**: Not divisible by 29.

4. **Divisibility by 8**:
- \( 8 = 2^3 \)
- The expression \( 6 \) is divisible by \( 2 \), but we need to check if \( 7^{29} + 7^{28} + \ldots + 1 \) is divisible by \( 4 \).
- Since \( 7 \equiv -1 \mod 8 \), \( 7^{29} + 7^{28} + \ldots + 1 \equiv (-1)^{29} + (-1)^{28} + \ldots + 1 \equiv -1 + 1 + \ldots + 1 \) (where the number of terms is \( 30 \)).
- This gives us \( 15 \) terms of \( 1 \) and \( 1 \) term of \( -1 \), resulting in \( 15 - 1 = 14 \), which is divisible by \( 4 \).
- Thus, \( 6(7^{29} + 7^{28} + \ldots + 1) \) is divisible by \( 8 \).
- **Conclusion**: Divisible by 8.

### Final Answer
The expression \( 49^{15} - 1 \) is completely divisible by **8**.