The greatest positive integer for which `13 ^(k)+1` is a factor of the sum `13^(71) + 13 ^(70) + 13 ^(69) +….+ 13+1`

Options

A. 35
B. 36
C. 71
D. 72

Explanation

To solve the problem, we need to find the greatest positive integer \( k \) such that \( 13^k + 1 \) is a factor of the sum \( S = 13^{71} + 13^{70} + 13^{69} + \ldots + 13 + 1 \).

### Step-by-step Solution:

1. **Identify the Sum**:
The sum \( S \) can be recognized as a geometric series. The series can be expressed as:
\[ S = 1 + 13 + 13^2 + \ldots + 13^{71} \]

2. **Use the Formula for the Sum of a Geometric Series**:
The formula for the sum of a geometric series is:
\[ S_n = \frac{a(r^n - 1)}{r - 1} \]
where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms. Here, \( a = 1 \), \( r = 13 \), and \( n = 72 \) (since the series goes from \( 13^0 \) to \( 13^{71} \)).
Thus, we have:
\[ S = \frac{1(13^{72} - 1)}{13 - 1} = \frac{13^{72} - 1}{12} \]

3. **Factor the Expression**:
We can factor \( 13^{72} - 1 \) using the difference of squares:
\[ 13^{72} - 1 = (13^{36} - 1)(13^{36} + 1) \]
Therefore, we can express \( S \) as:
\[ S = \frac{(13^{36} - 1)(13^{36} + 1)}{12} \]

4. **Determine the Factors**:
We need to find \( k \) such that \( 13^k + 1 \) divides \( S \). Notably, \( 13^{36} + 1 \) is a candidate since it appears in the factorization of \( S \).

5. **Set Up the Equation**:
For \( 13^k + 1 \) to be a factor of \( 13^{36} + 1 \), we can equate:
\[ 13^k + 1 = 13^{36} + 1 \]
This implies:
\[ k = 36 \]

6. **Conclusion**:
The greatest positive integer \( k \) for which \( 13^k + 1 \) is a factor of \( S \) is:
\[ \boxed{36} \]