The greatest positive integer k, for which `81^(k) + 1` is a factor of the sum `9^(350) + 9^(346) …… 9^(4) + 9^(2) + 1`

Options

A. 81
B. 85
C. 87
D. 88

Explanation

To find the greatest positive integer \( k \) for which \( 81^k + 1 \) is a factor of the sum \( 9^{350} + 9^{346} + \ldots + 9^4 + 9^2 + 1 \), we can follow these steps:

### Step 1: Rewrite the sum
The sum can be expressed as:
\[ S = 9^{350} + 9^{346} + 9^{342} + \ldots + 9^4 + 9^2 + 1 \]
This is a geometric series with the first term \( a = 1 \) (which is \( 9^0 \)) and the common ratio \( r = 9^2 = 81 \).

### Step 2: Identify the number of terms
To find the number of terms in the series, we note that the exponents decrease by 4 each time starting from 350 down to 0. The last term is \( 9^0 \), and the exponents are:
\[ 350, 346, 342, \ldots, 2, 0 \]
The number of terms can be calculated as:
\[ \text{Number of terms} = \frac{350 - 0}{4} + 1 = 88 + 1 = 89 \]

### Step 3: Use the formula for the sum of a geometric series
The sum of a geometric series is given by:
\[ S_n = a \frac{r^n - 1}{r - 1} \]
Here, \( n = 89 \), \( a = 1 \), and \( r = 81 \):
\[ S = \frac{81^{89} - 1}{81 - 1} = \frac{81^{89} - 1}{80} \]

### Step 4: Factor \( 81^{89} - 1 \)
We can factor \( 81^{89} - 1 \) using the difference of squares:
\[ 81^{89} - 1 = (81^{44} - 1)(81^{44} + 1)(81^{1} + 1)(81^{2} + 1) \ldots (81^{88} + 1) \]

### Step 5: Identify \( 81^k + 1 \) as a factor
We want to find the largest \( k \) such that \( 81^k + 1 \) divides \( S \). The term \( 81^{44} + 1 \) is a factor of \( 81^{89} - 1 \) and we can see that \( 81^k + 1 \) will divide \( 81^{44} + 1 \) if \( k \) is less than or equal to 44.

### Step 6: Conclusion
The greatest positive integer \( k \) such that \( 81^k + 1 \) divides \( S \) is \( k = 44 \).

### Final Answer
Thus, the greatest positive integer \( k \) is:
\[ \boxed{44} \]