Question

The value of the greatest positive integer \( k \), such that \( 49^k + 1 \) is a factor of \( 48(49^{125} + 49^{124} + \dots + 49^2 + 49 + 1) \) is

• A. 32
• B. 63
• C. 65
• D. 60

Hint:

For questions involving \( x^n - 1 \), remember the factorization hierarchy: \( x^{2m} - 1 = (x^m - 1)(x^m + 1) \). The factor \( x^m + 1 \) appears immediately at the first step of difference-of-squares expansion where the power is halved.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:

We need to simplify the given expression using the formula for the sum of a geometric progression (GP) and then analyze its factors.
Step 2: Key Formula or Approach:

The sum of a geometric series is \( S_n = a\frac{r^n - 1}{r - 1} \). The term \( x^n - y^n \) is divisible by \( x - y \). The term \( x^n - y^n \) contains \( x^{n/2} + y^{n/2} \) as a factor if \( n \) is even. 
Step 3: Detailed Explanation:

Let the given expression be \( E \). \[ E = 48(1 + 49 + 49^2 + \dots + 49^{125}) \] The term in the bracket is a GP with first term \( a=1 \), common ratio \( r=49 \), and number of terms \( n=126 \) (powers from 0 to 125). \[ \text{Sum} = \frac{49^{126} - 1}{49 - 1} = \frac{49^{126} - 1}{48} \] Substitute this back into the expression for \( E \): \[ E = 48 \left( \frac{49^{126} - 1}{48} \right) = 49^{126} - 1 \] We are looking for \( k \) such that \( 49^k + 1 \) is a factor of \( 49^{126} - 1 \). Recall the identity \( a^2 - b^2 = (a - b)(a + b) \). Let \( a = 49^{63} \). Then: \[ 49^{126} - 1 = (49^{63})^2 - 1^2 = (49^{63} - 1)(49^{63} + 1) \] From this, it is clear that \( 49^{63} + 1 \) is a factor of \( 49^{126} - 1 \). Therefore, \( k = 63 \) works. We must check if a larger \( k \) is possible. The factors of the form \( 49^k + 1 \) for \( 49^{126}-1 \) arise from the factorization \( x^{2m} - 1 = (x^m-1)(x^m+1) \). The term \( x^m+1 \) is a factor. Here \( 2m = 126 \implies m=63 \). Further splitting \( 49^{63}-1 \) or \( 49^{63}+1 \) would yield factors with lower powers or odd powers not in the form \( 49^k+1 \) with \( k>63 \). Thus, 63 is the greatest such integer. 
Step 4: Final Answer:

The greatest positive integer \( k \) is 63.