Question:
The greatest positive integer k, for which 49k + 1 is a factor of the sum
49125 + 49124 + ..... + 492 + 49 + 1, is:
The greatest positive integer k, for which 49k + 1 is a factor of the sum
49125 + 49124 + ..... + 492 + 49 + 1, is:
49125 + 49124 + ..... + 492 + 49 + 1, is:
Show Hint
Sum the geometric series to get \((49^{126}-1)/48\), then factor \(49^{126}-1\) as \((49^{63}-1)(49^{63}+1)\).
Updated On: Oct 31, 2026
- 32
- 60
- 63
- 65
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The Correct Option is C
Solution and Explanation
The sum is \(S=1+49+49^2+\dots+49^{125}=\dfrac{49^{126}-1}{48}=\dfrac{(49^{63}-1)(49^{63}+1)}{48}\).
Since \(49\equiv1\pmod{48}\), 48 divides \(49^{63}-1\), so \(\dfrac{49^{63}-1}{48}\) is an integer and \(S=\dfrac{49^{63}-1}{48}\cdot(49^{63}+1)\). Hence \(49^{63}+1\) divides S, so k = 63 works.
No larger k works: if \(49^k+1\) divides S then it divides \(49^{126}-1\); because \(49^k\equiv-1\), the order of 49 modulo \(49^k+1\) is 2k, which must divide 126, so k divides 63. The greatest such k is 63 (65 does not divide 63).
Answer: (C) 63
Since \(49\equiv1\pmod{48}\), 48 divides \(49^{63}-1\), so \(\dfrac{49^{63}-1}{48}\) is an integer and \(S=\dfrac{49^{63}-1}{48}\cdot(49^{63}+1)\). Hence \(49^{63}+1\) divides S, so k = 63 works.
No larger k works: if \(49^k+1\) divides S then it divides \(49^{126}-1\); because \(49^k\equiv-1\), the order of 49 modulo \(49^k+1\) is 2k, which must divide 126, so k divides 63. The greatest such k is 63 (65 does not divide 63).
Answer: (C) 63
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