The value of the greatest positive integer k, | vedclass

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Question

(B) Let $S = 49^{125} + 49^{124} + \ldots + 49 + 1$. This is a geometric series with $126$ terms,where the first term $a = 1$ and common ratio $r = 49

Vedclass · Accepted Answer

$63$

Vedclass · Answer

(B) Let $S = 49^{125} + 49^{124} + \ldots + 49 + 1$. This is a geometric series with $126$ terms,where the first term $a = 1$ and common ratio $r = 49$.Using the sum formula $S = \frac{a(r^n - 1)}{r - 1}$,we get $S = \frac{49^{126} - 1}{49 - 1} = \frac{49^{126} - 1}{48}$.The expression given is $48 	imes S = 48 	imes \frac{49^{126} - 1}{48} = 49^{126} - 1$.We want to find the greatest $k$ such that $49^k + 1$ divides $49^{126} - 1$.Note that $49^{126} - 1 = (49^{63} - 1)(49^{63} + 1)$.Since $49^k + 1$ must divide $(49^{63} - 1)(49^{63} + 1)$,we check $k = 63$.If $k = 63$,$49^{63} + 1$ is clearly a factor of $49^{126} - 1$.For $k > 63$,$49^k + 1$ cannot divide $49^{126} - 1$ because $49^k + 1 > 49^{126} - 1$ for $k > 126$,and for $63 Thus,the greatest positive integer $k$ is $63$.

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

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