Question

The greatest positive integer $k$, for which ${49}^k+1$ is a factor of the sum ${49}^{125}+{49}^{124}+\cdots +{49}^2+49+1,$ is

• A. $32$
• B. $60$
• C. $65$
• D. $63$

Answer 1

The correct option is D $63$

$1+49+{49}^2+\cdots +{49}^{125}=\frac{{49}^{126}-1}{49-1}=\frac{({49}^{63}+1)({49}^{63}-1)}{48}=\frac{({49}^{63}+1)\left(\right(1+48{)}^{63}-1)}{48}$
$=\frac{({49}^{63}+1)(1+48I-1)}{48}$ ; where $I$ is an integer.
$=({49}^{63}+1)I$
Greatest positive integer is $k=63$