JEE MAIN - Mathematics (2023 - 13th April Evening Shift - No. 18)
The remainder, when $$7^{103}$$ is divided by 17, is __________
Answer
12
Explanation
$7^{103}=7 \times 7^{102}$
$$ \begin{aligned} & =7 \times(49)^{51} \\\\ & =7 \times(51-2)^{51} \end{aligned} $$
Remainder = $7 \times(-2)^{51}$
$$ \begin{aligned} & =-7\left(2^3 \cdot(16)^{12}\right) \\\\ & =-56(17-1)^{12} \end{aligned} $$
Remainder $=-56 \times(-1)^{12}=-56+68=12$
$$ \begin{aligned} & =7 \times(49)^{51} \\\\ & =7 \times(51-2)^{51} \end{aligned} $$
Remainder = $7 \times(-2)^{51}$
$$ \begin{aligned} & =-7\left(2^3 \cdot(16)^{12}\right) \\\\ & =-56(17-1)^{12} \end{aligned} $$
Remainder $=-56 \times(-1)^{12}=-56+68=12$
Comments (0)
