The remainder, when $7^{103}$ is divided by $23$, is equal to:
Step-by-Step Solution
Key Concept: Reduce powers of $7$ modulo $23$ by recognising $7^{3}=343=345-2\equiv -2\pmod{23}.$ Then $7^{103}=7\cdot(7^{3})^{34}=7\cdot(-2)^{34}=7\cdot 2^{34}.$
$7^{103}=7\cdot(7^{3})^{34}=7\cdot 343^{34}.$
$343=345-2=23\cdot 15-2\Rightarrow 343\equiv -2\pmod{23}.$ So
$$7^{103}\equiv 7\cdot(-2)^{34}=7\cdot 2^{34}\pmod{23}.$$
Compute $2^{34}\pmod{23}$. By Fermat, $2^{22}\equiv 1$; $2^{34}=2^{22}\cdot 2^{12}\equiv 2^{12}=4096.$
$4096=178\cdot 23+2\Rightarrow 2^{12}\equiv 2\pmod{23}.$
$7^{103}\equiv 7\cdot 2=14\pmod{23}.$
Correct Answer: 4