The remainder when $7^{103}$ is divided by $25$ is

We need to find the remainder when $7^{103}$ is divided by $25$.

We can write $7^{103}$ as $7 \cdot 7^{102} = 7 \cdot (7^2)^{51}$.

We know that $7^2 = 49 \equiv -1 \pmod{25}$.

Raising both sides to the power of $51$, we get:

$(7^2)^{51} \equiv (-1)^{51} \pmod{25}$

$7^{102} \equiv -1 \pmod{25}$

Multiplying both sides by $7$, we get:

$7^{103} \equiv -7 \pmod{25}$

To find the positive remainder, we add $25$ to $-7$:

$-7 + 25 = 18$

Thus, the remainder when $7^{103}$ is divided by $25$ is $18$.

Answer: $18$