1. We need to find the remainder when $7^{103}$ is divided by $23$, i.e. compute $7^{103} \pmod{23}.$

2. Since $23$ is prime, we can use Fermat's Little Theorem: $a^{22} \equiv 1 \pmod{23} \quad \text{for } a \not\equiv 0 \pmod{23}.$ So, $7^{22} \equiv 1 \pmod{23}.$

3. Reduce the exponent $103$ modulo $22$: $103 = 22\cdot 4 + 15.$ Hence, $7^{103} = 7^{22\cdot 4 + 15} = (7^{22})^4\cdot 7^{15} \equiv 1^4\cdot 7^{15} = 7^{15} \pmod{23}.$

4. Now compute powers of $7$ modulo $23$ step by step: $7^2 = 49 \equiv 3 \pmod{23}$ because $49-46=3$.

Then, $7^4 \equiv 3^2 = 9 \pmod{23}.$

Also, $7^8 \equiv 9^2 = 81 \equiv 12 \pmod{23}$ because $81-69=12$.

5. Write $7^{15} = 7^8\cdot 7^4\cdot 7^2\cdot 7.$ Therefore, $7^{15} \equiv 12\cdot 9\cdot 3\cdot 7 \pmod{23}.$

6. Simplify step by step: $12\cdot 9 = 108 \equiv 16 \pmod{23}$ because $108-92=16$.

$16\cdot 3 = 48 \equiv 2 \pmod{23}$ because $48-46=2$.

$2\cdot 7 = 14 \pmod{23}.$

So, $7^{103} \equiv 14 \pmod{23}.$

7. Therefore, the remainder is $14$.

8. Checking options:

• A: $9$ ❌
• B: $6$ ❌
• C: $14$ ✅
• D: $17$ ❌

Thus, the correct option is C.