The remainder when \(121^{103}\) is divided by 101 is calculated using Fermat's Little Theorem, which applies since 101 is a prime number. By simplifying \(121\) modulo \(101\), it is found that \(121 \equiv 20 \pmod{101}\). Further calculations show that \(121^{101} \equiv 20 \pmod{101}\) and \(121^2 \equiv 97 \pmod{101}\). Ultimately, combining these results leads to \(121^{103} \equiv 21 \pmod{101}\). Therefore, the final answer is that the remainder is 21.