JEE Main 2025 — 29 January, Evening Shift — Question 58

We want the remainder when $7^{103}$ is divided by $23.$ Since $23$ is a prime number, by Fermat's Little Theorem,

$$7^{22} \equiv 1 \pmod{23}.$$ $$103 = 22 \times 4 + 15$$ $$\Rightarrow 7^{103} = (7^{22})^{4} \cdot 7^{15} \equiv 1^{4} \cdot 7^{15} \equiv 7^{15} \pmod{23}.$$

Now reduce $7^{15}$ modulo $23$:

$$7^{2} = 49 \equiv 3 \pmod{23}$$ $$7^{4} \equiv 3^{2} = 9 \pmod{23}$$ $$7^{8} \equiv 9^{2} = 81 \equiv 12 \pmod{23}$$ $$7^{15} = 7^{8} \cdot 7^{4} \cdot 7^{2} \cdot 7$$ $$\equiv 12 \cdot 9 \cdot 3 \cdot 7 \pmod{23}$$ $$= 2268 \equiv 14 \pmod{23}.$$ $$\boxed{\text{Remainder} = 14}$$