The remainder, when $$7^{103}$$ is divided by 23 , is equal to :

We can use the Fermat’s Little Theorem i.e. $$a^{p-1} \equiv 1 \pmod{p}$$

Here, $$a = 7$$ and $$p = 23$$. Since 23 is prime and 7 is not a multiple of 23, we can apply the theorem:

$$7^{22} \equiv 1 \pmod{23}$$

Now, we have to break down 103 as a multiple of 22. 

$$103 = (22 \times 4) + 15$$

$$\therefore\ $$ $$7^{103}=7^{22\times4+15}=\left(7^{22}\right)^{^4}\times7^{15}$$

We know that $$7^{22} \equiv 1 \pmod{23}$$.

$$7^{103} \equiv 1^4 \cdot 7^{15} \equiv 7^{15} \pmod{23}$$

Now, we need to calculate: $$7^{15}$$ mod 23

We know $$7^2=49\ ≡\ 3$$ (mod 23)

Similarly,  $$7^4=49\ =3^2≡9$$(mod 23)

$$7^8=49\ =9^2=81≡12$$(mod 23)

Now, $$7^{15}=7^8\times7^4\times7^2\times7$$

Substituting the values: $$12\times9\times3\times7$$

$$12\times9=108≡16$$ (mod23)

$$16\times3=48≡2$$ (mod23)

$$2\times7=14$$ (mod23)

Hence, the final answer is 14.