The number obtained by taking the last two digits of $$5^{2024}$$ in the same order is:

To extract the last two digits of any positive integer, reduce the number modulo $$100$$ because only the remainder upon division by $$100$$ affects the tens-and-units place.

Hence we need $$5^{2024} \bmod 100$$.

Start with the smallest powers of $$5$$ and inspect their residues modulo $$100$$:
$$5^1 = 5 \equiv 5 \pmod{100}$$
$$5^2 = 25 \equiv 25 \pmod{100}$$

Multiply once more to see whether the residue stabilises:
$$5^3 = 5^2 \cdot 5 = 25 \cdot 5 = 125 \equiv 25 \pmod{100}$$

The product $$25 \cdot 5$$ leaves the same remainder $$25$$ when divided by $$100$$ because $$125 - 25 = 100$$. Therefore each additional factor of $$5$$ keeps the residue at $$25$$:

For every $$n \ge 2$$,
$$5^{n} = 5^{n-1} \cdot 5 \equiv 25 \cdot 5 \equiv 25 \pmod{100}$$

Since $$2024 \gt 2$$, we are in this steady-state regime:
$$5^{2024} \equiv 25 \pmod{100}$$

Thus the last two digits of $$5^{2024}$$, written in the same order, are

25