the remainder is

$7^{103}\equiv{49}^{51}\times 7\pmod {25}\equiv{(-1)}^{51}\times 7\pmod{25}\equiv-7\pmod{25}\equiv18\pmod{25}$

Instead of analyzing $7^{103} \pmod {25}$ , let's analyze $7^{n} \pmod {50}$ for small $n$ .

$7^{1} \pmod {50} \equiv 7 \pmod {50}$

$7^{2} \pmod {50} \equiv -1 \pmod {50}$

$7^{3} \pmod {50} \equiv -7 \pmod {50}$

$7^{4} \pmod {50} \equiv 1 \pmod {50}$

$7^{5} \pmod {50} \equiv 7 \pmod {50}$

$7^{6} \pmod {50} \equiv -1 \pmod {50}$

$7^{7} \pmod {50} \equiv -7 \pmod {50}$

$7^{8} \pmod {50} \equiv 1 \pmod {50}$

There is a clear pattern. $7^{4k+n} \pmod {50} \equiv 7^{n} \pmod {50}$ , where $k \in \Z^{+}$ .

From this pattern, we can conclude that

$7^{103} \pmod {50} \equiv 7^{3} \pmod {50}$ .

We can calculate that $7^{3} = 343$ , and $343 \pmod {25} = \boxed {18}$ .

(This is my first $\LaTeX$ solution, please cut me some slack if my formatting is improper)