Ramanujan's Number at the top!

Since the base of the given number is multiple of 50 , we set a simple congruence (for our ease) , $50\equiv 1 \mod 7$

Now raising the powers of both sides of congruence as follows :

$((50^{7})^{7})^{7}=50^{343}\equiv 1 \mod 7$

And at this point , We take Ramanujan's Number !

i.e , $50^{343.1729}\equiv 1 \mod 7$

Now multiplying the whole congruence , by $50^{2}=2500$

We get , $50^{593049}\equiv 2500\mod 17500$ and hence required remainder is

${\color{#20A900} 2{\color{#20A900} 5{\color{#20A900} 0{\color{#20A900} 0}}}}$

By Fermat, we have $125000^{197682}\equiv 1 \pmod{7}$ , since the exponent is divisible by 6. Multiplying through with $125000=50\times 2500$ , we find that $125000^{197683}\equiv 125000\equiv \boxed{2500} \pmod{17500}$

Normal remainder property I felt instead of using any theorems as such.

We can simplify the denominator with numerator value to find remainders in general.

$125000^{197683}/17500 = 50^{197683}/7 = 1$ .

125000/17500 = 50/7 which gives remainder as 1 and 1 power anything is always 1.

So, actual answer will be 1 * 2500 = 2500.