I'm struggling when solving the simultaneous linear congruences $$x\equiv 3 \pmod{101^{1000}}$$ and $$x\equiv 3 \pmod{7^{200}}$$ where the moduli are very large. I haven't got an issue when solving more reasonably sized moduli.
Could I solve this by reducing them to $x\equiv 3 \pmod{101}$ and $x\equiv 3 \pmod7$? I did this and I got $x\equiv 3 \pmod{707}$ using the Chinese Remainder Theorem. Could I somehow use this result to be $mod7^{200}101^{1000}$ or have I approached this problem completely wrong?
Thanks in advance.
Actually, in this case the equation can be solved easily. Just use the fact that $$101^{1000} \mid x-3$$ $$7^{200} \mid x-3 $$ Because $\text{gcd}(101^{1000},7^{200}) = 1$, these two equations are equivalent with $$101^{1000}.7^{200} \mid x-3$$ So your solutions are all the numbers that bear this property.