Sunday, April 30, 2017

elementary number theory - Solving simultaneous linear congruences.

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.

No comments:

Post a Comment

analysis - Injection, making bijection

I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...