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.
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