Yeah i know these: 5^2003=3 mod7 5^2003=4 mod11 5^2003=8 mod13 and 1001=7x11x13 but i couldnt find that 5^2003=y(mod1001) can anyone express clearly how to conclude this problem by using chinese remainder theorem? thanks in advance.
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Originally Posted by eyke Yeah i know these: 5^2003=3 mod7 5^2003=4 mod11 5^2003=8 mod13 and 1001=7x11x13 but i couldnt find that 5^2003=y(mod1001) can anyone express clearly how to conclude this problem by using chinese remainder theorem? thanks in advance. I get answer 125 by Euler's Theorem. 5^2003 = 5^(1000*2+3) = [5^(1000*2)][5^3] 5 and 1001 are relative primes by Euler's Theorem, we know 5^1000 = 1 (mod 1001) [5^(1000*2)][5^3] (mod 1001) = 1 * 5^3 (mod 1001) =125 hope this is helpful
Originally Posted by eyke Yeah i know these: 5^2003=3 mod7 5^2003=4 mod11 5^2003=8 mod13 You have: . The congruences can be written as , . This gives and , this combines into . We can write, and . Therefore,
Originally Posted by ThePerfectHacker You have: . The congruences can be written as , . This gives and , this combines into . We can write, and . Therefore, that's the answer i needed. thanks.
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