x=2 (mod 5), 2x=3 (mod 7), 3x=4 (mod 11)
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hello, Originally Posted by Sally_Math x=2 (mod 5), 2x=3 (mod 7), 3x=4 (mod 11) now use the chinese remainder theorem.
Originally Posted by Moo hello, now use the chinese remainder theorem. ok i know that u start this way:- using a=b+km (for eq. 1) which is x=2+5k then substituting into eq. 2, u get 2(2+5k)=3 (mod 7) then we get 4+20k=3 (mod 7) and then that's where i get stuck so please help me understand the way we finish it.
Originally Posted by Sally_Math x=2 (mod 5), 2x=3 (mod 7), 3x=4 (mod 11) The congruence, is equivalent to which becomes . The congruence, is equivalent to which becomes . Thus, you have to solve, Combine the last two congruences and so, This is equivalent to, We get, Combine,
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