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The following system of linear congruences in its given form cannot be solved using the Chinese Remainder Theorem. Can you help me transform the system sufficiently such that the Chinese Remainder Theorem can be applied, without actually solving the system?

$$x \equiv 1 \pmod {15}$$ $$2x \equiv 11 \pmod {21}$$ $$4x \equiv −6 \pmod {35}$$

(I can reduce $x \equiv 1 \pmod {15}$ to $x \equiv 1 \pmod 5$ and $x \equiv 1 \pmod 3$.
But I'm not sure how to reduce the other two congruences to end up with mod $3,5$ and $7$?)

Thanks in advance

  • 1
    Do you know how to simply solve $2x\equiv 11\pmod {21}$ as $x\equiv A\pmod {21}$ for some $A$?2012-12-27
  • 0
    gcd(2,21)=1 and 1 divides 11, so there is 1 in-congruent solution modulo 21. x=16 works, so x≡16(mod21) ?2012-12-27
  • 0
    I think it is difficult not to see a solution in the effort to transform the system! But there may be more than one solution.2012-12-27
  • 0
    I know that x=16 works, but I am trying to reduce the system to modulo 3,5 and 7 rather than 15,21 and 35 because gcd(15,21,35) is not equal to 1. Is this a correct approach?2012-12-27
  • 0
    Yes. That is correct. It works because in both case $x\equiv 1 \ \ (3)$. So, you have consistency. If the same happens on the third equation then you are o.k. I think it works there too because the answer is aganin $16$.2012-12-27

2 Answers 2