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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

  • 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

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The general "strategy" is just to transform $2x\equiv 11\pmod{21}$ into $x\equiv 121\pmod{21}=16\pmod{21}$, where $11$ is $2$'s inverse. From here you can use Chinese reminder theorem.

So solving this we have $x\equiv 1\pmod{3},x\equiv 2\pmod{7},x\equiv 1\pmod{5}$Now we use classical algorithm, we have $70*1+21*1+15*2=70+21+30=121\equiv 16\pmod{105}$.

  • 0
    Well, OP should be able to fill in the details if he knows Chinese reminder theorem.2012-12-27
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$x \equiv 1 \pmod{15}$ $2x \equiv 11 \pmod{21}$ This means $x \equiv a \pmod{105}$ Note that possible values of $a$ consistent with $x \equiv 1 \pmod{15}$ are $\{1,16,31,46,61,76,91\}$. Hence, $2a \in \{2,32,62,92,122,152,182\}$. Now we are given that $2a \equiv 11 \pmod{21}$. Hence, $2a = 32$ is the only possible solution. Hence, $x \equiv 16 \pmod{105}$ Further, you have $2x \equiv -3 \pmod{35}$, which is consistent with $x \equiv 16 \pmod{105}$. Hence, we get that $x \equiv 16 \pmod{105}$