I have the question: Solve simultaneously the three congruences 3x≡4(mod 7), 4x≡5(mod 8) and 5x≡6(mod 9). I have found solutions to the first one and last one, which are x≡6(mod 7) and x≡3(mod 9). However, the middle one is giving me some trouble as using the division rule I would have x≡1.25(mod 2), which would not make sense with the other congruences.
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1Congruences are only defined for integers. – Randall Mar 28 '20 at 21:16
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1As you have noticed, $4x\equiv 5\pmod 8$ is impossible. Indeed, if $8,|,(4x-5)$ for some integer $x$ then we could deduce $4,|,5$, a contradiction. – lulu Mar 28 '20 at 21:16
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Also, what is the "division rule"? – Randall Mar 28 '20 at 21:16
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1If there are integers $x,y$ with $4x+8y=5$ then $5$ is even, contradiction. – Bill Dubuque Mar 28 '20 at 21:44