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 March 13th, 2013, 10:30 AM #1 Newbie   Joined: Mar 2013 Posts: 1 Thanks: 0 congruence Hello everyone, My task is to solve this: $\begin{cases} 2x\equiv 42 \pmod{10}\\4x\equiv 42 \pmod{12}\\24x\equiv 42 \pmod{14}\end{cases}$ My attempt to solve: $2x\equiv 42 \pmod{10}, 2|10$ so I can reduce $x\equiv 21 \pmod{5} \Leftrightarrow x\equiv 1 \pmod{5}$ so I can reduce $x=5k+1$ for some $k \in \mathbb{Z}$ $4(5k+1)\equiv 42 \pmod{12} \Leftrightarrow 20k+4\equiv 42 \pmod{12} \Leftrightarrow 20k\equiv 38 \pmod{12} \Leftrightarrow 8k\equiv 2 \pmod{12} \Leftrightarrow 4k\equiv 1 \pmod{6}$ $gcd(4,6)=2,gcd(4,1,6)=1$ are different, so no solution exist. I'm not sure what's next but I can use $x$ from first to third congruence: $24\equiv 42 \pmod{14}$ I know that $2|14$so I can reduce to: $12x\equiv 21 \pmod{7} \Leftrightarrow 3x\equiv 0 \pmod{7} \Leftrightarrow x=7n$ so what's next, what is the answer?
 March 13th, 2013, 02:30 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,932 Thanks: 1126 Math Focus: Elementary mathematics and beyond Re: congruence 2x ? 42 (mod 10) ?
 March 13th, 2013, 02:38 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,472 Thanks: 2039 As 4|12, but not 42, $4x\equiv 42\,\pmod{12}$ has no solutions.

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