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How can we solve $2x\equiv 18\ (\operatorname{mod} 50)$? I'm not sure what to do when the item being modded on the right is not $1$.

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3 Answers 3

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Hint : you can rewrite this equation as $2x-18 = 50k$ for some $k\in \mathbb{Z}$

this is what the modular equation translates to. now solve for x in terms of k to get your solutions

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  • $\begingroup$ x = 25k + 9. But how do I know what k is? $\endgroup$
    – Doug Smith
    Dec 17, 2012 at 17:15
  • $\begingroup$ try plugging in any integer for $k$. then plug your value for $x$ into the original equation to see that its satisfied. it works for any $k$ so long as its an integer. $\endgroup$ Dec 17, 2012 at 17:16
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    $\begingroup$ @DougSmith , for any integer value of $\,k\,$ you get an answer (and, thus, there's an infinite number of answers): $$k=0\Longrightarrow x=9\,,\,x=-2\Longrightarrow x=-91\,,\,k=1\Longrightarrow x=34\,\ldots etc.$$ $\endgroup$
    – DonAntonio
    Dec 17, 2012 at 17:18
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    $\begingroup$ It is worth observing that this method of solution proceeds by finding the one solution to $x\equiv 9 \mod 25$ and this becomes two distinct solutions modulo 50. Not least because this is a phenomenon which happens in other mathematical contexts - where a number of objects (here solutions modulo 50) in some way "sit over" a single object in a simpler context (here solutions modulo 25). $\endgroup$ Dec 17, 2012 at 17:26
  • $\begingroup$ @AndréNicolas surely you typo'd that, seems it should be $x \equiv 9 \mod 25$ $\endgroup$ Dec 17, 2012 at 17:27
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Hint (pretty huge, but whadda...):

$$2x=18\pmod{50}\Longleftrightarrow 2x=18+50k\,\,,\,k\in\Bbb Z\Longleftrightarrow x=9+25k\Longleftrightarrow\ldots$$

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This means that $2x-18=50k$ for some integer $k$. Hence $x=25k+9$.

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