# Solve $2x\equiv 18\ (\operatorname{mod} 50)$

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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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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x = 25k + 9. But how do I know what k is? – Doug Smith Dec 17 '12 at 17:15
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. – MSEoris Dec 17 '12 at 17:16
@DougSmith , for any integer value of $\,k\,$ you get an answer (and, thus, there's an infinite number of answers): $$k=0\Longrightarrow x=9\,,\,x=-2\Longrightarrow x=-91\,,\,k=1\Longrightarrow x=34\,\ldots etc.$$ – DonAntonio Dec 17 '12 at 17:18
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). – Mark Bennet Dec 17 '12 at 17:26
@AndréNicolas surely you typo'd that, seems it should be $x \equiv 9 \mod 25$ – MSEoris Dec 17 '12 at 17:27

$$2x=18\pmod{50}\Longleftrightarrow 2x=18+50k\,\,,\,k\in\Bbb Z\Longleftrightarrow x=9+25k\Longleftrightarrow\ldots$$
This means that $2x-18=50k$ for some integer $k$. Hence $x=25k+9$.