Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $1$.

share|improve this question

3 Answers 3

up vote 5 down vote accepted

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

share|improve this answer
    
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
1  
@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
1  
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

Hint (pretty huge, but whadda...):

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

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.