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I am trying to solve the following question:

Determine $n$ between $0$ and $19$ such that $(2311)(3912) = n \mod 20$.

Where I am now:

(2311) = 11(mod 20)

(3912) = 12(mod 20)

Multiply the 11X12 to get 132;

So: (2311)(3912) = 132(mod 20)

My question is, do I need to further simplify this result or is my solution above in a generally acceptable form. Sorry if this is a really lame question, I am still learning this subject.

user401806
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1 Answers1

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Hint:

$$\begin{cases}2,311=11\pmod{20}\\{}\\ 3,912=12\pmod{20}\end{cases}\;\;\;\implies 2,311\cdot3,912=11\cdot12\pmod{20}=\ldots$$

DonAntonio
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  • A good hint that the OP should get $n=12$. – Juniven Acapulco Dec 25 '16 at 05:37
  • I got to the point of (2311)(3912) = 132(mod 20); but do I need to do anything else to this answer such as further simplification if possible or this is enough? – user401806 Dec 25 '16 at 16:49
  • What residue do you get when you divide $;132;$ by $;20;$ ? That is your best answer...and this is plain definition. Perhaps you should go over this in your otes. – DonAntonio Dec 25 '16 at 18:47
  • So further simplification is needed until the remainder term is in the lowest value for the modulo number? I am trying to find a generalized approach to this type of problem. Thanks!! – user401806 Dec 25 '16 at 19:02
  • @user401806 That's the usual point, though for any formal use you can use any representative of a residue class. Many authors require the smallest such residue which is non-negative. – DonAntonio Dec 25 '16 at 21:56
  • Great feedback! Thanks so much! – user401806 Dec 25 '16 at 22:25