Science:Math Exam Resources/Courses/MATH312/December 2008/Question 03 (c)

MATH312 December 2008

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Question 03 (c)
Determine all integers x that satisfy the following system of linear congruences. $\displaystyle {\begin{aligned}x&\equiv 3\mod {6}\\x&\equiv 1\mod {10}\end{aligned}}$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
It's a trap! Be careful about using the Chinese Remainder Theorem here since 6 and 10 are not coprime. That being said, the same type of idea can be used to get a solution so long as you isolate for 0 and factor at the key step where you cannot invert a non coprime element.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The conditions $\displaystyle {\begin{aligned}x&\equiv 3\mod {6}\\x&\equiv 1\mod {10}\end{aligned}}$ give us that there is a positive integer s such that $\displaystyle {\begin{aligned}x&=3+6s.\end{aligned}}$ Substituting into the second equation yields $\displaystyle {\begin{aligned}3+6s&\equiv 1\mod {10}.\end{aligned}}$ Isolating for 0 gives $\displaystyle {\begin{aligned}2(3s+1)&\equiv 0\mod {10}.\end{aligned}}$ Hence, we have that 5 must divide the term in the bracket, that is $\displaystyle {\begin{aligned}3s+1\equiv 0\mod {5}\end{aligned}}$ This gives $\displaystyle {\begin{aligned}3s&\equiv -1\mod {5}.\end{aligned}}$ Multiplying both sides by 2 gives $\displaystyle {\begin{aligned}s&\equiv -2\equiv 3\mod {5}.\end{aligned}}$ Hence $\displaystyle s=3+5t$ for some integer t. Thus, $\displaystyle {\begin{aligned}x&=3+6s=3+6(3+5t)=3+18+30t=21+30t\end{aligned}}$ and hence all solutions are given by $\displaystyle x\equiv 21\mod {30}$.

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