Science:Math Exam Resources/Courses/MATH312/December 2013/Question 02 (a)
Work in progress: this question page is incomplete, there might be mistakes in the material you are seeing here.
• Q1 (a) • Q1 (b) • Q2 (a) • Q2 (b) • Q3 (a) • Q3 (b) • Q4 (a) • Q4 (b) • Q5 (a) • Q5 (b) • Q6 (a) • Q6 (b) • Q7 (a) • Q7 (b) •
Question 02 (a) 

Find all solutions to the congruence 
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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint. 
Hint 1 

You might be tempted to just say that the solutions are but you must be careful! The modulus 55 is not prime. There is more work to do here. 
Hint 2 

Notice that reducing the given equation modulo 5 and 11 gives the system of congruences
Formulating the question like this should be clear that the Chinese Remainder Theorem would be really useful here. 
Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

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. We follow the solution as outlined in the hints. Notice that reducing the given equation modulo 5 and 11 gives the system of congruences
Solving these gives
As 5 and 11 are coprime, we may apply the Chinese remainder theorem. The above congruences can be solved as follows. For the case
we see that there is an integer m such that . If we further suppose that
we can see by combining these two equations that and so we see that . Writing for some integer n, we have that
and reducing modulo 55 yields . If instead in the second congruence we suppose that
we have that and so we see that . Multiplying both sides by 2 gives and so for some integer n. Hence, we have that
and reducing modulo 55 yields . Now we go back to the first congruence and use instead the congruence given by
we see that there is an integer m such that . If we further suppose that
we can see by combining these two equations that and so we see that . Multiplying by on both sides yields . Writing for some integer n, we have that
and reducing modulo 55 yields . If instead in the second congruence we suppose that
we have that and so we see that . Multiplying both sides by 2 gives and so for some integer n. Hence, we have that
and reducing modulo 55 yields . Thus, the complete set of solutions is given by
