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

MATH312 December 2008

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[hide] Question 03 (a)
Determine, using no moduli other than 111 in your final answer, all integers x that satisfy the following linear congruence. $\displaystyle 21x\equiv 6\mod {111}$

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!

[show] Hint
First, isolate for 0, factor then argue that one of the terms has to be divisible by 37. Solve that congruence then lift back to 111.

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.
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[show] Solution
Proceeding as suggested in the hint, we have $\displaystyle 3(7x-2)\equiv 0\mod {111}$ Now, as $\displaystyle 111=3\cdot 37$ , we see that we must have $\displaystyle 7x-2\equiv 0\mod {37}$ if the above congruence is to hold. Now, isolating gives $\displaystyle 7x\equiv 2\mod {37}$ We can crunch through the Euclidean algorithm to find the inverse of 7 but here I will use a clever time saving trick. The right hand side is equivalent to $\displaystyle 7x\equiv -35\mod {37}$ Thus, multiplying by the inverse of 7 on both sides yields $\displaystyle x\equiv -5\equiv 32\mod {37}$ . Thus, modulo 111, the solutions are $\displaystyle x\equiv 32,69,106\mod {111}$ seen by adding multiples of 37 to 32.