Science:Math Exam Resources/Courses/MATH220/April 2011/Question 06
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Question 06 

Let . Prove that if and only if is divisible by . 
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Hint 

Recall that the following are equivalent:

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Solution 1 

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Please rate my easiness! It's quick and helps everyone guide their studies. Let us first assume that . This is equivalent to the statement that or that there exists some integer so that or that . In such a case, we have that and so since is an integer, it follows that . So suppose now that ; that is, there is some integer so that . From this we can conclude that Shifting the first of those terms to the other side we obtain that Since is an integer, the result is proven. 
Solution 2 

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Please rate my easiness! It's quick and helps everyone guide their studies. Let us first show that if n is congruent to 3 mod 5, then 3n+1 is divisible by 5, that is, 3n+1 is congruent to 0 mod 5. Since we know that We can substitute this in the expression of 3n+1 and obtain that Which proves that 3n+1 is divisible by 5. Now, let's prove the converse, that is that if 3n+1 is divisible by 5, then n is congruent to 3 mod 5. We start with the fact that is 3n+1 is divisible by 5, then it is congruent to 0 mod 5 and write Which we can rewrite as We would like to get some information about n, not 3n but division in modular arithmetic is tricky. In this case, since 3 and 5 are coprime, we know it can be done: the inverse of 3 mod 5 is 2 since 23=6 which is 1 mod 5. So we multiply both sides of the equation by 2 and obtain And so since 6 is 1 mod 5 and 8 is 3 mod 5 we can conclude that Which shows that n is congruent to 3 mod 5 and finishes our proof. 