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

MATH220 April 2011

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Question 06
Let $n\in \mathbb {Z}$. Prove that $n\equiv 3{\pmod {5}}$ if and only if $3n+1$ is divisible by $5$.

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
Recall that the following are equivalent: $a\equiv b{\pmod {5}}$ $5\mid a-b$ There is some integer $k$ so that $5k=a-b$.

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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Let us first assume that $n\equiv 3{\pmod {5}}$. This is equivalent to the statement that $5\mid (n-3)$ or that there exists some integer $k$ so that $n-3=5k$ or that $n=5k+3$. In such a case, we have that $3n+1=3(5k+3)+1=5(3k)+10=5(3k+2)$ and so since $3k+2$ is an integer, it follows that $5\mid (3n+1)$. So suppose now that $5\mid (3n+1)$; that is, there is some integer $\ell$ so that $5\ell =3n+1$. From this we can conclude that $5\ell =3n+1\iff 5(2\ell )=6n+2=5n+n+5-3=5(n+1)+n-3$ Shifting the first of those terms to the other side we obtain that $5(2\ell -n-1)=n-3$ Since $2\ell -n-1$ is an integer, the result is proven.

Solution 2

Found a typo? Is this solution unclear? Let us know here. 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 $\displaystyle n\equiv 3{\pmod {5}}$ We can substitute this in the expression of 3n+1 and obtain that $\displaystyle {\begin{aligned}3n+1&\equiv 3(3)+1{\pmod {5}}\\&\equiv 10{\pmod {5}}\\&\equiv 0{\pmod {5}}\end{aligned}}$ 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 $\displaystyle 3n+1\equiv 0{\pmod {5}}$ Which we can rewrite as $\displaystyle {\begin{aligned}3n&\equiv -1{\pmod {5}}\\&\equiv 4{\pmod {5}}\end{aligned}}$ 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 2$\cdot$3=6 which is 1 mod 5. So we multiply both sides of the equation by 2 and obtain $\displaystyle 6n\equiv 8{\pmod {5}}$ And so since 6 is 1 mod 5 and 8 is 3 mod 5 we can conclude that $\displaystyle n\equiv 3{\pmod {5}}$ Which shows that n is congruent to 3 mod 5 and finishes our proof.

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