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

MATH220 April 2011

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

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[show] 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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[show] 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.

[show] 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.