n is even if and only if n leaves remainder 0, 2, 4, 6 or 8 when divided by 10 "Let n be any given positive integer. Prove that n is even if and only if n leaves remainder 0, 2, 4, 6 or 8 when divided by 10".
Am I correct in thinking that with regards to the "if and only if" statement, to prove that a statement is true, we have to prove that both parts are true? Would anyone be able to give me some advice/guidance as to where to begin with this?
 A: Let 
$$E=\{0,2,4,6,8\}$$
then we have:
$$\text{$n$ leaves remainder $0, 2, 4, 6$ or $8$ when divided by $10$}\iff n=10q+r,\quad r\in E$$


*

*If $n=10q+r,\quad r\in E$ then $n$ is even as sum of two even numbers (the if  condition)

*If $n$ is even then the unit number $r$ of $n$ in it's decimal writing belongs to $E$ and then $n-r$ is a multiple of $10$ so $n=10q+r,\quad r\in E$ (the only if condition)
A: When you ask "we have to prove both parts are true?", I can only answer "sort of": 


*

*Yes, if by "both parts" you mean "both directions of implication are true."


That is, what you are being asked to prove is the following:

Let n be any given positive integer. 



*

*$\Rightarrow:\;$ Prove that if $n$ is even, then $ n$ leaves remainder $0, 2, 4,
   6,\;\text{or}\; 8$ when divided by $10$".

*$\Leftarrow:\;$ Prove that if $n$ leaves a remainder of $0, 2, 4, 6, \text{ or } 8$
when divided by $10$, then $n$ is even.
Now use the tips already given to prove each of the above implications.
A: Yes, you have to prove not only that even numbers leave those remainders on being divided by 10, you have to prove that odd numbers don't.
Consider some integer $n$. Then the number $2n$ is even, right? If $2n = 10m + r$, where $0 \leq r < 10$, what is $r$? Must be $r \in \{0, 2, 4, 6, 8\}$. The number $2n + 1$ is odd, and therefore $2n + 1 = 10m + r + 1$ and $(r + 1) \in \{1, 3, 5, 7, 9\}$.
To be extra sure, look at it in a different direction. Suppose we're given $10m + r$ or $10m + r + 1$ and we have to figure out what $n$ is in $2n$. Since $10 = 2 \times 5$, it follows that $n = \frac{10m}{2} + \frac{r}{2} = 5m + \frac{r}{2}$.
Also try to see if you can find a number that leaves a remainder of $0, 2, 4, 6, 8$ when divided by 10 yet is odd.
A: I'm going to assume that by "both parts" you mean something like this:

If and only if PART ONE then PART TWO.

This means that if PART ONE is true, then PART TWO is true as well, but if PART ONE is false, then PART TWO is also false. There is no way for PART TWO to be true if PART ONE is false. Because otherwise you should say "if" instead of "if and only if," and maybe also "the converse is not necessarily true."
Consider for example Wilson's theorem:

If and only if $n$ is prime then $(n - 1)! \equiv -1 \mod n$.

This means that you will never find a composite $n$ such that $(n - 1)!$ leaves a remainder of $n - 1$ when divided by $n$. Contrast this to Fermat's little theorem:

If $n$ is prime, and $a$ is some integer coprime to $n$, then $a^{n - 1} \equiv 1 \mod n$.

Notice that it doesn't say "if and only if." That's because you can find composite numbers such that $a^{n - 1}$ leaves a remainder of 1 when divided by $n$. For example, try $a = 3$ and $n = 91 = 7 \times 13$.
You've been tasked with proving that

If and only if $n \equiv 0 \textrm{ or } 2 \textrm{ or } 4 \textrm{ or } 6 \textrm{ or } 8 \mod 10$ then is $n$ even.

So you have to prove that if $n$ is even it leaves one of those remainders then it is even and that if it does not leave one of those remainders then it is not even. You have to demonstrate, for example, that a number can't leave a remainder of 7 when divided by 10 yet be even.
I do understand that it might seem silly to have to prove something so obvious and basic. But long before Andrew Wiles could prove Fermat's conjecture, he had to learn to prove simple things like this.
A: Hint: When $10$ divide $n=2k,k\in\mathbb {Z^+}$ so we have $k\ge 5$. We can consider $k=5+i$ where $i\in\mathbb Z^+$. So $$n=2(5+i)=10+2i,~i\in\mathbb Z^+$$
