Modular arithmetic How can the equation 2x = 4 have solutions x=2 ,5 under operation multiplication mod 6 ? What is modular arithmetic and how does this operation asserts two answers to this single equation ?
 A: Modulo arithmetic is this, every integer $a \in \Bbb Z$ can be written in the form $a = bm + r$ were $m, b, r \in \Bbb Z$ (division theorem) so now we can define modulo arithmetic mod $b$ as a map which takes an integer $a = bm + r$ to $r$. Provided $a \ge b$. 
In simple terms when ever an integer is larger than or equal; the modulus, here $6$, we replace that integer with the remainder. If the integer is equal to $6$ or a factor of $6$ then the remainder is zero, and we replace the integer with 0. Thus in modulo $6$ there are only 6 "numbers" $0, 1, 2, 3, 4, 5$.
Now we can try $2, 5$ solutions.
$$2 \times 5 = 10$$ but $10 \bmod 6$ we have $$10 = 1 \times 6 + 4$$ here $r = 4$ and so we replace $10$ with $4$.
You can see that $2 \times 2 = 4$ because it is less than $6$ we don't replace it and obviously it is also a solution.
Beyond this if and when you learn about groups, it turns out modulo arithmetic $\bmod b$ can be used to define finite abelian groups of order $b$ typically denoted $\Bbb Z_b$ or in your case $\Bbb Z_6$. Additionally then there are commutative rings, and fields. 
A: "What is modular arithmetic" is a big question.  There's a lot to it, but the first thing is that "arithmetic mod 6" means that $6$ is considered to be $0$.  (Note that clocks work "mod 12".)  So here, $2\times 5 = 10 = 4+6 = 4+0 =4$, so $5$ is a solution to the equation.  So are $11$ and $17$, and so on.
A: Modular arithmetic is something you do everyday without necessarily being aware of it. For example: suppose it takes you seven hours to travel a certain distance, and you have to travel twice that distance starting at 3 p.m. today. At what time tomorrow will you get to your destination? That's arithmetic modulo 12. It's sometimes called "clock arithmetic."
In modular arithmetic, addition and multiplication are exactly the same but with the crucial difference that when the result is greater than the modulus, you divide the result by the modulus and work with the remainder. So in the travel example I gave just now, you calculate $3 + 2 \times 7 = 17$, which is greater than 12, so 17 wraps around to 5. This suggests you'll arrive at your destination at 5 a.m. tomorrow.
Now suppose your clock only has six hours, and you subdivide your day into a.m.1, a.m.2, p.m.1 and p.m.2. You depart on your trip at 3 p.m.1, you'll arrive at your destination at 5 a.m.1.
For your $2x \equiv 4 \pmod 6$ problem, you could see it as: if $x$ is how many hours it takes you to get to some destination or other, and you start your journey at hour zero (a.m.1, a.m.2, p.m.1 or p.m.2), what can $x$ be so that for twice the distance you arrive at hour 4 on the dial? $x$ can be 2 or 5. Or 8 or 11, or 14 or 17, etc., but those just wrap around.
The modulus doesn't have to be a multiple of 6. It can be any integer greater than 1. This is an extremely useful concept in number theory.
