Given $m$, $a$ and $b$ are very big numbers, how do you calculate $ (a*b)\pmod m$ ?
As they are very big number I can not calculate $(a*b)$ directly. So I need another method.
[Math] Find modulo of multiplication of two number
big numbersmodular arithmetic
Related Solutions
You can always compute $a^n$ by repeated squaring and multiplication, e.g., to get $a^{13}$, you square $a$, multiply the result by $a$, square twice, and multiply again by $a$. The exact sequence of operations is visible in the binary expansion of $n$.
You can always prevent the numbers from getting too big by reducing, using the modulus, whenever the numbers exceed the modulus.
There are two notions of "mod", but I'm afraid that your description is incorrect for both.
What you describe, "how many times a number goes into another number while leaving a remainder" is the following: If you are dividing $b$ by $a$, and you write $b = qa + r$ with $0\leq r \lt |a|$, then $r$ is called the "remainder", and $q$ is called the quotient. What you describe is $q$, the quotient.
Now, to the two notions of "modulo":
Modulo operation
This is very common in Computer Science. Here, $\bmod$ is a binary operation (like $+$, $-$, $\times$).
The precise definition varies from language to language. A common one is the following: if $a$ and $b$ are integers, and $a\neq 0$, then $b\bmod a$ (pronounced exactly like it's written) is defined to be the remainder when dividing $b$ by $a$. That is, if we write $b=qa + r$ with $0\leq r\lt|a|$, then $b\bmod a$ is defined to be $r$. Therefore, $7\bmod 3 = 1$ (because $7=2\times 3 + 1$), $43\bmod 13 = 4$ (because $42 = 3\times 13 + 4$), and $1001\bmod 13 = 0$ (because $1001 = 77\times 13 + 0$). This "mod" does indeed only give you nonnegative numbers.
There are other versions of the modulo operator; some, return the "residue class of smallest absolute value" (for example, modulo $3$ will return either $-1$, $0$, or $1$, rather than $0$, $1$, or $2$; and modulo $7$ will return $-3$, $-2$, $-1$, $0$, $1$, $2$, or $3$). Others, as robjohn mentions in comments, return a remainder of the same sign as $a$, others of the same sign as $b$.
Modulo relation
This is more common in mathematics. Given an integer $n$, we define a relation called "congruent modulo $n$" as follows: we say that two integers $a$ and $b$ are "congruent modulo $n$", written $a\equiv b \pmod{n}$, if and only if $b-a$ is a multiple of $n$. Note that $a$, $b$, and $n$ can be positive, negative, or zero.
This is what you are seeing: $-8\equiv 7\pmod{5}$ is true, because $7-(-8)=15$ is a multiple of $5$. $2\equiv -3\pmod{5}$ because $-3-2 = -5$ is a multiple of $5$. $-3\equiv -8\pmod{5}$ because $-8-(-3) = -5$ is a multiple of $5$.
Note that this is a relation: you don't get a number "out of it", like you do with the modulo operation above; this is a statement about whether two numbers are related.
Connection between the two notions
The simplest connection between the two concepts is the following:
Theorem. Let $a$ and $b$ be integers, and let $n\neq 0$ be a nonzero integer. Then $$a\equiv b\pmod{n}\textit{ if and only if } a\bmod n = b\bmod n.$$
That is: $a$ and $b$ are congruent modulo $n$ if and only if they leave the same remainder when divided by $n$.
Proof. Suppose that $a\equiv b\pmod{n}$, and write $a=qn + r$, $b=pn+s$, with $0\leq r\lt |n|$, $0\leq s\lt |n|$. Then $a\equiv b\pmod{n}$ if and only if $(qn+r)-(pn+s) = (q-p)n + (r-s)$ is a multiple of $n$; this occurs if and only if $r-s$ is a multiple of $n$, which occurs if and only if $|r-s|$ is a multiple of $n$.
But $0\leq |r-s|\lt|n|$; the only multiple of $n$ that has absolute value smaller than $|n|$ is $0$. So $r-s$ is a multiple of $n$ if and only if $|r-s|=0$, if and only if $r=s$. Since $r=a\bmod n$ and $s=b\bmod n$, we conclude that $a\equiv b\pmod{n}$ if and only if $a\bmod n = b\bmod n$. $\Box$
Best Answer
If $a,b,m$ are integers then you can use following property :
$a \equiv a_1 \pmod m \land b \equiv b_1 \pmod m \Rightarrow a\cdot b \equiv a_1 \cdot b_1 \pmod m$