I am studying congruence classes and having a hard time. I have two questions about the example below.
1) In the first step why do they replace $(2 * 3 * 4)$ with $2$ ?
2) In each step, how and why are they reducing for example $60$ $mod$ $11$ to $5$ $mod$ $11$ and $35$ $mod$ $11$ to $2$ $mod$ $11$?
On
Maybe you didn't learn the definitions yet ?
$a\equiv b\pmod{n}$ means that $a-b$ is a multiple of $n$ (in other words : there exists an integer $k$ such that $b=ka$).
Now $2\times3\times4=24$ and $24-2=22=2\times11$, hence $2\times3\times4\equiv 2\pmod{11}$
Similarly, we see that $60-5=5\times11$ and $35-2=3\times11$, hence $60\equiv5\pmod{11}$ and $60\equiv5\pmod{11}$
That is because congruences are compatible with addition and mulriplication ,exactly like equality:
If $a\equiv b $ and $a'\equiv b'\mod m$, then
Thus for instance, since $4!=24\equiv 2\mod 11$, we have $$6!=4!\cdot5\cdot6\equiv(2\cdot 5)\cdot6=10\cdot 6=60\equiv 5\mod11.$$
The last step is because a number is equivalent, by definition, to its remainder ($5$) upon Euclidean division by $11$.