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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 $2$ $mod$ $11$?

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    Note that $2\times 3\times 4 =24 = 2\cdot 11 + 2$. Similarly $60=5\cdot 11 + 5$ and $35=3\cdot 11 + 2$. Modular arithmetic, we say that two numbers $x$ and $y$ are equivalent modulo $n$, written as $x\equiv y\pmod{n}$ if and only if $x-y=nk$ for some integer $k$. – 2017-02-27

2 Answers 2

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That is because congruences are compatible with addition and mulriplication ,exactly like equality:

If $a\equiv b $ and $a'\equiv b'\mod m$, then

  • $a+a'\equiv b+b'\mod m$,
  • $aa'\equiv bb'\mod m$,
  • $ka\equiv kb\mod m$,
  • $a^r\equiv b^r\mod m$.

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 $11$.

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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 $60\equiv5\pmod{11}$