number theory - Substitutions in Modular Arithmetic

I've just learned modular arithmetic today, and am really struggling to understand a certain theorem.

The theorem given to us states the following:

Let m $\in \mathbb{N}$. For any integers a, b, c, and d, if $a \equiv b \pmod m$ and $c \equiv d \pmod m$ then,

1. $a+c \equiv b+d \pmod m$


2. $a-c \equiv b-d \pmod m$


3. $ac \equiv bd \pmod m$

In the next section, the notes state the following: "We can use properties of congruence to prove the (familiar) rule that an
integer is divisible by 3 if and only if the sum of its decimal digits is divisible
by 3. The key is to observe that $10 \equiv 1 \pmod 3$ and so by Theorem 5.10.3 [theorem stated above]
you can change 10 to 1 wherever it occurs. Remember that $3|n$ if and only
if $n \equiv 0 \pmod 3$."

Next, it goes through the proof it was talking about at the beginning of the first quote:

Suppose $n=d_k \cdot b_k + d_{k-1}\cdot b_{k-1} + \dots + d_1\cdot b + d_0$ where $d_k, d_{k-1},\dots, d_0$ are the digits of $n$. Also assume that $3|n$. We now have the following:

\begin{align}
n \equiv 0 \pmod 3 &\iff d_k\cdot 10^k + d_{k-1}\cdot 10^{k-1} + \dots + d_1\cdot 10 + d_0 \equiv 0 \pmod 3\\
&\iff d_k \cdot 1^k + d_{k-1}\cdot 1^{k-1} + \dots + d_1 \cdot 1 + d_0 \equiv 0 \pmod 3
\end{align}
since $10 \equiv 1 \pmod 3$.

I don't quite understand how any parts of the theorem stated above allows for substitution.

Thanks for any help.

Answer

We are not substituting anything. In an intuitive manner (though not completely rigorous), you can think of numbers congruent in a certain modulo as being equal. Hence since $10\equiv 1$ modulo $3$, we have
$$ d_k10^k+d_{k-1}10^{k-1}+\cdots+d_010^0= d_k1^k+\cdots+d_01^0=d_k+\cdots+d_0 \pmod 3.$$
Note that the only difference that occurred is that we changed all the $10$s which occurred in the expression to $1$s [to emphasize the thinking that $10$ and $1$ are really the same element, I've used $=$ in place of $\equiv$]. The reason we do this is because they are congruent mod $3$.

If you want a more formal explanation, note that regardless of $x$ and $n$, if $x=a$, then $ x^2=x\cdot x=a\cdot a=a^2$ modulo $n$ (this is part 3 of your theorem). By induction it is also true that $x^k=a^k$. Hence we can take linear combinations (this is parts 1 and 2 of your theorem) of $(1,x,x^2,\ldots,x^k)$ and get that the result is congruent to the same linear combination of $a$s. In other words, if $p$ is a polynomial and $x\equiv a$ modulo $n$, then $p(x)\equiv p(a)$ modulo $n$ as well. Now, apply this to your problem with $x=10$, $a=1$ and $n=3$. Do you see how everything works out?