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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.

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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?

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When writing a number in base ten, we understand each digit to be some multiple of a power of ten. This is what the "suppose..." section is trying to generalize. Lets look at a specific example:

$$456 = 4\cdot10^2 + 5\cdot10^1 + 6\cdot10^0$$

Now that the number is expressed as a sum of products, we can apply the theorems. For example, take the fifty part of four hundred and fifty six:

Let \begin{align} a&=5\\ b&=5\\ c&=10^1\\ d&=1 \end{align}

By the third theorem, since $5\equiv5\pmod 3$ and $10^1\equiv 1\pmod 3$, it follows that $5\cdot10^1 \equiv 5\cdot1 \pmod 3$. Note that all powers of ten are equivalent to one modulo three. You can make the same argument for $400\equiv 4\pmod 3$. Therefore:

$$456 = 4\cdot10^2 + 5\cdot10^1 + 6\cdot10^0 \equiv 4 + 5 + 6 \pmod 3$$

The real trick is being able to make this argument for an arbitrary number with an unknown number of digits. Can you show $10^n \equiv 1^n \equiv 1$?

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