Question about the formula for switching bases of numbers

Question

To express the base $10$ number $5213$ in base $7$, all I have to do is simply divide by $7$ as follows:

$5213 \div7=744$ with remainder $5$

$744\div 7=106$ with remainder $2$

$106\div7=15$ with remainder $1$

$15 \div7=2$ with remainder $1$

Hence $5213$ in base $7$ is $21125$.

Now the formula for this operation is as follows:

$N=a_nr^n+a_{n-1}r^{n-1}+...+a_2r^2+a_1r+a_0$ where $N$ be the given number, and $r$ the radix of the proposed base.

To find the values $a_0,a_1,a_2...a_n$, divide $N$ by $r$, then the remainder is $a_0$, and the quotient is:

$a_nr^{n-1}+a_{n-1}r^{n-2}+...+a_2r+a_1$

Then continue the operation until all required digits have been acquired.

My question is if you divide $N$ by $r$, then the formula becomes thus:

$\dfrac{N}{r}=a_nr^{n-1}+a_{n-1}r^{n-2}+...+a_2r+a_1+\dfrac{a_0}{r}$ and obviously the remainder is not simply $a_0$ but rather $\dfrac{a_0}{r}$. But in the above calculation with $5213$ the remainder is $a_0$, so I'm confused about what's going on here.

Answer 1

When you divide $N$ by $r$, obtaining $q$ with remainder $x$, it means $N = q r + x$.

The remainder $x$ is not $x/r$.

Answer 2

We have

$$N = a_n r^n + \ldots + a_1 r^1 + a_0 = A_1 r + a_0$$

Which gives us that the remainder when dividing $N$ by $r$ is $a_0$. Then, we remove that remainder from $N$ and then divide by $r$ (knowing that what we get will still be an integer):

$$\begin{eqnarray} \frac{N - a_0}{r} & = & \frac{A_1 r}{r} \\ & = & A_1 \\ & = & a_n r^{n-1} + \ldots + a_1 & = & A_2 r + a_1 \end{eqnarray}$$

For example, when you convert $5213$ into base $7$ what's happening is the following:

$$\begin{eqnarray} 5213 & = & \square \times 7 + 5 \\ (5213 - 5) \div 7 & = & 744 \\ & = & \square \times 7 + 2 \\ (744 - 2) \div 7 & = & 106 \\ & = & \square \times 7 + 1 \end{eqnarray}$$

and so on.