When learning to count, we were taught the digits $0, 1, 2, 3, 4, 5, 6, 7, 8,$ and $9$. We were told that the next number in the sequence is $10$, which consists of the digit $1$ written next to $0$. This is the Hindu-Arabic numeral system, which represents a number using the place-value notation. The number $321$ is interpreted as $3 \times 10^2 + 2 \times 10^1 + 1 \times 10^0$. These calculations are done in base 10, which is partially explained by having 10 fingers (including thumbs) which we use to count.

The use of place-value notation in base 10 is not standard throughout history. In fact, many of the great historical civilizations used their own form of number systems. For example, the Babylonains used a sexagesimal (base 60) numeral system, which allowed them to simply represent several fractions. Due to their influence, we now have 60 seconds in a minute, 60 minutes in an hour, and 360 degrees in a circle.

How does number base representation work? The **base** is the number of unique digits (including zero), which are used in the representation. In this post, $(X)_b$ will denote the number $X$ in base $b$. If a base is not given, it is assumed to be 10.

Suppose that Brilli the ant uses a base 6 system, as she has 6 legs for counting. What would the number $(321)_6$ mean to us?

Working from the right, we see that there is 1 unit of $6^0 = 1$, and then $2$ units of $6^1=6$, and finally 3 units of $6^2$. Hence, we can calculate that

$(321)_6 = 3 \times 6^2 + 2 \times 6^1 + 1 \times 6^0 = 121.$

More generally, we can convert a number in base $b$ into our decimal system via

$(b_n b_{n-1} \dots b_1 b_0 )_b = b_n \times b^n + b_{n-1} \times b^{n-1} + \ldots + b_1 \times b^1 + b_0 \times b^0.$

How do we reverse this process? What would our number $(321)_{10}$ mean to Brilli the ant? We need to find numbers $b_i$, each taking value $0, 1, 2, 3, 4$ or $5$. It can be tedious to guess and check, so we use the following algorithm. We start by performing division by $b$, and record the remainder. We then take the dividend and divide that by $b$ again, and record the remainder. This process continues till the dividend is finally 0. We then read the remainders that have been obtained. As an explicit example, to calculate $321$ in base 6, we perform the steps:

$\begin{array}{l | lll} 321 & 321& = 53 \times 6 &+ 3\\ 53 & 53 &= 8 \times 6 &+ 5 \\ 8 & 8 &= 1 \times 6 &+ 2 \\ 1 & 1 &= 0 \times 6 &+ 1 \\ \end{array}$

As such, the algorithm tells us that $(1253)_6 = 321$.

## 1. Convert $(123)_9$ into base 3.

Solution: We will work through base 10. First, we convert $(12345)_9$ into base 10. From the above, we know that

$(123)_9 = 1 \times 9^2 + 2\times 9^1 + 3 \times 9^0 = 102.$

Next, we use the algorithm to convert to base 3.

$\begin{array} {l | lll} 102 & 102 & = 34 \times 3 &+ 0 \\ 34 & 34 & = 11 \times 3& + 1\\ 11 & 11 & = 3 \times 3 &+ 2 \\ 3 & 3 & = 1\times 3 &+ 0 \\ 1 & 1 & = 0 \times 3 & + 1. \\ \end{array}$

Hence $(10210)_3 = 102 = (123)_9$.

## 2. Why does this algorithm to convert into base $b$ work?

Solution: Suppose we want to convert a number $N$ into base $b$. If $N = (b_n b_{n-1} \ldots b_1 b_0 )_b$, then we know that $N - b_0 = b ( b_n \times b^{n-1} + b_{n-1} \times b^{n-2} + \ldots + b_1 \times b^0 )$ is a multiple of $b$. But since $0 \leq b_0 < b$, this tells us that $b_0$ must be the remainder when $N$ is divided by $b$.

We are now interested in $\frac{ N - b_0} { b}$. Similarly, we obtain that $b_1$ must be the remainder when $\frac{N-b_0} { b}$ is divided by $b$. We repeat this process till we terminate at $0$.

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