I'm reading Discrete Mathematics by Kevin Ferland and I'm stuck with exercise 33 of Chapter 0: Write the octal number equivalent of 59.
I'm following this procedure:
59/8 | 3
7/8 | ??
By using an online calculator, the result should be 73, however how can the reminder of 7/8 be 7?
On
Determining the remainder here given by $\frac 78$ is inappropriate.
What we do have is that $\dfrac{59}8 = 7$ with a remainder $3$.
So $$\underbrace{59_{10}}_{59} \underbrace{=}_{=} \underbrace{\color{blue}{\bf 7}\times 8^1}_{56} \underbrace{+}_{+}\underbrace{\color{red}{\bf 3}\times 8^0}_{3}= \color{blue}{\bf 7}\color{red}{\bf3}_{8}$$
When you divide $7$ by $8$ you get an integer quotient of $0$ and a remainder of $7$: $7=0\cdot8+7$. This is the unique representation of $7$ in the form $8q+r$, where $q$ and $r$ are integers, and $0\le r<8$.