Octal System Conversion So I am trying to change base ten 1,398 to an octal equivalent. I had assumed that I would find it by taking 1,398 and changing it like base-four. So 1,398 is $1*8^3 + 3*8^2 +9*8^1 + 8*8^0*$ or $512+192+72+8=784$ but I guess the answer is 2566. Where am I wrong?
 A: Well, you want to express $1398$ is the octal base $\{0,\dots,7\}$. That means more precisely that you want to find coefficients $a_0,\dots,a_n$ such that
$$1398=\sum_{i=0}^na_i\cdot 8^i$$
The string $a_0a_1\dots a_n$ is then called the octal representation of $1398$, which is sometimes written as $1398={(a_0a_1\dots a_n)}_8$.
In your concrete example, you can actually see by a little trial and error, that
$$1398=2\cdot 8^3+5\cdot8^2+6\cdot 8^1+6\cdot 8^0$$
and thus $1398=2566_8$.

If you want a more precise method to determine the coefficients for a number $n$ in base $8$, you can follow this:


*

*Start with $i=0$.

*Set $a_i = n\;\mathrm{mod}\;8$, the remainder of $n$ after division by $8$.

*Set $n = \lfloor n / 8\rfloor$.

*Increment $i$ and continue at 2.


Note, that all the things I wrote generalize to different bases, I've just tried to tailor it more to your particular problem.
A: To methodically perform the conversion:
$1398 \mod 8 = 6$  ($8^0$ place octal digit)
$\lfloor 1398 / 8 \rfloor = 174$
$174 \mod 8 = 6$ ($8^1$ place octal digit)
$\lfloor 174 / 8 \rfloor = 21$
$21 \mod 8 = 5$ ($8^2$ place octal digit)
$\lfloor 21 / 8 \rfloor =2 $
$2 \mod 8 =2$ ($8^3$ place octal digit)
$\lfloor 2/8\rfloor =0$ 
So $1398_{10} = 2566_8$
