Conversion to base $9$ Where am I going wrong in converting $397$ into a number with base $9$?
$397$ with base $10$ $$= 3 \cdot 10^2 + 9 \cdot 10^1 + 7 \cdot 10^0$$  To convert it into a number with base $9$ 
$$3 \cdot 9^2 + 9 \cdot 9^1 + 7 \cdot 9^0 = 243 + 81 + 7  = 331$$
But answer is $481$?????  
 A: $481_{[9]} = 4\,\cdot\,81 +8\,\cdot9+1\,\cdot\,1 = 324 + 72 + 1 = 397_{[10]}$
The conversion you presented is incorrect, you converted $397_{[9]}$ into $331_{[10]}$, which is not what you want.
To convert it properly:
$397/9 = 44$ (remainder = $1$)
$44/9 = 4$ (remainder = $8$)
$4/9 = 0$ (remainder = $4$)
The remainders give the result: $\boxed{397_{[10]}=481_{[9]}}$
A: You cannot simply keep the same coefficients but change the $10$'s to $9$'s. What you must do is something like the following:
$$\begin{align}397&=3\times \color{red}{10}^2+9\times\color{red} {10}^1+7\times \color{red}{10}^0\\&=3\times(9+1)^2+9\times(9+1)+7\times 1\\&=3\times\color{blue}9^2+(3\times18+3)+\color{blue}9^2+\color{blue}9^1+(7)\\&=4\times \color{blue}9^2+(6\times9)+(10)+\color{blue}9^1\\&=4\times \color{blue}9^2+8\times \color{blue}9^1+1\times \color{blue}9^0\end{align}$$
What you did was simply change the tens in red to nines without doing any intermediate steps.
A: The problem is that
$$ (397)_{10} = 3 \cdot 10^2 + 9 \cdot 10^1 + 7 \cdot 10^0 \ne 3 \cdot 9^2 + 9 \cdot 9^1 + 7 \cdot 9^0 = (397)_9. $$
What we are trying to do is solve
$$ (397)_{10} = 3 \cdot 10^2 + 9 \cdot 10^1 + 7 \cdot 10^0 = a \cdot 9^2 + b \cdot 9^1 + c \cdot 9^0 = (abc)_9 $$
To find $a$ we take as many copies of $81$ from $397$ as we can:
$$ 397 - 4 \cdot 81 = 73. $$
Thus $a = 4$. Next, we take as many copies of $9$ from $73$ as we can:
$$ 73 - 8 \cdot 9 = 1, $$
so $b = 1$. Lastly, we take as many copies of $1$ from $1$ as we can:
$$ 1 - 1 \cdot 1 = 0. $$
Hence $c = 1$. Therefore the answer is $$(397)_{10} = (481)_9.$$
A: As I explained in 2011, one easy way is to write the number in Horner form in the original base, then do the base conversion from the inside-out, e.g. below where $\rm\color{#c00}{red}$ means radix $9$ notation
$$\begin{align} 397 
\,&=\, (3\cdot 10\, +\, 9)\,10 +7\\
&=\, (\color{#c00}{3\cdot 11+10})10+7\\
&=\qquad\quad\ \  \color{#c00}{43\cdot 11}+7\\
&=\qquad\qquad\ \ \ \color{#c00}{473}+7\\
&=\qquad\qquad\ \ \ \color{#c00}{481}\end{align}$$
A: Here is a layout of the conversion algorithm (successive Euclidean divisions)
$$\begin{array}{ccrcrc*{10}{c}}
&&44&&4 \\
9&\Bigl)&397&\Bigl)&44&\Bigl)& \color{red}{\mathbf 4}\\
&&\underline{36}\phantom{7}&&\underline{36}\\
&&37&&\color{red}{\mathbf 8} \\
&&\underline{36} \\
&&\color{red}{\mathbf 1}
\end{array}$$
This is based on Horner's scheme:
\begin{align}
[397]_{10}&=9\cdot 44+ \color{red}1=9(9\cdot 4+\color{red}8)+\color{red}1=9^2\cdot\color{red}4+9\cdot\color{red}8+1\cdot \color{red}1.
\end{align}
