# 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 $481$?????

• You started with a number in base $9$ and converted it to base $10$ rather than the reverse. Commented Aug 28, 2017 at 15:51
• To convert to base $9$ from base $10$, you could find the highest power of $9$ (e.g. $9,81,729$ ...) that is less than your number, then the largest multiple of that power that is still less than your number will be your left-most digit. So, in the case of $397_{10}$, the largest power of $9$ that will fit is $81$ (since $729$ is too large), and you can fit four $81$s: $4 \cdot 81=324< 397$ (five is too many, since $5 \cdot 81 = 405 > 397$. This means that the leftmost digit must be $4$. We are left with $397 - 4 \cdot 81 =73$. Now, find the largest multiple of $9$ that's less than $73$, etc. Commented Aug 28, 2017 at 15:56
• The operation you've done is convert $(397)_9$ to base $10$. Commented Aug 28, 2017 at 16:04
• One obvious sanity check is that if converting from a bigger base to a smaller number you should get a larger answer and visa versa. You can easily check your answer by converting it back to decimal to see if you get what you started with. Commented Aug 28, 2017 at 16:20

$481_{[9]} = 4\,\cdot\,81 +8\,\cdot9+1\,\cdot\,1 = 324 + 72 + 1 = 397_{[10]}$

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: $\boxed{397_{[10]}=481_{[9]}}$

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.

• I love that the three answers right now are all using different methods to convert bases :) Commented Aug 28, 2017 at 16:03
• I had not seen this method before but it's a nice approach. I would have used repeated division and took the remainders as @DanielCunha did because that's the way I was taught to convert decimal to binary but it works with any base. Commented Aug 28, 2017 at 16:08
• @WarrenHill I sometimes use this approach, especially when the bases are so close together (like $9$ and $10$ are only $1$ apart, so you only pick up small numbers with the bracket expansion).
– Dave
Commented Aug 28, 2017 at 16:10
• @Warren We can do this much more efificiently if we exploit the recursive Horner form of radix notation - see my answer. Commented Aug 28, 2017 at 16:41

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 $$(397)_{10} = (481)_9.$$

• Nice answer, +1! I think you should mention that $9^3 = 729>397$ is too large to be considered, which is why you are starting from $9^2 = 81$, instead of a higher power of $9$. Cheers Commented Aug 28, 2017 at 16:02

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}

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}

• But no division is needed if we use Horner optimally - see my answer. Commented Aug 28, 2017 at 16:39
• That''s fine but it's no so easy in practice to do operations in another base, especially if you have to carry digits. Commented Aug 28, 2017 at 17:14