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This has more to do with place value.

Consider the image:

The key is the fact that $3^2=9$

  • All the value below $3^2$ ($3^0$ and $3^1$) has to be accommodated in $9^0$

    which also means

  • all the values in $9^0$ have to be accommodated in the two places below $3^2$ ($3^0$ and $3^1$)

Similarly for any other places.


This rule also applies to conversion between binary and octal, or binary and hexadecimal, or conversion between any base $k$ and base $k^n$

In general, for conversion of a number $N$ in any base $k$ to base $k^n$, each digit of $N_{k^n}$ get converted to n digits of $N_k$.

For example:

$102201_3$

=$1.3^5+0.3^4+2.3^3+2.3^2+0.3^1+1.3^0$

=$(1.3^1+0)3^4+(2.3^1+2)3^2+(0.3^1+1)3^0$

=$3.9^2+8.9^1+1.9^0$

$\therefore102201_3$=$[10_3|22_3|01_3]_9=381_9$

So, $(2110201102220)_3$ = $[02_3|11_3|02_3|01_3|10_3|22_3|20_3]_9=2421386_9$

 

https://math.stackexchange.com/questions/1627453/why-can-we-convert-a-base-9-number-to-a-base-3-number-by-simply-converting-e

 

edited by

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