I think this one is pretty simple, I've tried to emphasize that in the presentation below. The key is that numbers are read in groups of 2 and each group gets the same conversion process. There is only a single special rule that is discussed at the end:

1) | 7,654,321 |
Start with an arbitrary number. | |||

2) | [07]_{ } | [65]_{ } | [43]_{ } | [21]_{ } |
From left to right group numbers in sets of 2. |

3) | [07]_{3} | [65]_{2} | [43]_{1} | [21]_{0} |
We'll add subscripts for book keeping. |

4) | [07]_{3} | [60+5]_{2} | [40+3]_{1} | [20+1]_{0} |
Now expand the sets into 10's and 1's. |

5) | ([7])_{3} | ([60][5])_{2} | ([40][3])_{1} | ([20][1])_{0} |
Write expansions as seperate numbers. |

6) | ()_{3} | ()_{2} | ()_{1} | ()_{0} |
Go ahead and convert to Ethiopic numbers. |

7) |
() + (3 * {}) |
() + (2 * {}) |
() + (1 * {}) |
() + (0 * {}) |
The subscripts now tell how many 's we need. |

8) |
() + (++) |
() + (+) |
() + () |
() + (0) | |

9) |
() + (+) |
() + () |
() + () |
() + (0) |
Reduce as per = + |

10) |
Group... | ||||

11) | Collect and we're done! | ||||

Note! Except for when we use subscript ``0'' (the far
right side) there is a rule that 1's in the 1's place are absorbed by an
, ,
or on the right. So if we
changed the ``5'' in [65] to a ``1'' in the above; the reduction would go:
as_{2} |
|||||

3) | [07]_{3} | [61]_{2} | [43]_{1} | [21]_{0} |
We'll add subscripts for book keeping. |

: | : | : | : | : |
: |

9) |
() + (+) |
() + () |
() + () |
() + (0) |
Reduce as per = + |

10) |
Group... |

The interesting consequence then is that
can ** only** appear in the one's place (the far right)!