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]2 to a ``1'' in the above; the reduction would go:
as |
|||||
| 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)!