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 |
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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)!