Expandindo `999999999` em potências de `13`
`(999999999)_{10} =`
`=1\cdot 13^{8} + 2\cdot 13^{7} + C\cdot 13^{6} + 2\cdot 13^{5} + 3\cdot 13^{4} + A\cdot 13^{3} + 1\cdot 13^{2} + 9\cdot 13^{1} + B\cdot 13^{0}`
`= (\text{12C23A19B})_{13}`
VALORES:
A | B | C |
10 | 11 | 12 |