2a26 + 5b7 + 67c = 2026 + 100 x a + 507 + 10 x b + 670 + c = 3203 + 100 x a + 10 x b + c
abc + 2000 = 100 x a + 10 x b + c + 2000
Vậy 2a26 + 5b7 + 67c > abc + 2000
b, a53 + 4b6 + 29c = 100 x a + 53 + 406 + 10 x b + 290 + c = 749 + 100 x a + 10 x b + c
abc + 750 = 100 x a + 10 x b + 10 x c + 750
Vậy a53 + 4b6 + 29c < abc + 750