Answer:

$1.00 = 100¢

29A + 39O = 1999

29A = 1999 ⇒ 0 ≤ A ≤ 68

39O = 1999 ⇒ 0 ≤ O ≤ 51

29A + 39O ≡ 9A + 9O ≡ 9(A + O) ≡ 9 (mod 10)

**Note**: Mod 10 means the remainder when the number is divided by 10.
Two numbers are equal mod 10 if after you divide them by 10, their remainders
are equal. Or when you subtract them, their difference is a multiple of 10.
15 and 5 are equal mod 10 since 15 - 5 is a multiple of 10.

∴ A + O ≡ 1 (mod 10)

Since 29 and 39 do not divide 1999, 29(A + O) < 29A + 39O < 39(A + O) ⇒ 51 < A + O < 68.

∴ A + O = 61.

1 | 1 | A | = | 61 | → | 1 | 1 | 61 | = | A | → | 1 | 1 | 61 | = | A | → | 1 | 1 | 61 | = | A | → | 1 | 0 | 38 | = | A | ||||||||||||||||||||||||||||||||||||

29 | 39 | O | 1999 | 29 | 39 | 1999 | O | 0 | 10 | 230 | O | 0 | 1 | 23 | O | 0 | 1 | 23 | O |

**Thus, A = 38 and O = 23 is the unique solution.**