Because A * 3 = E or E- 1 or E-2 (including carry), a may be equal to 1, 2,3 (if a is greater than 3, e is 2 digits, and since a is the first digit, it is not 0).

When a = 1: E may be equal to 3, 4 and 5 (substituting E*3 = 10n+A, 10n is the decimal number of E*3), it is impossible to find it.

When a = 2: E may be equal to 6,7,8 (substituting E*3 = 10n+A, 10n is the decimal number of E*3), it is impossible to find them.

When A = 3 and E = 9 (substitute E*3 = 10n+A, 10n is the decimal number of E*3), it cannot be found. Because A * 3 = E or E- 1 or E-2 (including carry), a may be equal to 1, 2,3 (if a is greater than 3, e is 2 digits).

When a = 1: E may be equal to 3, 4 and 5 (substituting E*3 = 10n+A, 10n is the decimal number of E*3), it is impossible to find it.

When a = 2: E may be equal to 6,7,8 (substituting E*3 = 10n+A, 10n is the decimal number of E*3), it is impossible to find them.

When A = 3 and E = 9 (substitute E*3 = 10n+A, 10n is the decimal number of E*3), it cannot be found.

Therefore, it is found that A has no solution, that is, the original problem has no solution.