The sum of the two digits of a two-digit number is 12, and the difference…

A two-digit number can be written as 10x + y, where x is the tens digit and y is the units digit. The digit-sum and digit-difference given in the question become two simultaneous equations in x and y — solving them shows which digit-values are possible, and if more than one valid pair exists, no single number is uniquely fixed by the conditions.

1. Let the tens digit be x and the units digit be y, so the number is 10x + y.

2. From the digit sum: x + y = 12.

3. From the digit difference: |x − y| = 6, so x − y = 6 or y − x = 6.

4. Case 1: x − y = 6 together with x + y = 12 gives x = 9 and y = 3, i.e. the number 93.

5. Case 2: y − x = 6 together with x + y = 12 gives x = 3 and y = 9, i.e. the number 39.

Both results hold up independently: for 39, the digits 3 and 9 sum to 12 and differ by 6; for 93, the digits 9 and 3 sum to 12 and differ by 6. So the two given conditions produce two equally valid two-digit numbers rather than one.

Since the number is not uniquely determined by the given conditions, the correct choice is None of these.