How many three-digit numbers are possible such that the difference between the original number and the number obtained by reversing the digits is 396 ? (no digit is repeated)

Let the three-digit number be 100a + 10b + c. The reversed number is 100c + 10b + a.

The difference is given as:
(100a + 10b + c) - (100c + 10b + a) = 396
99(a - c) = 396
a - c = 4

To satisfy the condition of being a three-digit number upon reversal, c cannot be 0. Possible pairs for (a, c) are (5, 1), (6, 2), (7, 3), (8, 4), and (9, 5). This gives 5 possible pairs.

Since no digit is repeated, the middle digit b can be any digit from 0 to 9 except the two digits used for a and c. Thus, there are 8 possible values for b (10 total digits - 2 used digits).

Total numbers = 5 pairs × 8 values = 40.