Let the three-digit number be represented as ABC where A, B, and C are the digits. According to the problem:
- The middle digit B is half of the leftmost digit A which implies B = A/2.
- The rightmost digit C is three times the middle digit B which implies C = 3B.
Hence, the number can be expressed in terms of A as follows:
ABC = 100A + 10B + C
Substitute B = A/2 and C = 3B = 3(A/2) into the expression:
- ABC = 100A + 10(A/2) + 3(A/2)
- = 100A + 5A + (3/2)A
- = 100A + 5A + 1.5A = 106.5A
Since A, B, and C are digits, A must be an integer.
The values for A can be 2, 4, 6, 8 (Such that B and C are also integers since A needs to be even for B and C to be whole numbers).
Considering the condition that reversing the digits gives a new number that is 198 greater:
- 100C + 10B + A = 100A + 10B + C + 198
- 99C = 99A + 198
- C = A + 2
Plugging C = 3B = A + 2 into the equation gives:
- B = (A + 2)/3
- For A = 8, B = (8 + 2)/3 = 10/3 which is not an integer.
- For A = 6, B = (6 + 2)/3 = 8/3 which is not an integer.
- For A = 4, B = (4 + 2)/3 = 6/3 = 2, B is an integer.
By substituting A = 4 and B = 2, calculate C:
Thus, the original number is 426 and the reversed number 624. Checking:
- The difference is 624 - 426 = 198 which is correct.
Therefore, the middle digit B is 2.