Certain 3-digit numbers following characteristics: 1. All the three digits are different. 2. The number is divisible by 7. 3. The number on reversing the digits is also divisible by 7. How many such 3-digit numbers are there?

Let the 3-digit number be 100a + 10b + c. Its reversed number is 100c + 10b + a. Condition 1: All three digits (a, b, c) are different. Also, a and c cannot be 0, as they are the first digits of 3-digit numbers. So a, c belong to {1, 2, ..., 9}, and b belongs to {0, 1, ..., 9}. Condition 2: The number (100a + 10b + c) is divisible by 7. Using the divisibility rule for 7 (100x + 10y + z is divisible by 7 if 2x + 3y + z is divisible by 7), we get: 2a + 3b + c must be divisible by 7. (Equation 1) Condition 3: The reversed number (100c + 10b + a) is also divisible by 7. Similarly, 2c + 3b + a must be divisible by 7. (Equation 2) Subtract Equation 2 from Equation 1: (2a + 3b + c) - (2c + 3b + a) must be divisible by 7. (a - c) must be divisible by 7. Since a and c are distinct digits from 1 to 9, the possible values for (a - c) are 7 or -7. Case 1: a - c = 7 This implies (a, c) can be (8, 1) or (9, 2). a) If (a, c) = (8, 1): Substitute into Equation 1 (2a + 3b + c is divisible by 7): 2(8) + 3b + 1 = 16 + 3b + 1 = 17 + 3b. For 17 + 3b to be divisible by 7, and knowing 17 leaves a remainder of 3 when divided by 7, we have: 3 + 3b must be divisible by 7, or 3(1 + b) must be divisible by 7. Since 3 is not divisible by 7, (1 + b) must be divisible by 7. For b to be a digit (0-9), (1 + b) = 7, which gives b = 6. The digits are (8, 6, 1). They are all different. The number is 861. (861 / 7 = 123). Its reverse is 168. (168 / 7 = 24). This is a valid number. b) If (a, c) = (9, 2): Substitute into Equation 1: 2(9) + 3b + 2 = 18 + 3b + 2 = 20 + 3b. For 20 + 3b to be divisible by 7, and knowing 20 leaves a remainder of 6 when divided by 7, we have: 6 + 3b must be divisible by 7, or 3(2 + b) must be divisible by 7. Thus, (2 + b) must be divisible by 7. For b to be a digit (0-9), (2 + b) = 7, which gives b = 5. The digits are (9, 5, 2). They are all different. The number is 952. (952 / 7 = 136). Its reverse is 259. (259 / 7 = 37). This is a valid number. Case 2: a - c = -7 (or c - a = 7) This implies (a, c) can be (1, 8) or (2, 9). a) If (a, c) = (1, 8): Substitute into Equation 1: 2(1) + 3b + 8 = 2 + 3b + 8 = 10 + 3b. For 10 + 3b to be divisible by 7, and knowing 10 leaves a remainder of 3 when divided by 7, we have: 3 + 3b must be divisible by 7, or 3(1 + b) must be divisible by 7. Thus, (1 + b) must be divisible by 7. For b to be a digit (0-9), (1 + b) = 7, which gives b = 6. The digits are (1, 6, 8). They are all different. The number is 168. (168 / 7 = 24). Its reverse is 861. (861 / 7 = 123). This is a valid number. b) If (a, c) = (2, 9): Substitute into Equation 1: 2(2) + 3b + 9 = 4 + 3b + 9 = 13 + 3b. For 13 + 3b to be divisible by 7, and knowing 13 leaves a remainder of 6 when divided by 7, we have: 6 + 3b must be divisible by 7, or 3(2 + b) must be divisible by 7. Thus, (2 + b) must be divisible by 7. For b to be a digit (0-9), (2 + b) = 7, which gives b = 5. The digits are (2, 5, 9). They are all different. The number is 259. (259 / 7 = 37). Its reverse is 952. (952 / 7 = 136). This is a valid number. We found 4 such numbers: 861, 952, 168, and 259. The final answer is B