
### Instruction: help me answer this question Let the 3-digit number be ABC, where C is at the unit's place, B at the tens place and A at the hundreds place. As the number has to be even, the digits possible at C are 2 or 4 or 6. That is, the number of possible digits at C is 3. Now, as repetition is allowed, the digits possible at B is 6. Similarly, at A, also, the number of digits possible is 6. Therefore, the total number of possible 3-digit numbers = 6 × 6 × 3 = 108 , think step by step
### Assistant: To find the total number of 3-digit numbers, we need to find the number of possible digits for each place and multiply them together. Here, the number of possible digits for C is 3, for B is 6 and for A is also 6. , The answer is 108
