To form even three-digit numbers using the digits $0, 1, 2, 3, 4, 5, 6, 7$, with the conditions that the hundreds and tens digits are greater than 1 and that repetition of digits is allowed, we can proceed step by step.

Step 1: Hundreds place

The hundreds digit must be greater than 1, so it can be any of the digits $2, 3, 4, 5, 6, 7$. Thus, there are 6 choices for the hundreds place.

Step 2: Tens place

Similarly, the tens digit must also be greater than 1. Thus, we have the same 6 choices for the tens place: $2, 3, 4, 5, 6, 7$.

Step 3: Units place

The units digit must make the number even, meaning the digit has to be one of the even digits. From the set {0, 1, 2, 3, 4, 5, 6, 7}, the even digits are $0, 2, 4, 6$. Thus, there are 4 choices for the units place.

Total number of even three-digit numbers

To calculate the total number of possible three-digit numbers, we multiply the number of choices for each digit: $6 \times 6 \times 4 = 144$

So, there are 144 even three-digit numbers that satisfy the given conditions.

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Related Questions:

1. How many odd three-digit numbers can be formed with similar conditions?
2. What if repetition of digits was not allowed? How many numbers would be possible?
3. How many total three-digit numbers (both even and odd) can be formed under the same conditions?
4. What are the possible combinations if only prime digits are allowed for the hundreds and tens places?
5. How many four-digit even numbers can be formed under similar conditions?

Tip:

When solving problems with digit restrictions, always break the task down step by step, analyzing each digit's constraints individually.