Key Concepts and Formulas

• Permutation: The number of ways to arrange r objects from a set of n distinct objects, denoted by $P(n, r)$ or $_nP_r$, is given by $P(n, r) = \frac{n!}{(n-r)!}$.
• Counting Principle: If there are m ways to do one thing and n ways to do another, then there are m × n ways to do both.
• Number greater than 10,000: For a number to be greater than 10,000, it must be a 5-digit number since we are only using the digits 0, 2, 4, 6, and 8.

Step-by-Step Solution

Step 1: Determine the possible digits for each place value.

Since we want to form numbers greater than 10,000 using the digits 0, 2, 4, 6, 8 without repetition, we need to form 5-digit numbers. Let's consider the possible choices for each digit place: ten-thousands, thousands, hundreds, tens, and units.

Step 2: Count the number of choices for the ten-thousands place.

The ten-thousands place cannot be 0, otherwise the number would not be a 5-digit number. Therefore, we have 4 choices (2, 4, 6, or 8) for the ten-thousands place.

Step 3: Count the number of choices for the thousands place.

After choosing a digit for the ten-thousands place, we have 4 remaining digits to choose from for the thousands place (including 0).

Step 4: Count the number of choices for the hundreds place.

After choosing digits for the ten-thousands and thousands places, we have 3 remaining digits to choose from for the hundreds place.

Step 5: Count the number of choices for the tens place.

After choosing digits for the ten-thousands, thousands, and hundreds places, we have 2 remaining digits to choose from for the tens place.

Step 6: Count the number of choices for the units place.

After choosing digits for the ten-thousands, thousands, hundreds, and tens places, we have only 1 remaining digit to choose from for the units place.

Step 7: Apply the counting principle to find the total number of such numbers.

The total number of 5-digit numbers greater than 10,000 that can be formed using the digits 0, 2, 4, 6, 8 without repetition is:

$4 \times 4 \times 3 \times 2 \times 1 = 96$

Common Mistakes & Tips

• Remember that the first digit (ten-thousands place) cannot be zero. This is a common mistake.
• Be careful not to repeat digits. The problem specifically states that digits are not allowed to repeat.
• The counting principle is crucial for solving permutation and combination problems. Make sure you understand how to apply it.

Summary

We are asked to find the number of 5-digit numbers greater than 10,000 that can be formed using the digits 0, 2, 4, 6, 8 without repetition. The ten-thousands place has 4 choices (2, 4, 6, or 8). Then the thousands place has 4 choices, the hundreds place has 3 choices, the tens place has 2 choices, and the units place has 1 choice. Multiplying these choices together gives us $4 \times 4 \times 3 \times 2 \times 1 = 96$. The question is asking for the last two digits of this number, so the answer is 96, thus the last two digits are 96.

The final answer is \boxed{96}.