I am a 4-digit even number. I am less than 2500. The digits in my hundreds and the tens place is different. The digits in my hundreds place are less than the digits in the one thousands place. The sum of my last 2-digits equal 9. I am _?

2,109

To find the answer, we need to break down the given information into smaller clues and eliminate possible options based on those clues.

1. The number is a 4-digit even number.
Since the number needs to be even, the last digit must be one of {0, 2, 4, 6, 8}.

2. The number is less than 2500.
From this clue, we can eliminate any number greater than or equal to 2500 and focus on numbers less than that.

3. The digits in the hundreds and tens place are different.
This means the hundreds digit cannot be the same as the tens digit.

4. The digits in the hundreds place are less than the digits in the thousands place.
This clue indicates that the hundreds digit must be smaller than the thousands digit.

5. The sum of the last two digits equals 9.
The options for the last two digits are: {09, 18, 27, 36, 45, 54, 63, 72, 81, 90}.

Based on the given clues, we can start by narrowing down the possibilities.

Since the number is less than 2500 and an even number, we can remove all numbers that have 1 or 3 in the thousands place as they will create numbers greater than 2500 when multiplied by 2.
To further narrow it down, we can remove numbers that have a hundreds digit greater than the thousands digit because the hundreds digit needs to be smaller.

Removing the options that violate these clues, we are left with the following possibilities for the hundreds and thousands digit: {1, 2}.

Now, we can combine these possibilities with the options for the last two digits: {09, 18, 27, 36, 45, 54, 63, 72, 81, 90}.

After checking through the remaining possibilities, we can see that the only number that satisfies all the given clues is 2016.

Therefore, the answer is 2016.

Are you missing some information?

2172, 2154, 2136, 1072, 1054, 1036?