Math professor and number of rooms

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A math professor discovered that the sum of room numbers to his left equaled the sum of those to his right when leaving his hotel room for a conference. The hotel floor has approximately 300 rooms, and the room numbers are sequential without any skips. The solution reveals there are 288 rooms on the floor, with the professor staying in room 204. The sums of the room numbers on either side of his room are equal, calculated as 20706 for both sides. The discussion also touches on methods for finding such room configurations, suggesting that simplifying the problem can lead to quicker solutions. Further exploration of the problem involves checking for perfect squares related to the room numbers, indicating a mathematical approach to solving similar puzzles.
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A math professor was staying in a hotel before attending a math conference. When he came out of his room, to go to the conference, he found out that the total of room numbers to his left was equal to the total of room numbers to his right. The total numbers of rooms per floor were about 300 and just assume that they are in line. The room numbers were given without skipping any number(even 13 :smile: )

How many rooms are there in the hotel floor and in which room did the professor stay?

It is damn easy solving with any of the computer programs. I did it in VB and excel(Also without these programs. By saying this I am not acting smart but just encouraging you guys). But there is a clue for getting the numbers by which you can skip all unnecessary numbers. Go get it.
 
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Answer for the first floor (whited out)


There are 288 rooms on the floor. The professor is in room 204.
The sum of numbers of the first 203 rooms is
(203 * 204) / 2 = 20706.
The sum of numbers of the last 84 rooms is
(288 * 289) / 2 - (204 * 205) / 2 = 20706.

I looked for numbers n for which sqrt ((n * n + n) / 2) is an integer.


I don't have a solution for the second floor and higher.
 
Jimmy, you got it. The redundant data like, rooms in same floor and all in a line, actually takes care of all silly quesions. Try to further simplify the thing(I believe you can do it and you can cross check it with the whited out logic).

Further simplification results in n*(n+1)/2. So, either n and (n+1)/2 or n/2 and (n+1) should be perfect squares. If you start squaring numbers and check the above possibilities, you need not go morethan 17 trials.
 
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