A problem from the 2003 Moscow Mathematical Olympiad:
A store has three floors, which are connected only by an elevator. At night the store is empty, and during the workday:
(1) Of the customers who enter the elevator on the second floor, half go to the first floor and half to the third floor.
(2) The number of customers who get out the elevator on the third floor is less than 1/3 the total number of customers who get out of the elevator.
Which is greater, the number of customers who go from the first floor to the second on a given workday, or the number who go from the first floor to the third?
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Denote the number of trips from the first floor to the second floor by n12, and so on. Now we can divide all the day’s elevator trips into three groups: trips to the third floor (n13 + n23), trips from the third floor (n31 + n32), and trips between the first and second floors (n12 + n21). By assumption, the first group makes up less than 1/3 of all trips. Now, if the store is empty at night, and if the floors communicate only by elevator, then the total number of customers who enter the third floor on a given day is equal to the number who leave it. This means that the first group above is the same size as the second. And it follows that the last group is the largest:
(n31 + n32) = (n13 + n23) < (n12 + n21)
If we substitute n21 = n23, we get n13 < n12: More customers go from the first floor to the second than from the first to the third.
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