Maths notation is often overly complex. Let’s see how this can be better:
This is a pretty cool and interesting question from AMC.
There are four lifts in a building. Each makes three stops, which do not have to be on consecutive floors on include the ground floor. For any two floors, there is at least one lift which stops on both of them. What is the maximum number of floors that this building can have?
Clearly, there are twelve lift stops altogether. Each pair of floors has a lift which connects them. Hence, from combinatorics , there are at most five floors.
Alternatively, we can consider that first lift stops on floors 1, 4, 5. Second lift stops on 2, 4, and 5. Then the third stops on, 3, 4, 5. Lastly, the fourth stops on 1, 2, 3.
Thus, there are at most five floors
The positive integers and satisfy
What is the value of ?
From our combinatorics knowledge, we have the following combinations of factors
Clearly, only fits our condition of the form and
Solving, we find that and .
I came across this interesting algebra/ arithmetic problem the other day.
Do you think this will always work?
Hint: We can use algebra to prove this cool sequence easily.
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