Thus, it is a min point.
There are no stationary points for this curve.
Evaluate with a calculator…
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
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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Saw this question on the internet and thought its rather interesting. A-levels students should be all buried other their workload, amidst the mock papers and ten year series. Hopefully this might help.
Hint: try and group the number together.
Many students go wow when I evaluate workings, without a GC. I’m not showing off, but it is because I don’t really carry a calculator with me. haha. So some students do ask me how I evaluate the square of numbers so quickly. I thought, in light of A-level’s coming, I should share something interesting.
Firstly, we trace back to a formula that we saw in primary school.
This formula is really going to be the core of us solving any square of numbers.
Next, we just need to split the number that you want to square effectively. So consider,
Some students will ask why not use instead. This is plausible, but we usually are better with addition than multiplication hah.