HSC Stuff > HSC Mathematics Extension 2
Mathematics Extension 2 Challenge Marathon
RuiAce:
We prove this, by demonstrating that in every set of \(n\) objects, all \(n\) of them have the same colour. Note that this is an equivalent statement, and thus is also valid to prove.
When \(n=1\) this is trivially true, because any set that has only has one object will only have one colour, so all objects must have the same colour.
Assume that every set with \(k\) objects have the same colour. Consider what happens when \(n=k+1\).
For any arbitrary set with \(k+1\) elements, we may label the elements in it, so that the set takes the form \( \{1,2,3,\dots,k,k+1\} \). A subset of this set would just be the first \(k\) elements, i.e. the set \( \{1,2,3,\dots,k\} \). Since this is one of the many sets that have \(k\) elements, from our inductive assumption all elements in this set must have the same colour.
Now, another subset of this set would be the last \(k\) elements, i.e. the set \( \{ 2,3,\dots,k,k+1\} \). Again, this is one of many sets that have \(k\) elements, so all the elements in this set must also have the same colour.
But these two sets have overlaps, so by consequence all elements in \( \{1,2,3,\dots,k,k+1\} \) must have the same colour. Hence, all elements in an arbitrary set with \(k+1\) elements all have the same colour.
So it follows by induction that in every set of \(n\) objects, all colours of the objects in said sets are the same, and hence every colour must be the same.
This is obviously wrong. But why?
RuiAce:
This may be hard to believe, but I promise this question is doable via only 4U methods.
(Won't provide hints unless someone actually attempts it.)
Opengangs:
About the Putnam CompetitionThe William Lowell Putnam Mathematical Competition (often just abbreviated to Putnam) is an annual Mathematics competition geared towards undergraduates, who all compete for prize money and scholarships to some of the most prestigious and highly regarded universities in the world. Examples include: Harvard University, MIT, Princeton, Carnegie Mellon, etc.
The test is divided up into halves (A and B). It's a typical 6 hour long paper (3 each), with only 12 questions. The maximum amount of marks awarded for each question is 10 marks. Even though there are 120 points to be awarded, the median is only around 1-5 points. The top ever score was 63/120 by students at Harvard University.
The very last question was a Question 6 from a previous Putnam paper, and the goal here is to walk you through the process by first depicting an easier way in approaching mathematical thinking.
TheAspiringDoc:
--- Quote from: RuiAce on January 21, 2018, 06:05:15 pm ---This may be hard to believe, but I promise this question is doable via only 4U methods.
(Won't provide hints unless someone actually attempts it.)
--- End quote ---
I’m gonna guess the integrand is an odd function and therefore the definite integral is 0?
RuiAce:
--- Quote from: TheAspiringDoc on May 16, 2018, 06:22:26 pm ---I’m gonna guess the integrated is an odd function and therefore the definite integral is 0?
--- End quote ---
You'll find that \(f(-x) \neq -f(x)\).
Edit: I had a go at this question again. I will post the answer, but not any solution yet.\[ \frac\pi{8072} \]
________________
Seeing as though this is being used as a challenge, I'll just explain a bit about what's going on with the notation.
--- Quote from: Ali_Abbas on May 16, 2018, 04:04:00 pm ---Not sure if this is a challenge question or not (depends on whether there's a simpler solution than mine which there probably is), but I'll post it here anyways.
I'm only like 85-90% sure my proof of part (i) is correct as it's quite lengthy and slightly complicated but I did put a lot of thought into it so hopefully it's fine.
--- End quote ---
\( \mathbb{N} \backslash \{ 2\} \) is just asking for all natural numbers, i.e. 0, 1, 2, 3, 4, 5, 6, ..., but except 2. So if you read on, we're really just assuming \(n\) is a prime number not equal to 2.
Part i is easy enough to visualise. But be careful how you argue it........that might be harder than actually doing parts ii and iii. In part ii, basically you know that \( \sum \) means to add all the terms. \( \prod\) means to multiply them instead.
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