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UKMT Intermediate Mathematical Olympiad 2016

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Original post by Renzhi10122
Yes please :smile:


Done tell me what you think and how did you take this in y11
Original post by TauBilly
I'm assuming that's low though right?

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they asked for us to write our imc scores on it, maybe they use that
Original post by 123Master321
Done tell me what you think and how did you take this in y11


So I got 45 in year 11, I somehow remember that I got 3,10,10,10,10,2 (made a dumb mistake in q1, thought something was prime, turned out it was divisible by 17 or so)... got an invite to Leeds camp, but so did 2 others at my school, and only 1 of us could go, and that 1 wasn't me :frown:
Seems like quite a fun paper though.
Original post by Renzhi10122
So I got 45 in year 11, I somehow remember that I got 3,10,10,10,10,2 (made a dumb mistake in q1, thought something was prime, turned out it was divisible by 17 or so)... got an invite to Leeds camp, but so did 2 others at my school, and only 1 of us could go, and that 1 wasn't me :frown:
Seems like quite a fun paper though.


Sorry to sound nosy but what was your imc score that year
Original post by 123Master321
Sorry to sound nosy but what was your imc score that year


Hmm, I can't remember my exact score, but I got one of the last questions wrong, so 128? I'm not exactly sure how much the question was worth and what the penalty was.
Original post by ben167
I did the Hamilton. For 1 i got 840 (by multiplying out the single digit primes N is a factor of 210). For 2 I just used pythagoras' theorem on the quarter circle and continued from there. For 3 I got 41 (although I think this is wrong). For 4 I got 2 root 3 and for 5 i got 4 possibilities (the integers must be consecutive). What about you?

3 is closer to 14. I can't remember the actual answer though. It isn't 41 I'm pretty sure.
5 was 8 as there were options such as 1,2,4,6,7 (correction, 10, forgot 1,2,4,5,8 1,4,5,7,8. That's a few marks lost)
In question 6, you ignore 5 and 0 as possible ends of numbers and then you get the units by trying to choose a set of three or four that multiply to give a unit of 4. The tens are by doing to the power of three or four until you find the section it's between. Then you do the midpoint of that section to find what that is, find out you need the higher group in both the 3 numbers and 4 numbers. The possibilities are 26, 27, 28, 29, or 77, 78, 79. The second group is correct and gives 474474.
(edited 8 years ago)
How did people do q2 on maclaurin. I had to use the fact that the sum of the first n odd numbers is n^2. Do you think that will pass as common knowledge
I turned it into a prism and looked at it from a side view plan view etc to calculate the perimeter. Just said the area was 36 by adding up each row

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Original post by TauBilly
I turned it into a prism and looked at it from a side view plan view etc to calculate the perimeter. Just said the area was 36 by adding up each row

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Im sorry to tell you this but it was asking for this pattern to hold generally for any such shape which followed the constraints, not just the one in the diagram, which is why you needed the sum of the first n odd numbers

I mean otherwise it would be a very trivial problem indeed
Original post by 123Master321
How did people do q2 on maclaurin. I had to use the fact that the sum of the first n odd numbers is n^2. Do you think that will pass as common knowledge


Not too sure if it will, since it is both easy to prove and makes up an important part of he proof. Maybe if you derived it from an arithmetic series it would be fine, but if you just state that fact on its own you might lose a mark. I'm not entirely sure how they mark Maclaurin so I might be wrong but I did try proving it to be safe.
Oh no! Could you tell me what the constraints were please?

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Original post by 123Master321
Im sorry to tell you this but it was asking for this pattern to hold generally for any such shape which followed the constraints, not just the one in the diagram, which is why you needed the sum of the first n odd numbers

I mean otherwise it would be a very trivial problem indeed


That's really annoying. Its not even that hard to show in the general case as well
Original post by SushiBunny
Not too sure if it will, since it is both easy to prove and makes up an important part of he proof. Maybe if you derived it from an arithmetic series it would be fine, but if you just state that fact on its own you might lose a mark. I'm not entirely sure how they mark Maclaurin so I might be wrong but I did try proving it to be safe.


Yeah its annoying because I know the proof as well... Do you think they will penalise heavily

By the way I found this on the intermediate mentoring scheme so I guess this gives me some hope:
Screen Shot 2016-03-13 at 14.04.00.png49.7KB
Original post by TauBilly
Oh no! Could you tell me what the constraints were please?

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Just adding more and more rows to the thing
Original post by 123Master321
Just adding more and more rows to the thing


Yeah I realise what they were now. I always make one mistake not reading the question...

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Original post by TauBilly
Yeah I realise what they were now. I always make one mistake not reading the question...

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That is annoying, I made a point to fully read all questions
Original post by 123Master321
Yeah its annoying because I know the proof as well... Do you think they will penalise heavily

By the way I found this on the intermediate mentoring scheme so I guess this gives me some hope:


Ooh, cool problem. Couldn't you just say that the average of the 100 numbers must be 100^99 though and not worry about adding squares?

But yeah, it really shouldn't be too big a deal, missing out proving that the odd numbers add to n^2. You'll still get 10-, I'm pretty sure.
Original post by SnakeyEm
3 is closer to 14. I can't remember the actual answer though. It isn't 41 I'm pretty sure.
5 was 8 as there were options such as 1,2,4,6,7 (correction, 10, forgot 1,2,4,5,8 1,4,5,7,8. That's a few marks lost)
In question 6, you ignore 5 and 0 as possible ends of numbers and then you get the units by trying to choose a set of three or four that multiply to give a unit of 4. The tens are by doing to the power of three or four until you find the section it's between. Then you do the midpoint of that section to find what that is, find out you need the higher group in both the 3 numbers and 4 numbers. The possibilities are 26, 27, 28, 29, or 77, 78, 79. The second group is correct and gives 474474.


Could you expand on your explanation of solution 6? I.e. could you explain what you did and why? I did Hamilton as well and I'm sure I got everything correct, but this one. I got a solution to this, but I'd like to see to see how you did it to compare my solution to yours (I assume yours is correct)

Thanks
Original post by SushiBunny
Ooh, cool problem. Couldn't you just say that the average of the 100 numbers must be 100^99 though and not worry about adding squares?

But yeah, it really shouldn't be too big a deal, missing out proving that the odd numbers add to n^2. You'll still get 10-, I'm pretty sure.


That is an interesting way of thinking about it! Oh and by the way solutions are out https://www.ukmt.org.uk/pdfs/SolnsMaclaurin16.pdf can you look at the solution for question 2's area, to me it doesnt seem mathematically rigorous, is it?
Original post by 123Master321
That is an interesting way of thinking about it! Oh and by the way solutions are out https://www.ukmt.org.uk/pdfs/SolnsMaclaurin16.pdf can you look at the solution for question 2's area, to me it doesnt seem mathematically rigorous, is it?


It seems fine to me, even though it is just an outline. It's quite easy to make it rigorous, at least. Essentially you write 3 =2+1, 5=3+2, 7=4+3, ..., 2i+1=(i+1)+i. Then you pair up the terms again in reverse order - e.g. (n-1)+1, (n-2)+2, ..., (n-i)+i. Now you have n sets of numbers that sum to n.
If you really wanted you could use sigma notation, but I don't know how to LaTeX it. They probably didn't want to include that in the official solution so they went with a more intuitive pictorial way that does the same thing.


Looking at the official solutions, I'm annoyed I didn't see the quick way they gave for question 1. Apart from that I'd say I'm satisfied I got basically the same solutions for 2, 5 and the alternative they gave for 4. I have no idea how my solutions for 3 and 6 will go down though.
(edited 8 years ago)

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