Monthly Mathematics Competition Notification Watch

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llduncanll
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#21
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#21
(Original post by IntegralNeo)
see the results on 18th September
:confused:
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SsEe
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#22
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#22
Hmmm. I'm interested to see who'll win.

For the record. I found answers to all except question 1 including a generalization of the adding up of number combinations one but decided not to enter due to my not answering Q1.
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burje-t
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#23
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#23
Neo?

Did anyone solve problem 1?
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Fermat
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#24
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#24
(Original post by burje-t)
Neo?

Did anyone solve problem 1?
Yes
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Fermat
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#25
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#25
My solution to problem 1.

1)
Pigeonhole Principle
If there are n pigeons and k pigeonholes and k < n then at least one of the pigeonholes must contain 2 or more pigeons.

We have 74 matches played over 47 days with at least one match per day. Let xi be the number of matches played up and including day no. i, i = 1, 2, … , 47.
Then we can write,
1 <= x1 < x2 < x3 < … < x47 = 74 --------------------------------(1)

Required to prove
That there are subscripts i and j such that xi + 19 = xj.

This implies that there are exactly 19 matches played over the period of the days i+1, i+2, … , j.
Using inequality (1), we can write,
20 <= x1 +19 < x2 +19 < x3 +19 < … < x47 +19 = 93.
What we have now are 94 numbers, x1, x2, …, x47, x1+19, x2+19, … ,x47+19 which have among them 93 different values.
Hence, by the Pigeonhole Principle, two of them must have the same value. Since none of x1, x2, …, x47 are the same, and none of x1 +19, x2+19, …, x47+19 are the same, then one of x1, x2, …, x47 must be the same as one of x1 +19, x2+19, …, x47+19, i.e. there is an i and a j giving,
xi + 19 = xj. Q.E.D.
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Euler
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#26
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#26
(Original post by burje-t)
Neo?

Did anyone solve problem 1?
only two solutions were submitted, namely by fermat and sephonline and both 10/10..well done!
i dont know when any of the mods gonna update the esults post...
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dvs
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#27
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#27
I was bested by question 1.

Well done Fermat & sephonline.
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llduncanll
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#28
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#28
Code:
llduncanll	3	2,3,4,5		  40 		(Gold)
hahahaha i rule :cool:
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llduncanll
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#29
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#29
(Original post by llduncanll)
hahahaha i rule :cool:
although Fermet owns my ass, quite literally :eek:
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Fermat
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#30
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(Original post by llduncanll)
although Fermet owns my ass, quite literally :eek:
Code:
Fermat     5      1,2,3,4,5    50      (Golder)
*chortle* I ruler
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burje-t
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#31
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#31
Nice solution there fermat. I think I actually understood it.

Here's a thought. It looks like people either got 10p or 0p on a question (with a few exceptions). Don't you think 10 is a bit much? 2 or 3 points per question seem more reasonable (eg 3p for complete answer, 2p for a small mistake and 1p for correct method but wrong answer/answer without working etc)? I imagine it would be easier to grade the answers that way.
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Simba
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#32
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Yay I scored 35 :cool: , I'm only 14 years old but I love doing maths hehe, I made a tiny mistake on question 5, and couldn't figure out how to do question 1, but the rest I did all correct !!! Congratulations everyone !!!
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Fermat
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#33
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#33
(Original post by burje-t)
Nice solution there fermat. I think I actually understood it.

Here's a thought. It looks like people either got 10p or 0p on a question (with a few exceptions). Don't you think 10 is a bit much? 2 or 3 points per question seem more reasonable (eg 3p for complete answer, 2p for a small mistake and 1p for correct method but wrong answer/answer without working etc)? I imagine it would be easier to grade the answers that way.
Hi burje,
grading this competion is not simple. IntegralNeo and I talked about this earlier on in this thread - see page 1. The current grading system, with 10p/question and gold, silver, etc is just something to start off with. And depending on the results of this competion and the entries made to it, and comments made about it, this grading can be adjusted. The more competitions we have, and the more results, the better able we will be able to grade competion attempts.
First of all, what is the purpose/aim of these compettions. That should be defined. They are open to all comers - hopefully to encourage people (of all levels) not only to do maths but also to enjoy it. So perhaps the questions shouldn't be too difficult - but not too easy for fear of not providing a challenge to others of greater capability.
And what about the points system for individual questions? 10 points too much? Maybe. How should it be awarded? Points for method, style, ingenuity, ... ? Should the points awarded be different depending on what level a person is at? e.g. level 5 have to put more effort into presentation than level 1. Perhaps we can do it a bit at a time. Make (small) modifications to whatever the current grading system is and see how things go with it?
The more input we get abouit a grading system, the better an idea we will have.

Although I keep on saying "we", this is integralneo's competition and the final decision is up to him. By "we", I just mean "us" in general
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burje-t
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#34
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#34
This set-up with medals for different levels looks quite good. It was mostly for Neo's sake I suggested less points per question. I mean, how can you differ a 6p solution from a 7p one? I propose a 2 point system, something like this (much like the one Mika used):

2p. Complete Solution - Correct answer and evidence of working (that is, evidence that the solution is not found on the internet or purely using calculator etc). Alternatively, complete proof.
1p. Partial Solution - Either correct answer with wrong/no working or wrong answer but correct working. Alternatively, partial proof.

Then you could have a 2* (two-star) or something for especially neat and beatiful solutions. These solutions could then be posted for public envy.
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Euler
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#35
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#35
In reply to Fermat and burje-t.

By looking at the results almost everyone got Gold so perhaps the boundary needs to shift up...(ur thoughts plz)

2nd point about the 10pt system i mark it this way:

0-totally wrong
1-3 - correct method in mind but do it the wrrong way
4-6 - incorrect answer but right method
7-9 - complete solution with few steps missing/arithmetic slip but correct answer (somewhat ambigous)
10 - Complete solution with each step clearly stated and correct answer (elegant solution which anyone will follow with some maths familiarity)

the varation in each strand depends on to what extent that strand has been fulfilled e.g. in 4-6 strand the person with right method all along but incorrect answer due to last 2 steps wrong will get 6 and someone wiht half method wrong gets 4...(ur thoughts on this as well plz!)

and finally ur thoughts on the level system i.e. alevel are on level 3, etc...

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Bezza
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#36
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#36
Hmm, I can't understand why I only got 38 - the only slight mistake I can see is that I only put one of the solutions for q3 and I also had a typo. My solution was:

3) Let first number in arithmetic series be a, then 3 terms are a, a+11, a+22. For geometric series, 3 terms are a-6, a+10, 2a+44. As this is a geometric series, there's a common ratio so (a+10)/(a-6) = (2a+44)/(a+10)
(a+10)^2 = 2(a+22)(a-6)
a^2 + 12a - 364 = 0
(a-14)(a+6)=0
Since looking for positive integers for a, a=14 so 3 terms of arithmetic series are 14, 25, 36 and geometric series are 8, 24, 72 (common ratio is 3)

I somehow managed to type 6 rather than 26! (Can't see why this loses 2 marks!) I left out the other possible series as the question asks "Determine the positive real numbers which form the arithmetic sequence" so the solution with -26 as the first term doesn't fit this criteria!
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Euler
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#37
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#37
(Original post by Bezza)
Hmm, I can't understand why I only got 38 - the only slight mistake I can see is that I only put one of the solutions for q3 and I also had a typo. My solution was:

3) Let first number in arithmetic series be a, then 3 terms are a, a+11, a+22. For geometric series, 3 terms are a-6, a+10, 2a+44. As this is a geometric series, there's a common ratio so (a+10)/(a-6) = (2a+44)/(a+10)
(a+10)^2 = 2(a+22)(a-6)
a^2 + 12a - 364 = 0
(a-14)(a+6)=0
Since looking for positive integers for a, a=14 so 3 terms of arithmetic series are 14, 25, 36 and geometric series are 8, 24, 72 (common ratio is 3)

I somehow managed to type 6 rather than 26! (Can't see why this loses 2 marks!) I left out the other possible series as the question asks "Determine the positive real numbers which form the arithmetic sequence" so the solution with -26 as the first term doesn't fit this criteria!
u didnt lose 2 marks for this question the question indeed asked for postive integer values. However u lost 2 marks for question 2 as u did not provide me with an equation which the question asked for...

(Original post by Bezza)
2) AB is a segment of y = 3x/2 + 3/2, BC is a segment of y = -x + 14, AC is a segment of y=3.
At A y=3 and 3 = 3x/2 + 3/2, x=1 so A(1,3)
At B 3x/2 + 3/2 = -x + 14, so x=5, y = -5 + 14 = 9 so B(5,9)
At C y=3, 3 = -x + 14, x = 11 so C(11,3)

Let P be point (X,Y)
Using PA=PB, (X-1)^2 + (Y-3)^2 = (X-5)^2 + (Y-9)^2
Simplifies to 8X + 12Y = 96, 2X + 3Y = 24

Using PA=PC, (X-1)^2 + (Y-3)^2 = (X-11)^2 + (Y-3)^2
Simplifies to 20X = 120, X = 6, sub into other eqn 3Y = 24-12 = 12, Y=4 so P is (6,4)
where is the equation as asked?
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Bezza
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#38
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#38
(Original post by IntegralNeo)
where is the equation as asked?
oops! Obviously didn't read the question properly :rolleyes: I didn't spot that and thought I'd been harshly marked - sorry!
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Euler
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#39
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#39
(Original post by Bezza)
oops! Obviously didn't read the question properly :rolleyes: I didn't spot that and thought I'd been harshly marked - sorry!
its alright...now u know y primary teachers used to stress "always read the question properly"
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Fermat
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#40
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#40
(Original post by IntegralNeo)
...
By looking at the results almost everyone got Gold so perhaps the boundary needs to shift up...(ur thoughts plz)
Yes, I think the boundary marks should be upped. make level 1 Gold the same as level 2 currently is, then increase by 10 all the medal marks in level 2.
For level 3 up the medal grades by 1 mark so that the 5th question would have to be attempted at least to get a Gold. For example,

1 - Pre GCSE (Gold - 20, Silver - 17, Bronze - 15, Honorary otherwise)
2 - GCSE (Gold - 30, Silver - 25, Bronze - 20, Honorary otherwise)
3 - Advanced (Gold - 41, Silver - 36, Bronze - 31, Honorary otherwise)
4 - Undergraduate (Gold - 45, Silver - 40, Bronze - 35, Honorary otherwise)
5 - Graduate (Gold - 48, Silver - 45, Bronze - 43, Honorary otherwise)

Now see what happens

(Original post by IntegralNeo)
2nd point about the 10pt system i mark it this way:

0-totally wrong
1-3 - correct method in mind but do it the wrrong way
4-6 - incorrect answer but right method
7-9 - complete solution with few steps missing/arithmetic slip but correct answer (somewhat ambigous)
10 - Complete solution with each step clearly stated and correct answer (elegant solution which anyone will follow with some maths familiarity)

the varation in each strand depends on to what extent that strand has been fulfilled e.g. in 4-6 strand the person with right method all along but incorrect answer due to last 2 steps wrong will get 6 and someone wiht half method wrong gets 4...(ur thoughts on this as well plz!)
This seems ok but with a few comments.
Uh, what's the difference between
1-3 - "wrong way" but "correct method"
and
4-6 - "right way" but "incorrect answer"
Could you clarify this a bit more neo?
Also, there is variation of three marks in each of the middle three strands.
Are you happy that no matter who was marking answers, they would all follow the marking scheme by giving the same award of marks within each strand. In other words - could the allocation of marks within a strand be subjective depending on the individual marker.
The smaller the range in any strand - as you have given - the more consistent would marking be by different markers - I think

(Original post by IntegralNeo)
and finally ur thoughts on the level system i.e. alevel are on level 3, etc...

:confused:
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