Bmo 2023/24

This is the first time doing BMO1 (year 11). I found it to be slightly more difficult than previous past papers. I answered half of Q1, fully answered Q2,Q4 and Q5.
What do you reckon the grade boundaries for distinction will be?

Reply 10

Question: For Q4, if half of my proof was based on an assertation that was wrong, but I end up obtaining one of the 2 possible values for n, how many marks would I lose?
Another question: How would the markers mark Q5, the answer is quite easy to obtain (250) but how rigorous does your proof need to be?
Another question: For Q2, is it enough to show that if 2 numbers in the sequence differed by k, then the next number will never be consecutive unless k is 1?
Another question: For Q1, if I didn't get an answer, but I noticed the fact that if you shift one of the letters to the left, all the letters that have been passed by that letter can't move (e.g. OLYMSPIAD, S is shifted to the left so PIAD cannot move), how many marks would I get for that?

Sorry for dumping these questions on you guys, but as I said, this is the first BMO I have ever done. I did do the hamilton last year, but I'm not sure if they mark it in the same way. I would also appreciate it if someone explained to me how BMO1 is marked.

(edited 5 months ago)

Reply 11

dan3214

Original post by kmannnn

Question: For Q4, if half of my proof was based on an assertation that was wrong, but I end up obtaining one of the 2 possible values for n, how many marks would I lose?
Another question: How would the markers mark Q5, the answer is quite easy to obtain (250) but how rigorous does your proof need to be?
Another question: For Q2, is it enough to show that if 2 numbers in the sequence differed by k, then the next number will never be consecutive unless k is 1?
Another question: For Q1, if I didn't get an answer, but I noticed the fact that if you shift one of the letters to the left, all the letters that have been passed by that letter can't move (e.g. OLYMSPIAD, S is shifted to the left so PIAD cannot move), how many marks would I get for that?

Sorry for dumping these questions on you guys, but as I said, this is the first BMO I have ever done. I did do the hamilton last year, but I'm not sure if they mark it in the same way. I would also appreciate it if someone explained to me how BMO1 is marked.

For marking - They use a 0+ 10- mark-scheme, This is the 2022 (round 1) report's description of how they mark:

"What we are looking for are full solutions to problems. This involves identifying a suitable strategy, explaining why your strategy solves the problem, and then carrying it out to produce an answer or prove the required result. In marking each question, we look at the solution synoptically and decide whether the candidate has a viable overall strategy or not. An answer which is essentially a solution will be awarded near maximum credit, with marks deducted for errors of calculation, flaws in logic, omission of cases or technical faults. On the other hand, an answer which does not present a complete argument is marked on a ‘0 plus’ basis; up to 4 marks might be awarded for particular cases or insights."

So I'd guess you will have got 2-4 marks for q1.

For question 5, to get full or near full marks you'll need to prove 250 flaws can be achieved, and that it is the smallest amount obtainable.

I dunno about your answer to question 2, I used strong induction for my proof. They look at differences in the video solutions, so you've got that going for you.

Reply 12

Original post by dan3214

For marking - They use a 0+ 10- mark-scheme, This is the 2022 (round 1) report's description of how they mark:

"What we are looking for are full solutions to problems. This involves identifying a suitable strategy, explaining why your strategy solves the problem, and then carrying it out to produce an answer or prove the required result. In marking each question, we look at the solution synoptically and decide whether the candidate has a viable overall strategy or not. An answer which is essentially a solution will be awarded near maximum credit, with marks deducted for errors of calculation, flaws in logic, omission of cases or technical faults. On the other hand, an answer which does not present a complete argument is marked on a ‘0 plus’ basis; up to 4 marks might be awarded for particular cases or insights."

So I'd guess you will have got 2-4 marks for q1.

For question 5, to get full or near full marks you'll need to prove 250 flaws can be achieved, and that it is the smallest amount obtainable.

I dunno about your answer to question 2, I used strong induction for my proof. They look at differences in the video solutions, so you've got that going for you.

Thanks, that was very helpful.

Could you answer my question about Q4 and half of my proof being wrong? Would I receive a 0+ mark or a 10- mark?

Reply 13

dan3214

Original post by kmannnn

Thanks, that was very helpful.

Could you answer my question about Q4 and half of my proof being wrong? Would I receive a 0+ mark or a 10- mark?

I think it depends on how rigorous and well-reasoned the rest of your answer is. Also the reason why you didn't get both answers for n will mean alot I imagine, if you didnt get them because your method and reasoning was ultimately flawed then it'll be a 0+, but if you had a solid method which should have worked if not for maybe a wrong assumption or mistake or two then I imagine it'll be 10-. Just my thoughts though its my first bmo too. I really hope the boundaries are quite generous this year.

Reply 14

Original post by dan3214

I think it depends on how rigorous and well-reasoned the rest of your answer is. Also the reason why you didn't get both answers for n will mean alot I imagine, if you didnt get them because your method and reasoning was ultimately flawed then it'll be a 0+, but if you had a solid method which should have worked if not for maybe a wrong assumption or mistake or two then I imagine it'll be 10-. Just my thoughts though its my first bmo too. I really hope the boundaries are quite generous this year.

Yeah,

What I did was notice that n2^n + 1 has to be an odd square, not an even square, so n2^n + 1 = (2k+1)^2
Minus 1 from both sides, then completing the square yields the equation: n2^n = 2k(2k+2)
Divide by 4: n2^(n-2) = k^2 + k
Rearrange to get k^2 + k - n2^(n-2) = 0
What I should have done is notice that n2^(n-2) can only be 2 or 6 for k to be an integer, so n can be 2 or 3. This is because the quadratic is only factorizable if n2^(n-2) is 2 or 6.
Instead what I did was complete the square and find k in terms of n, from which I was able to find 1 possible value for n (3).

Reply 15

Original post by dan3214

I think it depends on how rigorous and well-reasoned the rest of your answer is. Also the reason why you didn't get both answers for n will mean alot I imagine, if you didnt get them because your method and reasoning was ultimately flawed then it'll be a 0+, but if you had a solid method which should have worked if not for maybe a wrong assumption or mistake or two then I imagine it'll be 10-. Just my thoughts though its my first bmo too. I really hope the boundaries are quite generous this year.

Yeah same, I hope the grade boundary for distinction is a bit lower than usual (high 10s/low 20s).

Reply 16

Juni2704

I've heard from some acquaintances that they felt questions this year seem much easier. It appears that the grade boundaries may be quite high this year.

(edited 5 months ago)

Reply 17

Original post by Juni2704

I've heard from some acquaintances that they felt questions this year seem much easier. It appears that the grade boundaries may be quite high this year.

I think you can argue both ways. Question 1, supposedly being the most accessible, needed an actual proof which makes it harder, however Q2 and Q5 were surprisingly easy. Also, in my opinion, last year's paper was easier so the grade boundaries might go down.

(edited 5 months ago)

Reply 18

Juni2704