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# BMO Banter watch

1. (Original post by DFranklin)
Just grinding through options is doable (and what I think most people who get it out under time pressure end up doing - it's what I did).

There's some discussion of this question here: http://www.thestudentroom.co.uk/show...=475982&page=2
Thanks for that.
(Original post by DeanK22)
that is a quadratic diophantine equation that you can solve. Have you come across quadratic diophantine equations before?

infact, here is a site with step-by-step method with the above problem;

I've seen some basic Pell equations, but I didn't realise we could extend the idea like that. Cheers.
2. doing BMO 1 1996.

f(1)=1996 f(1)+f(2)+...+f(n)=n^2.f(n)

find f(1996) i get
Spoiler:
Show
2/1997
can anyone verify?
3. Agreed.
4. (Original post by DFranklin)
Agreed.
yay

when answering i didn't justify that it was the sum of reciprocal triangle numbers, would i have been penalised for this?
5. (Original post by Totally Tom)
yay

when answering i didn't justify that it was the sum of reciprocal triangle numbers, would i have been penalised for this?
Probably.
6. (Original post by Totally Tom)
yay

when answering i didn't justify that it was the sum of reciprocal triangle numbers, would i have been penalised for this?
This was quite a nice question. I'm wondering how you do it using the sum of reciprocal triangular numbers though.
7. (Original post by GHOSH-5)
This was quite a nice question. I'm wondering how you do it using the sum of reciprocal triangular numbers though.
make triangle number partial fraction and stuff cancels leaving the sum to n to equal 1-1/n
8. (Original post by Totally Tom)
make triangle number partial fraction and stuff cancels leaving the sum to n to equal 1-1/n
Ah right. I just went with:
Spoiler:
Show

Generalising this:

But we get lots of things cancelling...

Setting k = 1995, n = 1996, we get the answer.
9. (Original post by GHOSH-5)
Ah right. I just went with:
Spoiler:
Show

Generalising this:

But we get lots of things cancelling...

Setting k = 1995, n = 1996, we get the answer.
nice, i'm just going through papers looking for accessible questions atm strictly avoiding any geometry
10. (Original post by Totally Tom)
nice, i'm just going through papers looking for accessible questions atm strictly avoiding any geometry
ditto
11. Hi,

Thought I might use my existing thread, rather than start another. Just attempting Q2, BMO-1, 1998, and I think I've managed to prove that the remainder is 0, any chance someone could check, as I'm a little unsure?
(Original post by Question 2, BMO-2, 1998)
Let , . For define to be the remainder of when it is divided by 100. What is the remainder when

is divided by 8?
Attempt
We find that [Property 1] where is the nth Fibonacci number ( and for ).

Edit: we can ignore the fact that is the remainder of on division by 100 as, suppose is that remainder, and so for some k; it is clear that .

Let us show that has a period of 6 in modulo 4 arithmetic. The first 6 Fibonacci numbers are 1, 1, 2, 3, 5, 8. Their remainders on division by 4 are 1, 1, 2, 3, 1, 0 respectively. Due to the recurrence defining Fibonacci numbers, it is clearly true that, for all k

.

Notice that this implies that for all , and therefore that for all such n, .

What we have shown about Fibonacci numbers also implies that (due to Property 1) and also that .

Note that there are 1332 terms of the form or , and so their quadratic residues modulo 8 sum to 1332, and there are 333 terms of the form so their quadratic residues modulo 8 sum to 1332, and all the other terms' quadratic residues (terms of the form leave no remainder on division by 8, so adding everything up, we get

.
Thanks in advance for any help
12. A brute force computation says the remainder is definitely 0 (the sum before division is 6594784).

The one thing I don't like about what you've done is "Due to the recurrence defining Fibonacci numbers, it is clearly true that, for all k ..." Reading it, I'm not sure why you think it's true - that is, I can think of several ways of proving it's true, but I don't know which method you actually have in mind. I don't know enough about BMO marking to say if you'd get away with it though.
13. (Original post by DFranklin)
A brute force computation says the remainder is definitely 0 (the sum before division is 6594784).

The one thing I don't like about what you've done is "Due to the recurrence defining Fibonacci numbers, it is clearly true that, for all k ..." Reading it, I'm not sure why you think it's true - that is, I can think of several ways of proving it's true, but I don't know which method you actually have in mind. I don't know enough about BMO marking to say if you'd get away with it though.
Cheers.

Hmm, maybe I ought to make it more rigorous. I suppose we could prove it* inductively. It is true for the first case as above, and suppose it is true for some . Therefore:

And so by induction, the statement is true. Would that be enough?

*It being the following for :

.
14. Yes, that would be fine. It feels a bit overkill, but I didn't see anything markedly better/clearer myself.

What you did before might have been OK - it's in the "it's obvious how to prove this if you are competent" category.
15. (Original post by DFranklin)
Yes, that would be fine. It feels a bit overkill, but I didn't see anything markedly better/clearer myself.

What you did before might have been OK - it's in the "it's obvious how to prove this if you are competent" category.
Thanks again
16. Just had a quick question, so thought I might ask it here; for Q3 BMO-1 2006 [the combinatorics question], could anyone confirm that the answer is 2520?

17. I seem to recall that being the answer - haven't actually redone the question to make sure, but I have a reasonable memory for these things. (and from "2520" I could guess which question you meant, so that's a good sign).
18. (Original post by DFranklin)
I seem to recall that being the answer - haven't actually redone the question to make sure, but I have a reasonable memory for these things. (and from "2520" I could guess which question you meant, so that's a good sign).
Cheers I'm a little confused with Q4 of the same paper. I've attached a drawing of what I think the situation is like, but for the tangent at P to touch T at some point Q, surely it's obvious after this that S and T have the same radius; hence PQ = AP immediately? I think I'm missing something or assuming something I shouldn't
Attached Images

19. Don't ask me about BMO geometry (or indeed anything that involves drawing diagrams...)
20. The line PQ is tangent to the smaller circle at Q.

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