2018 International Mathematical Olympiad

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Here is a piece about a chess and math prodigy, entering UCLA at age of 14, rated 2491!

YouTube

And a third year math student, Julia Ruscher:

YouTube

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I wonder if Terence plays chess also?

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In your opinion? Which is worth something between Jack and sh%t, and Jack's out of town

You just keep showing your extreme ignorance.

If you persist in labelling my , actually perfectly reasonable solution, as 'barely correct' you are showing a basic lack of knowledge.

Your best plan would be to stick to the 'my solution is a neater one than yours' argument. That bit IS true,, sadly for you, I never disputed that.

I simply offered an alternative solution.

And no one can take that away from me.

Talking of which, th e sadly obsessed lap dog is back, maybe you can teach him how to change the subject of formulae in tricky things like V = I/R

Oh yes, your continued refusal to answer/comment on the saddo who thumbed down my first post on here is noted.

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Edit - it was my 2nd post on the thread that was mysteriously thumbed down by person unknown

How is something 'barely correct'

surely it is either correct, or it is incorrect

Originally posted by @blood-on-the-tracks
I see you edited

Do I?

If you would like to produce a counter example where 1/(14n + 3) or
3 minus that expression CAN be simplified (choose your 'n' from infinity integers), then I will withdraw.

Do I need to explain why 1 over an integer cannot be cancelled?

Or that n/(an +/- 1) cannot be simplified, a being any integer ?

Ho ...[text shortened]... eed full explanation and mine does?

Now then, this thumbs down. Ok. I will ask

Was it you?

The algebra is correct. However, it is not at all clear to me that your argument works. You've rewritten the equation roughly as follows:

F(raction) = (ac - 1)/bc = (1/b)(a - 1/c)

Having factorized the equation and made a simplification in the first term it is not obvious you've got any right to make a claim about the reducibility of F, as it was originally written, based on the irreducibility of 1/c.

Lets try

the 1/2 is irrelevant, as that is 'used up' if you recombine the '3 - 1/(14n + 3)' to produce the original expression. Try it.

Now concentrate on the '3 - 1/(14n + 3)'

14n + 3 is an integer, so let us replace it with N

3 - 1/N = (3N-1)/N

i referenced this earlier. I assume you are familiar with the fact that (3N -1)/N is in its simplest form? Cannot be cancelled?

3 x an integer minus 1 (or plus 1) over the integer will never cancel.

(Indeed, a x an integer plus/minus 1 ove r the integer ( a = any integer also) will not cancel either)

Hope that helps a little

I would , of course, have explained all of this also to the learned sages who mark the Maths Olympiad, and hope to score 1/7

Originally posted by @blood-on-the-tracks
Lets try

the 1/2 is irrelevant, as that is 'used up' if you recombine the '3 - 1/(14n + 3)' to produce the original expression. Try it.

Now concentrate on the '3 - 1/(14n + 3)'

14n + 3 is an integer, so let us replace it with N

3 - 1/N = (3N-1)/N

i referenced this earlier. I assume you are familiar with the fact that (3N -1)/N ...[text shortened]... plained all of this also to the learned sages who mark the Maths Olympiad, and hope to score 1/7

I checked the algebra before I posted and do not need to "try it". You've now got a fraction (3N - 1)/N. Suppose N is odd. Then (3N - 1) is even and the factor of 2 you've discarded divides it. You insist that this factor is irrelevant on the grounds that it's 'used up', the problem is this language, it really isn't clear what you mean.

You need to clearly state that you added a factor of 2 to the numerator in the initial rewriting, so that the cancellation with the external term is just an artifact of the method. The reader shouldn't be required to decode your intention, I had to even after your repeated clarification. There's a similar difficulty with Duchess's method in that she invokes Bezout's identity but insists on not naming it, even in her later clarifying post.

Originally posted by @deepthought
[b]I checked the algebra before I posted and do not need to "try it". You've now got a fraction (3N - 1)/N. Suppose N is odd. Then (3N - 1) is even and the factor of 2 you've discarded divides it. You insist that this factor is irrelevant on the grounds that it's 'used up', the problem is this language, it really isn't clear what you mean.

It is to me . I am not a Maths teacher, I just explain as best I can what is patently obvious to me. Sorry if that falls short of your expectations, but , frankly I don't care.

I would agree with your criticism of D64's method to a point, but again, despite her/he being in full aggressive mode at the moment, I would say that in a forum in which Maths is being discussed at some reasonable level of competence, it isn't necessary.

I fully understood the proof D64 presented a couple of pages ago and, I think, complemented the said poster.

You sometimes wonder why you bother