HSC 2015 MX2 Marathon ADVANCED (archive) (1 Viewer)

glittergal96 · Jun 13, 2015

Re: HSC 2015 4U Marathon - Advanced Level

simpleetal said:
well done though, I think it was difficult for a question 1 though

I don't think it's that bad as far as competition problems go, but I guess Q1s are often easier.

Were there other interesting problems in this contest?

RealiseNothing · Jun 13, 2015

Re: HSC 2015 4U Marathon - Advanced Level

glittergal96 said:
I don't think it's that bad as far as competition problems go, but I guess Q1s are often easier.

Were there other interesting problems in this contest?

Yer I don't think it's that bad of a Q1 either. The method you have to use (assuming an arbitrary person sees at least 18 etc) is obvious, the main problem would be reaching the contradiction.

It's still better than my year's one where the inequality question didn't even work.

glittergal96 · Jun 13, 2015

Re: HSC 2015 4U Marathon - Advanced Level

RealiseNothing said:
It's still better than my year's one where the inequality question didn't even work.

Really? What was the question?

simpleetal · Jun 13, 2015

Re: HSC 2015 4U Marathon - Advanced Level

glittergal96 said:
I don't think it's that bad as far as competition problems go, but I guess Q1s are often easier.

Were there other interesting problems in this contest?

of course, but as you said q1s are typically easier lol.
I think q6 was quite interesting as well, which involved the use of the chinese remainder theorem but my latexing skills arent good enough for me to post the question. Perhapsthere is someone else on this forum who has done the test and is able to share the question?

simpleetal · Jun 13, 2015

Re: HSC 2015 4U Marathon - Advanced Level

btw realisenothing, how many points did u score? I believe you scored a HD in that comp? I just want to get a general idea of the cut off marks lol.

glittergal96 · Jun 13, 2015

Re: HSC 2015 4U Marathon - Advanced Level

simpleetal said:
of course, but as you said q1s are typically easier lol.
I think q6 was quite interesting as well, which involved the use of the chinese remainder theorem but my latexing skills arent good enough for me to post the question. Perhapsthere is someone else on this forum who has done the test and is able to share the question?

Was it really that tex-heavy? Can you describe it unambiguously without latex?

2azn4u · Jun 13, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Here's the chinese remainder theorem question from this year's senior unsw maths comp paper.

I tried doing it by contrapositive when I couldn't figure out how to do it by contradiction. Haven't yielded any results from it yet.

RealiseNothing · Jun 14, 2015

Re: HSC 2015 4U Marathon - Advanced Level

simpleetal said:
btw realisenothing, how many points did u score? I believe you scored a HD in that comp? I just want to get a general idea of the cut off marks lol.

You don't get given your exact score only the stats etc.

If I had to guess my score, I'd say around 19 or 20/25 (it is out of 30, but since the inequality question didn't work I'm not sure how they marked that?)

I know out of the 5 questions that did work I got 3 of them completely, and partial solutions for the remaining 2.

This was for a mid-HD (I was 8th out of year 12's and around 15th overall from memory)

RealiseNothing · Jun 14, 2015

Re: HSC 2015 4U Marathon - Advanced Level

glittergal96 said:
Really? What was the question?

simpleetal · Jun 14, 2015

Re: HSC 2015 4U Marathon - Advanced Level

RealiseNothing said:
You don't get given your exact score only the stats etc.

If I had to guess my score, I'd say around 19 or 20/25 (it is out of 30, but since the inequality question didn't work I'm not sure how they marked that?)

I know out of the 5 questions that did work I got 3 of them completely, and partial solutions for the remaining 2.

This was for a mid-HD (I was 8th out of year 12's and around 15th overall from memory)

ah okay. Do you have any idea what the general cut off mark for HD is then? For senior? Like for a paper of average difficulty

Drsoccerball · Jun 14, 2015

Re: HSC 2015 4U Marathon - Advanced Level

RealiseNothing said:

$x_1 \geq y_1$

$\therefore \frac{1}{x_1} \leq \frac{1}{y_1}$

$\therefore \frac{1}{x_1} +\frac{1}{x_2} + \frac{1}{x_3} \leq \frac{1}{y_1} + \frac{1}{y_2} +\frac{1}{y_3}$

lel

InteGrand · Jun 14, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Drsoccerball said:
$x_1 \geq y_1$

$\therefore \frac{1}{x_1} \leq \frac{1}{y_1}$

$\therefore \frac{1}{x_1} +\frac{1}{x_2} + \frac{1}{x_3} \leq \frac{1}{y_1} + \frac{1}{y_2} +\frac{1}{y_3}$

lel

That inequality in the question didn't work lol

(Or maybe you were joking with that comment haha)

Drsoccerball · Jun 14, 2015

Re: HSC 2015 4U Marathon - Advanced Level

InteGrand said:
That inequality in the question didn't work lol

(Or maybe you were joking with that comment haha)

Yeah i was joking

but due to the initial condition why arn't i allowed to say that

InteGrand · Jun 14, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Drsoccerball said:
Yeah i was joking but due to the initial condition why arn't i allowed to say that

We don't need $x_1\geq y_1$ .

e.g. $x_1 = 0.1,y_1 = 0.2$ , then compensate by having the other $y_i$ (s) less than the $x_i$ (s) so the given conditions hold.

RealiseNothing · Jun 14, 2015

Re: HSC 2015 4U Marathon - Advanced Level

simpleetal said:
ah okay. Do you have any idea what the general cut off mark for HD is then? For senior? Like for a paper of average difficulty

I remember reading something like 1 in 4 people don't even get a whole question right (though this seems kinda high since I don't think the exam is THAT hard). This might have been for the junior paper though which would make more sense since they aren't as experienced.

Low 20's/30 seems about right for a HD-cut off I think, though I'm really not too sure.

glittergal96 · Jun 16, 2015

Re: HSC 2015 4U Marathon - Advanced Level

2azn4u said:
View attachment 32115
Here's the chinese remainder theorem question from this year's senior unsw maths comp paper.

I tried doing it by contrapositive when I couldn't figure out how to do it by contradiction. Haven't yielded any results from it yet.

I think this works:

$Suppose the claim is not true. \\ \\ Then for each $i$ there exists $n_i$ such that $a_i$ does not divide $F(n_i)$. This of course means that there is a prime factor $p_i$ of $a_i$ that does not divide $F(n_i)$. In fact, since $F(n_i+k)\equiv F(n_i) \mod k$ (by expanding out the terms of the former binomially, for example), we have an entire residue class $c_i\mod p_i$ on which $F(n)$ is indivisible by $p_i$. We choose $c_i=c_j$ whenever $p_i=p_j$. \\ \\We then have a collection of residue classes modulo distinct primes, which by the Chinese remainder theorem has non-empty intersection. That is, there exists some positive integer $m$ such that $p_i$ does not divide $F(m)$, for any $i$. But since the problem statement tells us that $a_i|F(m)$ for some $i$, this is a contradiction.\\ So the claim is true!$

glittergal96 · Jun 16, 2015

Re: HSC 2015 4U Marathon - Advanced Level

RealiseNothing said:

Lol, that was seriously the question as printed?

That is so bad...

Sy123 · Jun 18, 2015

Re: HSC 2015 4U Marathon - Advanced Level

$\\ $Prove that the fractional part of$ \ \sqrt{4n^2 + n} \ $is less than$ \ \frac{1}{4}$

RealiseNothing · Jun 18, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
$\\ $Prove that the fractional part of$ \ \sqrt{4n^2 + n} \ $is less than$ \ \frac{1}{4}$

Let $\sqrt{4n^2+n} = A$

Then:

$2n = \sqrt{A^2-n} < A < \sqrt{A^2+3n+1} = 2n+1$

So A lies 'n' more than an integer, and 3n+1 less than the next integer. Since the square root function is concave down, any chord lies beneath the curve, hence the result.

Sy123 · Jun 18, 2015

Re: HSC 2015 4U Marathon - Advanced Level

RealiseNothing said:
Let $\sqrt{4n^2+n} = A$

Then:

$2n = \sqrt{A^2-n} < A < \sqrt{A^2+3n+1} = 2n+1$

So A lies 'n' more than an integer, and 3n+1 less than the next integer. Since the square root function is concave down, any chord lies beneath the curve, hence the result.

Which chord is this? I just let the fractional part be 'r' and found a quadratic equation that implied r < 1/4 (after doing what you did first)

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