HSC 2014 MX2 Marathon ADVANCED (archive) (1 Viewer)

Davo_01 · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

hit patel said:
I was surprised when you said you are in year 10. Then I realised the you had a post right after it hahahahah. You are intelligent for your age probably in atleast the 2nd percentile in your age group of NSW from what I can see. If you dont mind me asking how much did you get for hsc cos thats what to uni's matter more than intelligence and the actual ability.

I got 97 for 3u and 4u and overall 99.4, which im super happy with but i was aiming much higher leading towards hsc (long story). And thanks!

Not just returning the compliment but you seem very capable yourself!

Sy123 · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

VBN2470 said:
How many pairs $(a, b)$ of positive integers are there such that $a$ and $b$ are factors of $6^6$ and $a$ is a factor of $b$ ?

$\\ 6^6 = 3^6 \cdot 2^6$

To find the number of solutions (a,b) that satisfies the above conditions, then we look at the number of ways we can pick from 6 3s and 6 2s, such that when we pick our second number we are picking from the remaining numbers to multiply on to b.

What I mean is:

3, 3, 3, 3, 3, 3

2, 2, 2, 2, 2, 2

Say we pick some combination of the above 6 to be a: 3 * 3 * 3 * 2 * 2

We have 3 3s left and 4 2s left. To get b, we pick any combination of the rest of the numbers, i.e. the remaining 3s and 2s. And multiply this remaining combination with a, this is how b is a factor of a.

So we have (warning dodgy combinatorics, I am confident in the concept of my solution, but not the calculation)

One sec, fixing dodgy combinatorics

hit patel · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Davo_01 said:
I got 97 for 3u and 4u and overall 99.4, which im super happy with but i was aiming much higher leading towards hsc (long story). And thanks! Not just returning the compliment but you seem very capable yourself!

Whow. Those are great marks but yes I understand that you expected for more. Thanks for the compliment hahaha but capability doesnt usually mean success

. But once again, great results nonetheless. Congrats.

seanieg89 · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Davo_01 said:
Woah! Thats a crazy smart solution! Had no clue how you would use cube roots of unity.
one thing though, i think there should be an x in front of PQR?

Thanks.

Yep, that's right...the last term should be -3xPQR.

Sy123 · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $To find the number of natural pairs$ \ (a,b) \ $such that$ \ a,b \ $are factors of$ \ 6^6 \ $and$ \ a \ $is a factor of$ \ b \\ $Consider the prime factorisation of$ \ 6^6 = 3^6 \cdot 2^6 \\ $Some factor of$ \ 6^6 \ $can be represented as some combination of the products of some subset of 6$ \ 3s \ $and 6$ \ 2s \\ \\ $Consider some arbitrary factor$ \ a \ $of the number with$ \ k \ $prime factors consisting only of$ \ 3 , 2 \\ $To find the number of combinations possible, we give the analogous problem, to put$ \ k \ $identical objects into containers labelled$ \ 3 \ $and$ \ 2 \\ $This turns out to be$ \ \frac{(k+1)!}{k!} = (k+1)$

$\\ $Thus there is some$ \ (k+1) \ $of$ \ a \ $with$ \ k \ $prime factors$ \\ \\ $To construct$ \ b \ $we take the remaining$ \ (12-k) \ $factors, and apply the same method, then with this new number, we multiply this by$ \ a \ $to yield$ \ b \\ \\ $Thus it follows that the multiplier of$ \ a \ $to yield$ \ b \ $i.e.$ \ \frac{b}{a} \ $has$ \ 12-k \ $prime factors, thus there are$ \ (13-k) \ $such$ \ \frac{b}{a} \\ \\ $Thus, to find the total amount of pairs, we simply sum this scenario across$ \ k \in [0,12] \\ \\ $Therefore the final answer should be$ \\ \Rightarrow \sum_{k=0}^{12} (k+1)(13-k) = 455$

Sy123 · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Davo_01 said:
I get only 1?

If we take the trivial case:

a = N

b = 2N

c = 3N

d = 6N

then:

the sum is 2/N, and this is an integer for N= 1 and N=2, so there must be more than one solution

jyu · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

VBN2470 said:
How many functions $f(x)=ax^2+bx+c$ are there with the property that, for all $x$ , $f(x)=f(-x)=f(x^2)$

$f(x)=c$

Davo_01 · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
If we take the trivial case:

a = N

b = 2N

c = 3N

d = 6N

then:

the sum is 2/N, and this is an integer for N= 1 and N=2, so there must be more than one solution

yeh so my answer is 2 now, can you check my reasoning i posted above

Sy123 · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Davo_01 said:
yeh so my answer is 2 now, can you check my reasoning i posted above

On your case where the numbers add up to 1

You say: "my next guess would be a=3, b=4, c=5 and d=6 which gives sum 0.95, while any other combination should give an even smaller number as we must pick larger numbers."

THis is nto true since I can pick say 1/2 + 1/3 + 1/6 + 1/7 = 8/7 > 1, and this is not covered in your above cases.

Davo_01 · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
On your case where the numbers add up to 1

You say: "my next guess would be a=3, b=4, c=5 and d=6 which gives sum 0.95, while any other combination should give an even smaller number as we must pick larger numbers."

THis is nto true since I can pick say 1/2 + 1/3 + 1/6 + 1/7 = 8/7 > 1, and this is not covered in your above cases.

yeh i just realised and found 3rd combination, im trying to fill in the holes but not finished, have you solved it?

Sy123 · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Davo_01 said:
yeh i just realised and found 3rd combination, im trying to fill in the holes but not finished, have you solved it?

I have not given it enough thought but it looks like a formal explanation using inequalities is the way to go.

Davo_01 · Apr 11, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
I have not given it enough thought but it looks like a formal explanation using inequalities is the way to go.

oh ok ima come back to this later, gonna move on to something else.

seanieg89 · Apr 12, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Davo_01 said:
Prove the following expression:
$(2+\sqrt{2})^n\cos{\frac{n\pi}{4}}=\sum\limits_{k=0}^{2n} \binom {2n} {k}\cos{\frac{k\pi}{4}}$

and similarly prove:
$(2+\sqrt{2})^n(\cos{\frac{n\pi}{4}} +\sin{\frac{n\pi}{4}})=\sum\limits_{k=0}^{2n} \binom {2n} {k}(\cos{\frac{k\pi}{4}} +\sin{\frac{k\pi}{4}})$

$(1+\textrm{cis}(\pi/4))^{2n}=((\textrm{cis}(\pi/8)+\textrm{cis}(-\pi/8))\textrm{cis}(\pi/8))^{2n}\\=(4\cos^2(\pi/8))^n\textrm{cis}(n\pi/4)=(2+\sqrt{2})^n\textrm{cis}(n\pi/4).\\ \\ $Equate real and imaginary parts to finish.$$

Sy123 · Apr 13, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $Using squeeze theorem, find the real function$ \ f \ $so that$ \ \lim_{n \to \infty} (n!)^{f(n)} \ $converges and is positive$$

RealiseNothing · Apr 13, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Using squeeze theorem, find the real function$ \ f \ $so that$ \ \lim_{n \to \infty} (n!)^{f(n)} \ $converges and is positive$$

$f(n) = \frac{1}{n^2}$ ?

RealiseNothing · Apr 13, 2014

Re: HSC 2014 4U Marathon - Advanced Level

What I got after a quick attempt was:

$(n!)^0 < (n!)^{\frac{1}{n^2}} < (n^n)^{\frac{1}{n^2}} = n^{\frac{1}{n}}$

Now:

$\lim_{n \to \infty} (n!)^0 = 1$

Also by AM-GM:

$n^{\frac{1}{n}} < \frac{n+(n-1)}{n} = 2-\frac{1}{n}$

$\lim_{n \to \infty} n^{\frac{1}{n}} < \lim_{n \to \infty} 2 - \frac{1}{n} = 2$

So:

$1 < \lim_{n \to \infty} (n!)^{\frac{1}{n^2}} < 2$

Do you want the actual value it converges to?

Sy123 · Apr 13, 2014

Re: HSC 2014 4U Marathon - Advanced Level

RealiseNothing said:
What I got after a quick attempt was:

$(n!)^0 < (n!)^{\frac{1}{n^2}} < (n^n)^{\frac{1}{n^2}} = n^{\frac{1}{n}}$

Now:

$\lim_{n \to \infty} (n!)^0 = 1$

Also by AM-GM:

$n^{\frac{1}{n}} < \frac{n+(n-1)}{n} = 2-\frac{1}{n}$

$\lim_{n \to \infty} n^{\frac{1}{n}} < \lim_{n \to \infty} 2 - \frac{1}{n} = 2$

So:

$1 < \lim_{n \to \infty} (n!)^{\frac{1}{n^2}} < 2$

Do you want the actual value it converges to?

Nope that's fine well done (though you might want to show that the sequence in n is increasing/decreasing to show it is not periodic)

My method involved getting an upper and lower bound for $\sum_{k=1}^n \ln k$ using upper/lower rectangles on y=ln(x), manipulating a bit and trying to see if you can get it to converge to something. (you get 1/xln(x))

seanieg89 · Apr 13, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $Using squeeze theorem, find the real function$ \ f \ $so that$ \ \lim_{n \to \infty} (n!)^{f(n)} \ $converges and is positive$$

There are so many lol. My favourite is

$f(n)=0.$

Sy123 · Apr 13, 2014

Re: HSC 2014 4U Marathon - Advanced Level

oops

VBN2470 · Apr 15, 2014

Re: HSC 2014 4U Marathon - Advanced Level

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