HSC 2015 MX2 Marathon (archive) (10 Viewers)

Sy123 · Jul 6, 2015

Re: HSC 2015 4U Marathon

$\\ $Prove that any number of the form$ \ 4n + 2 \ $for positive integers$ \ n \ $cannot be written as the difference of two square numbers$$

Sy123 · Jul 6, 2015

Re: HSC 2015 4U Marathon

Ekman said:
I'm assuming you would use a similar method but instead of using the numbers F_n, you would use the factors of those integers, and prove that the factors cannot be integers, hence proving they do not share common factors.

That wasn't what I did but that may work

kawaiipotato · Jul 6, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Prove that any number of the form$ \ 4n + 2 \ $for positive integers$ \ n \ $cannot be written as the difference of two square numbers$$

$Assume 4n + 2 = (a-b)(a+b) is true$

$The factors of 4n+2 differ by 2b (where b is an integer)$

$\therefore $ If a is even, b is odd, then a+b is odd and a-b is odd $$

$If a is odd, b is even, then a+b is odd and a-b is odd$

$If a is even, b is even, then a+b is even and a-b is even$

$If a is odd, b is odd, then a+b is even and a-b is even$

$If it was case 1 and 2, then the product$ $a^2 - b^2$ $is odd and cannot equal to 4n+2, as 4n+2 = 2P and cannot be true for case 1 and 2$

$$ If it was case 3 and 4, then the product $ [tex]$ a^2 - b^2 [/tex] $is even. However, (a-b)(a+b) must be divisible by four, as two even numbers multiplied together will have a factor of 4$

$4n + 2 cannot be expressed in the form 4Q, where Q is an integer, and does not work for case 3 and 4.$

$Hence, 4n+2 cannot be expressed as the difference of two squares$
Not sure if this is valid?

Sy123 · Jul 6, 2015

Re: HSC 2015 4U Marathon

kawaiipotato said:
$Assume 4n + 2 = (a-b)(a+b) is true$

$The factors of 4n+2 differ by 2b (where b is an integer)$

$\therefore $ If a is even, b is odd, then a+b is odd and a-b is odd $$

$If a is odd, b is even, then a+b is odd and a-b is odd$

$If a is even, b is even, then a+b is even and a-b is even$

$If a is odd, b is odd, then a+b is even and a-b is even$

$If it was case 1 and 2, then the product$ $a^2 - b^2$ $is odd and cannot equal to 4n+2, as 4n+2 = 2P and cannot be true for case 1 and 2$

$$ If it was case 3 and 4, then the product $ [tex]$ a^2 - b^2 [/tex] $is even. However, (a-b)(a+b) must be divisible by four, as two even numbers multiplied together will have a factor of 4$

$4n + 2 cannot be expressed in the form 4Q, where Q is an integer, and does not work for case 3 and 4.$

$Hence, 4n+2 cannot be expressed as the difference of two squares$
Not sure if this is valid?

It's a little unclear what you are saying in the middle but I see what you are trying to say, it is my method too

Sy123 · Jul 6, 2015

Re: HSC 2015 4U Marathon

$\\ $Show that if$ \ p, q \ $are both distinct prime numbers greater than 2, then$ \ pq \ $can be written as the difference of two positive squares, in exactly two different ways$ \\ \\ $Show that this cannot be the case if either$ \ p \ $or$ \ q \ $equals 2$.$

Sy123 · Jul 6, 2015

Re: HSC 2015 4U Marathon

$\\ $Find the number of ways of writing the number 675 as the difference of 2 squares$$

Sy123 · Jul 6, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Let$ \ F_n = 2^{2^n} + 1 \\ \\ $i) Show by mathematical induction that$ \ F_0F_1\dots F_{n-1} = F_n - 2 \ $for$ \ n \geq 1 \\ \\ $ii) Hence show that there is no$ \ F_i \ $and$ \ F_j \ $with$ \ i <j \ $such that there is an integer that divides into both$ \ F_i \ $and$ \ F_j$

$\\ $For (ii), suppose that there are in fact$ \ F_i \ $and$ \ F_j \ $with$ \ i < j \ $such that they have a common factor$ \ M > 1 \\ \\ $Then, by the identity in part (i)$ \ F_0F_1\dots F_i \dots F_{j-1} = F_j - 2 \\ \\ $By dividing both sides by$ \ M \ $then$ \ F_0 \dots F_{i-1}\frac{F_i}{M} F_{i+1} \dots F_{j-1} = \frac{F_{j}}{M} - \frac{2}{M} \\ \\ \frac{2}{M} = \frac{F_j}{M} - F_0\dots \frac{F_i}{M} \dots F_{j-1} \ (*)$

$\\ \frac{F_j}{M} \ $and$ \ \frac{F_i}{M} \ $are integers since$ \ M \ $is a factor of each, so$ \ \frac{F_j}{M} - F_0\dots \frac{F_i}{M} \dots F_{j-1} \ $is an integer, therefore$ \ \frac{2}{M} \ $is an integer, by$ \ (*)$

$\\ \frac{2}{M} \ $is an integer$ \ \Rightarrow \ M = 2$

$\\ $However, if$ \ M =2 \ $then$ \ F_j \ $is divisible by 2, meaning it is even, however$ \ F_j = 2\cdot 2^{2^j -1} + 1 \ $which is odd by definition$ \\ \\ $Therefore, if there exists an$ \ M \ $that divides$ \ F_i \ $and$ \ F_j \ $this implies that$ \ F_j \ $is both even and odd, which is a contradiction, therefore there is no common factor$ \ M \ $of$ \ F_i \ $and$ \ F_j \ $as required$$

Sy123 · Jul 6, 2015

Re: HSC 2015 4U Marathon

I'm getting a lot of these questions from:

http://mathsorchard.weebly.com/step-past-papers.html

http://www.physicsandmathstutor.com/step/

STEP papers are very valuable resources for MX2 since the questions are of advanced difficulty (relative to the HSC) and are of HSC style. Though there may be some topics outside the syllabus in these papers, if you're able to spot them and avoid them, then otherwise the questions are great (and sometimes the HSC and companies like CSSA rip questions out of these!)

Ekman · Jul 7, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Show that if$ \ p, q \ $are both distinct prime numbers greater than 2, then$ \ pq \ $can be written as the difference of two positive squares, in exactly two different ways$ \\ \\ $Show that this cannot be the case if either$ \ p \ $or$ \ q \ $equals 2$.$

Isn't this exactly the same as the previous question? Instead that you say pq is an odd number, because prime numbers greater than 2 are odd numbers, and so the only two way they can be written in the form of a^2 - b^2 is when a is odd and b is even and vice versa. However you cant write pq like that if either p or q is 2 as an even number multiplied by anything results in an even number.

Sy123 · Jul 7, 2015

Re: HSC 2015 4U Marathon

Ekman said:
Isn't this exactly the same as the previous question? Instead that you say pq is an odd number, because prime numbers greater than 2 are odd numbers, and so the only two way they can be written in the form of a^2 - b^2 is when a is odd and b is even and vice versa. However you cant write pq like that if either p or q is 2 as an even number multiplied by anything results in an even number.

Can you expand on your method? From what I can see you've proven that there are 2 possibilities, for a, b (that a is even b odd, a odd b even) but that doesn't mean there are exactly 2 solutions, for example 675 can be written in (i think it's 6) different ways of having a^2 - b^2, even though there are only 2 possibilities.

InteGrand · Jul 7, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Show that if$ \ p, q \ $are both distinct prime numbers greater than 2, then$ \ pq \ $can be written as the difference of two positive squares, in exactly two different ways$ \\ \\ $Show that this cannot be the case if either$ \ p \ $or$ \ q \ $equals 2$.$

$Without loss of generality, let $q>p>2$ (so $q$ and $p$ are both odd as they are primes, and so $pq$ is odd too). Let $a$ and $b$ be positive integers with $a>b$. Then $pq=a^2-b^2\Longleftrightarrow (a+b)(a-b)=pq$. Since the only way to multiply two positive integers together to get $pq$ is $q\times p$ or $pq\times 1$ (since the only factors of $pq$ are $1,p,q$ and $pq$, as $p$ and $q$ are prime and from the Fundamental Theorem of Arithmetic), $(a+b)(a-b)=pq$ is equivalent to one of the following distinct linear systems:$

$\left\{\begin{matrix}a+b=q\\a-b=p\end{matrix}\right. (1)$

$or$

$\left\{\begin{matrix}a+b=pq\\a-b=1\end{matrix}\right. (2)$

$(since $a+b>a-b$ and $pq>1$, $q>p$).$

$(These are clearly distinct as $p\neq 1$ etc.)$

$System $(1)$ can be solved to give $a=\frac{p+q}{2},b=\frac{q-p}{2}$, which is valid as both fractions are integers as $p$ and $q$ have the same parity, so their sums and differences are even. System $(2)$ can be solved to give $a=\frac{pq+1}{2},b=\frac{pq-1}{2}$, which is again valid as $pq$ is odd, so $pq\pm 1$ is even, and clearly $pq>1$, so both fractions are positive integers. Hence there are exactly two different ways, as required.$

$If one of $p$ or $q$ equals 2 (i.e. if $p=2$, as we let $q>p$), the above result is not true, as then $p$ and $q$ have differing parities (since $q$ will be an odd prime), so the numerators of the fractions above all will be odd, and hence the solutions are \emph{not} integers.$

Sy123 · Jul 8, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Find the number of ways of writing the number 675 as the difference of 2 squares$$

$675 = 3^3 5^2 = A^2 - B^2$

$\\ $The solutions to this equation are of the form$ \\ A - B = 3^{n}5^{m} \ $and$ \ A+B = 3^{3-n}5^{2-m} \ $for$ \ n = 0, 1, 2, 3 \ $and$ \ m = 0, 1, 2$

$\\ $Since$ \ A = \frac{3^{n} 5^m + 3^{3-n} 5^m}{2} \ $and$ \ B = \frac{3^{3-n} 5^{2-m} - 3^n 5^m}{2} \ $are integers for each combination of$ \ m,n \ $since$ \ 3^k5^j \ $is odd, and the addition of two odds is even and thus divisible by 2$$

$\\ $Moreover$ \ 3^{k_1}5^{j_1} = 3^{k_2}5^{j_2} \Rightarrow k_1 = k_2 , m_1 = m_2$

$\\ $So, cycling through any unique combinations of factors, the value is$ \ 4 \times 3 \ $since there are$ \ 4 \ $possibilities for$ \ n \ $and$ \ 3 \ $possibilities for$ \ m \\ \\ $However since we are looking for unordered combinations (so picking$ \ 3^15^2 \ $and$ \ 3^25^0 \ $is the same as the reverse), this means we must divide by$ \ 2 \\ \\ \therefore \frac{4\times 3}{2} = 6 \ $is the number of ways of writing$ \ 675 \ $as the difference of two squares$$

Sy123 · Jul 8, 2015

Re: HSC 2015 4U Marathon

$\\ $Let$ \ x_0 = 4 \ $and$ \ x_{n+1} = \frac{1}{5}(x_n^2 + 6)$

$\\ $i) Show that$ \ x_{n+1} > x_n \ $for integers$ \ n \geq 0$

$\\ $ii) Show that if$ \ x_0 < 3 \ $then it is not true that$ \ x_{n+1} > x_n \ $for integers$ \ n \geq 0$

porcupinetree · Jul 8, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Let$ \ x_0 = 4 \ $and$ \ x_{n+1} = \frac{1}{5}(x_n^2 + 6)$

$\\ $i) Show that$ \ x_{n+1} > x_n \ $for integers$ \ n \geq 0$

$\\ $ii) Show that if$ \ x_0 < 3 \ $then it is not true that$ \ x_{n+1} > x_n \ $for integers$ \ n \geq 0$

Here's my solution for part (i) and an idea for part (ii) - it's probably not written in the most mathematically concise way but it's something:

https://imgur.com/kldJ69o

SilentWaters · Jul 8, 2015

Re: HSC 2015 4U Marathon

Perhaps a more systematic variation of your response would mention two things:

(1) the quadratic $x_{n-1}-x_n$ is monotonically increasing above zero for x > 3, meaning that successive values of x starting from 4 get larger and larger. That is, we continue to move right along that quadratic with each new iteration of x from $x_0$ , guaranteeing the inequality in (i)

(2) in the range 2 < x < 3 the inequality breaks down, sure, but what about x = 2 and x < 2? Mention that the inequality turns into an equality for the former case, and in the latter case, the quadratic $x_{n-1}-x_n$ (the difference between successive values of x) monotonically decreases towards zero. That is, $x_{n+1}\rightarrow x_n$ as $x\rightarrow 2$ , until we again have an equality instead of an inequality.

Sy123 · Jul 8, 2015

Re: HSC 2015 4U Marathon

porcupinetree said:
Here's my solution for part (i) and an idea for part (ii) - it's probably not written in the most mathematically concise way but it's something:

https://imgur.com/kldJ69o

I may be wrong but I feel like there is some sort of circularity here:

It might be a misinterpretation but it seems you assume that x(0) is the lowest of the x(n) which from there allows you to prove that x(n+1) > x(n).
So it is either circular in that you assume x(n+1) > x(n) to get that min(x(n)) = x(0), or, you have independent reasons for knowing that min(x(n)) = x(0), in which case what would you say to prove this?

What if because of the 1/5 in front of the bracket that there is some value below x(0)?

The general gist of it is true however (the reasoning prior to this), and to complete the proof safely, I would recommend simply using proof by induction

Step 1 is easy, since we can calculate x(1) and show that x(1) > x(0).

Step 2 allows us the assumption that x(k) > x(k-1) > ... > x(0) = 4 > 3

Step 3, the inductive step, simply requires doing what you've already done, but then using Step 2 as justification for x(n) > x(0).

------

As for part 2, it is sufficient to show that $x_1 < x_0$ for $2 < x_0 < 3$

Sy123 · Jul 8, 2015

Re: HSC 2015 4U Marathon

$\\ $By using de Moivre's theorem and the Binomial Theorem, show that$ \\\\ \sum_{i=0}^n \binom{n}{i} \sin((2i+1)x) = 2^n \cos^n x \sin((n+1)x)$

Ekman · Jul 9, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $By using de Moivre's theorem and the Binomial Theorem, show that$ \\\\ \sum_{i=0}^n \binom{n}{i} \sin((2i+1)x) = 2^n \cos^n x \sin((n+1)x)$

$$ Let: $ z=\cos{x} + i\sin{x}$

$z^{n+1}\left(z+\frac{1}{z}\right)^n = 2^{n}\cos^n{x}(cis{x})^{n+1} = 2^{n}\cos^n{x}(\cos{(n+1)x} + i\sin{(n+1)x})$

$z^{n+1}\left(z+\frac{1}{z}\right)^n =\binom{n}{0}z^{1} + \binom{n}{1}z^{3} + \dots + \binom{n}{n-2}z^{2n-3} + \binom{n}{n-1} z^{2n-1} + \binom{n}{n} z^{2n+1} = \sum_{i=0}^n \binom{n}{i} cis{((2i + 1)x)}$

$\therefore \sum_{i=0}^n \binom{n}{i} cis{((2i + 1)x)} = 2^{n}\cos^n{x}(\cos{(n+1)x} + i\sin{(n+1)x})$

$\\$Equate all imaginary terms: $ \sum_{i=0}^n \binom{n}{i} \sin((2i+1)x) = 2^n \cos^n x \sin((n+1)x)$

Sy123 · Jul 23, 2015

Re: HSC 2015 4U Marathon

Axio · Jul 23, 2015

Re: HSC 2015 4U Marathon

Sy123 said:

i.

a=kvx
dv/dx =kx
integrating gives: v=k/2(x^2 + d^2)

v=dx/dt
integrating gives: x=dtan((2dkt+pi)/4)

as x->inf, (2dkt+pi)/4 -> pi/2. And rearranging gives required result

ii.

x=sqrt((U^2 -B^2)/e^-kt/B -1) +U^2), where B=d^2 -2U/k

as t->inf, x->d^2-2U/k

HSC 2015 MX2 Marathon (archive) (10 Viewers)

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