HSC 2015 MX2 Marathon (archive) (7 Viewers)

InteGrand · Jul 3, 2015

Re: HSC 2015 4U Marathon

Speed6 said:
There are a multiple of ways doing it right?

$Yes, we could also say $z_1z_2=0\Rightarrow |z_1z_2|=|0|\Rightarrow |z_1||z_2|=0 \Rightarrow$ at least one of $|z_1|,|z_2|$ must be 0, since two real numbers can only multiply to give 0 if at least one of them is 0. Since the only complex number with modulus 0 is 0 itself, at least one of $z_1,z_2$ must be 0. \blacksquare$

Drsoccerball · Jul 3, 2015

Re: HSC 2015 4U Marathon

New question !
$$ Given that $ cos(n\theta) =\frac{z^n +\frac{1}{z^n}}{2} $ and $ sin(n\theta) = \frac{z^n -\frac{1}{z^n}}{2i}$

$$ Hence find $ \int_0^{2\pi}{sin(n\theta)sin2\theta}d\theta$

kawaiipotato · Jul 3, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
New question !
$$ Given that $ cos(n\theta) =\frac{z^n +\frac{1}{z^n}}{2} $ and $ sin(n\theta) = \frac{z^n -\frac{1}{z^n}}{2i}$

$$ Hence find $ \int_0^{2\pi}{sin(n\theta)sin2\theta}d\theta$

Dunno how to use those complex identities but you could use the expansions:
$cos(a-b) = cosacosb+sinasinb$

$cos(a+b) = cosacosb-sinasinb$

$$ Subtracting,$

$cos(a-b) - cos(a+b) = 2sinasinb$

$\frac{cos(a-b) - cos(a+b)}{2} = sinasinb$

$Let a = n\theta, b = 2\theta$

$\therefore sin(n\theta)sin(2\theta) = \frac{cos(\theta(n-2)) - cos(\theta(n+2))}{2}$

$$ I = $ \frac{1}{2} \int_0^{2\pi} cos(\theta(n-2)) - cos(\theta(n+2)) d\theta$

$= \frac{1}{2} [\frac{sin(\theta(n-2))}{n-2} - \frac{sin(\theta(n+2))}{n+2}]_0^{2\pi}$

$= 0, $ when n is an integer, n \neq \pm 2 $$

Drsoccerball · Jul 3, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
New question !
$$ Given that $ cos(n\theta) =\frac{z^n +\frac{1}{z^n}}{2} $ and $ sin(n\theta) = \frac{z^n -\frac{1}{z^n}}{2i}$

$$ Hence find $ \int_0^{2\pi}{sin(n\theta)sin2\theta}d\theta$

$sin(n\theta) \cdoot sin(2\theta)= \frac{z^n -\frac{1}{z^n}}{2i} \cdot \frac{z^2-\frac{1}{z^2}}{2i}$
Expand and simplify to get the expression you got... ceebs doing it im busy atm

kawaiipotato · Jul 3, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
$sin(n\theta) \cdoot sin(2\theta)= \frac{z^n -\frac{1}{z^n}}{2i} \cdot \frac{z^2-\frac{1}{z^2}}{2i}$
Expand and simplify to get the expression you got... ceebs doing it im busy atm

wouldn't it be in terms of z? how would you integrate with respect to theta?

Drsoccerball · Jul 3, 2015

Re: HSC 2015 4U Marathon

kawaiipotato said:
wouldn't it be in terms of z? how would you integrate with respect to theta?

its in terms of z but recall : $cos(n\theta)= \frac{z^n+\frac{1}{z^n}}{2}$

VBN2470 · Jul 3, 2015

Re: HSC 2015 4U Marathon

$Suppose that $f$ is a continuous increasing (and hence invertible) function on $[a,b]$. Let $c=f(a)$ and $d=f(b)$. Show that if $a,b,c,d\geq{0}$ then \\\\ $\int_{c}^{d} f^{-1}(t)\text{ d}t=bd-ac-\int_{a}^{b}f(x)\text{ d}x$.$

Sy123 · Jul 3, 2015

Re: HSC 2015 4U Marathon

$\\ $Find the amount of positive integers that divide into$ \ 3^n 5^m \ $for non-negative integers$ \ m,n$

InteGrand · Jul 3, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Find the amount of positive integers that divide into$ \ 3^n 5^m \ $for non-negative integers$ \ m,n$

$(n+1)(m+1)$

braintic · Jul 3, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
$(n+1)(m+1)$

Repeat for 12^m 54^n

InteGrand · Jul 3, 2015

Re: HSC 2015 4U Marathon

braintic said:
Repeat for 12^m 54^n

$(m+3n+1)(2m+n+1)$

Drsoccerball · Jul 3, 2015

Re: HSC 2015 4U Marathon

can sum 1 explain

InteGrand · Jul 3, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
can sum 1 explain

$Write the number $N\geq 2$ as a product of its prime factors: $N=\left(p_1\right) ^{n_{p_1}}(p_2) ^{n_{p_2}}...(p_K) ^{n_{p_K}}=\prod _{j=1} ^{K}(p_j) ^{n_{p_j}}$, where $K$ is the number of distinct prime factors of $N$. (By the fundamental theorem of arithmetic, all integers $N\geq 2$ can be written in a unique way like this, up to ordering of the primes.) Then the number of positive factors of $N$, $f(N)$, is the number of all possible combinations of powers of its prime factors, since each factor of $N$ is made up from the prime factors of $N$ itself (otherwise there would be factors of $N$ which have prime factors that are not prime factors of $N$, a contradiction, since all the factors of factors of $N$ must be factors of $N$). Since we can choose from $0$ to $n_{p_j}$ of each prime factor $n_{p_j}$ (which is $(n_{p_j} +1)$ choices for each prime factor $p_j$), this makes $f(N)=\prod _{j=1} ^{K}(n_{p_j}+1)$.$

$E.g. to find the number of positive factors of $24$, write $24$ in terms of its prime factors: $24=2^\textbf{3} \times 3^\textbf{1}$. To form positive factors of $24$, we can take either 0,1,2, or 3 `2's, and multiply this with either 0 or 1 `3's, i.e. the factors of 24 are of the form: $2^{\text{0 OR 1 OR 2 OR \textbf{3}}}\times 3^{\text{0 OR \textbf{1}}}$. So there are 4 (=\textbf{3}+1) choices for the 2 and 2 (=\textbf{1}+1) choices for the 3, giving $4\times 2=8$ factors. The case where we take 0 of each prime factor gives the factor 1.$

Drsoccerball · Jul 4, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
$Write the number $N\geq 2$ as a product of its prime factors: $N=\left(p_1\right) ^{n_{p_1}}(p_2) ^{n_{p_2}}...(p_K) ^{n_{p_K}}=\prod _{j=1} ^{K}(p_j) ^{n_{p_j}}$, where $K$ is the number of distinct prime factors of $N$. (By the fundamental theorem of arithmetic, all integers $N\geq 2$ can be written in a unique way like this, up to ordering of the primes.) Then the number of positive factors of $N$, $f(N)$, is the number of all possible combinations of powers of its prime factors, since each factor of $N$ is made up from the prime factors of $N$ itself (otherwise there would be factors of $N$ which have prime factors that are not prime factors of $N$, a contradiction, since all the factors of factors of $N$ must be factors of $N$). Since we can choose from $0$ to $n_{p_j}$ of each prime factor $n_{p_j}$ (which is $(n_{p_j} +1)$ choices for each prime factor $p_j$), this makes $f(N)=\prod _{j=1} ^{K}(n_{p_j}+1)$.$

$E.g. to find the number of positive factors of $24$, write $24$ in terms of its prime factors: $24=2^\textbf{3} \times 3^\textbf{1}$. To form positive factors of $24$, we can take either 0,1,2, or 3 `2's, and multiply this with either 0 or 1 `3's, i.e. the factors of 24 are of the form: $2^{\text{0 OR 1 OR 2 OR \textbf{3}}}\times 3^{\text{0 OR \textbf{1}}}$. So there are 4 (=\textbf{3}+1) choices for the 2 and 2 (=\textbf{1}+1) choices for the 3, giving $4\times 2=8$ factors. The case where we take 0 of each prime factor gives the factor 1.$

.....................................

Drsoccerball · Jul 4, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
$Suppose that $f$ is a continuous increasing (and hence invertible) function on $[a,b]$. Let $c=f(a)$ and $d=f(b)$. Show that if $a,b,c,d\geq{0}$ then \\\\ $\int_{c}^{d} f^{-1}(t)\text{ d}t=bd-ac-\int_{a}^{b}f(x)\text{ d}x$.$

$I= \int_c^d{f^{-1}(t)}dt$

$$ By IBP, letting $ u=f^{-1}(t)$

$\therefore I= [xf^{-1}(t)]_{f(a)}^{f(b)} -\int_c^d{x(f^{-1}(t))'}dt$

$let u=f^{-1}x$

$bd-ca -\int_a^b{f(x)}dx$
[]

Sy123 · Jul 4, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
can sum 1 explain

First of all, $3^{n_1} 5^{m_1} = 3^{n_2} 5^{m_2} \Rightarrow n_1 = n_2 , m_1 = m_2$

Now, if we want to find a number that divides into $3^n5^m$ , we need to be able to construct this number by multiplying 3's and 5's. What I mean is, $\\ $if$ \ k \ $divides$ \ 3^n5^m \ $then$ \ k \ $is of the form$ \ 3^{p}5^q$

So, the problem then, can be translated into:

$\\ $Find the number of ways of picking any number of balls, from a group of$ \ n \ $blue balls and$ \ m \ $red balls$$

(at least for just a visualization). Note that the first line of this post is important because it means that any selection must be unique.

To do this problem now, see that there are (n+1) possible number of ways to pick n blue balls (either pick 0 balls, 1, 2, ..., n) and (m+1) possible number of ways to pick m red balls (0, 1, 2, ... , m). So the number of ways in total is (n+1)(m+1)

-----

Note that the translation into a problem about balls is purely to restate the problem in familiar terms, it is in no way essential to the solution.

Drsoccerball · Jul 4, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
First of all, $3^{n_1} 5^{m_1} = 3^{n_2} 5^{m_2} \Rightarrow n_1 = n_2 , m_1 = m_2$

Now, if we want to find a number that divides into $3^n5^m$ , we need to be able to construct this number by multiplying 3's and 5's. What I mean is, $\\ $if$ \ k \ $divides$ \ 3^n5^m \ $then$ \ k \ $is of the form$ \ 3^{p}5^q$

So, the problem then, can be translated into:

$\\ $Find the number of ways of picking any number of balls, from a group of$ \ n \ $blue balls and$ \ m \ $red balls$$

(at least for just a visualization). Note that the first line of this post is important because it means that any selection must be unique.

To do this problem now, see that there are (n+1) possible number of ways to pick n blue balls (either pick 0 balls, 1, 2, ..., n) and (m+1) possible number of ways to pick m red balls (0, 1, 2, ... , m). So the number of ways in total is (n+1)(m+1)

-----

Note that the translation into a problem about balls is purely to restate the problem in familiar terms, it is in no way essential to the solution.

If the question stated that all integers excluding itself would the answer be nm?

Drsoccerball · Jul 4, 2015

Re: HSC 2015 4U Marathon

im assuming this works with n amount of factors aswell?

InteGrand · Jul 4, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
If the question stated that all integers excluding itself would the answer be nm?

$No, it'd be $(n+1)(m+1)-1$, because you'd be getting rid of exactly one factor.$

Drsoccerball · Jul 4, 2015

Re: HSC 2015 4U Marathon

$\\ $Find the amount of positive integers that divide into$ \ 1^{a_1}2^{a_2}...n^{a_n} \ $for non-negative integers$ \$

$(a_1+1)(a_2+1)...(a_n+1)$ ?

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