Recent content by Blast1

Re: Probability The answer should be 1/3 for the version with 4 machines. P (both first two tests reveal faulty machines) = 1/6 as established by earlier posts. Note that we can also identify which two machines are faulty by also having the two tests reveal non faulty machines, meaning that the...

As a quick way of showing why the antiderivative cannot be minus x/(9sqrt(9-x^2)) + C, try differentiating this expression and see if you get the original expression in the question. The answer for this question is plus x/(9sqrt(9-x^2)) + C because they assumed that costheta = plus sqrt...

Lol the papers (both 3u and 4u) were easy in terms of level of thinking required to solve every question. But i still made some sillies damn

a^2+b^2=3c^2. $ Suppose there exists a natural solution (a,b,c) to this. Since the quadratic residue mod 3 is 0 or 1, the equation can only hold if a and b are both divisible by 3 (otherwise LHS would be 1 or 2 mod 3 whilst the RHS is 0 mod 3). So, a=3k, b=3j, c=3l,$ 9k^2+9j^2=3(9l^2). $Dividing...

This wording means that the question is now valid, before you said that any subinterval of length n contained a subinterval of k primes (pi(x)>k>0), which was clearly not true and hence I was confused. I would normally post my solution, but Seanieg89 has already done so and mine was pretty...

? In my previous post, I was addressing the second part of your question, not the first (I suppose I should've mentioned this). Basically, in your question you claim that if we have a subinterval of length n (has n elements), then there must exist a subinterval containing k primes for all k...

I don't think this is quite right? Just because the sum of a subset is divisible by n-x, it doesn't necessarily mean that the sum is n-x mod N? Or have I misinterpreted what you're saying? Here's what I had. Suppose that a_i are the elements of the set. Now, consider the sums...

Re: HSC 2016 4U Marathon $ (i) Notice that $ e^{x-1} \geq x $ for all x$ \geq1. $ Now, motivated by the fact that we need to remove the e somehow, we make the substitution $ x_i=\frac{na_i}{a_1+...+a_n}. $ Thus $ e^{\sum_{i=1}^{n}x_i-n} \geq x_1x_2...x_n. \Rightarrow 1 \geq...

Re: HSC 2016 4U Marathon - Advanced Level $Notice that $ 1= \frac{1}{2}+\frac{1}{3}+\frac{1}{6}. $ Applying the algorithm which I described in my last post,$ 1=\frac{1}{2}+\frac{1}{4}+ ... + \frac{1}{2^{n-2}}+\frac{1}{3(2^{n-3})}+\frac{1}{3(2^{n-2})}.$ So, $a_i=\frac{1}{2^{i}} \ $where$ \...

Re: HSC 2016 4U Marathon Fairly new here, I discovered these forums earlier this week lol. I'm still not quite used to the art of LaTexing :( (I'm using a site called codecogs to help me out for the time being)

Re: HSC 2016 4U Marathon - Advanced Level $Statement true for n=3, as 1= $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}.$ Assume that the statement is true for some n=k. So there are k distinct naturals which satisfy $ \frac{1}{a_1}+...+\frac{1}{a_n}=1. $ Thus, there are k distinct naturals all greater...

Re: HSC 2016 4U Marathon 1. e^{ix}=cisx. \ $By pairing vectors together$, cis(\frac{\pi}{180})$ with $cis(\frac{44\pi}{180}),....cis(\frac{22\pi}{180})$ with $cis(\frac{23\pi}{180}), $The resultant vector after adding the vectors in each pair gives us $ \sqrt 2cis(\frac{\pi}{8}). $So, the...