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Re: HSC 2016 4U Marathon - Advanced Level $The base cases are trivial. Assume statement is true for n=k. Suppose we now have k+1 coins. To get an odd number of heads, we can either have an odd number of heads in the first k coins and toss a tails for the k+1th coin, or get an even number of...

Re: HSC 2016 4U Marathon - Advanced Level When I said in my solution that you could show that the complex roots satisfying this condition had imaginary part =0, I meant exactly what you had outlined in your post (showing that theta=0)

Re: HSC 2016 4U Marathon - Advanced Level $Suppose the polynomial had a root$ \alpha $, then $\alpha ^2 \ $would be another root. This would lead to a chain of infinite roots, implying that the polynomial is a constant function, unless $\alpha^2=\alpha .$ the only such$ \alpha $ satisying...

$Notice that$ 2(\frac{1}{7\cdot 3^{7}}+\frac{1}{9\cdot 3^{9}}+\frac{1}{11\cdot 3^{11}}+....)< 2(\frac{1}{7\cdot 3^7}+\frac{1}{7\cdot 3^9}+\frac{1}{7\cdot 3^{11}}+....)= \frac{2}{7\cdot 3^7} \cdot(1+\frac{1}{3^2}+\frac{1}{3^4}+.....)=\frac{1}{7\cdot2^2\cdot3^5}

Re: 2015 permutation X2 marathon I'll assume that everyone knows how a knight moves in a chess board. Since the piece starts at the top left corner and has to travel to the bottom right in a 25 x 25 board, it must move right 24 squares, and down 24 squares. As there is a condition that the...

$Firstly, notice that $x\leq e^{x-1},$ via calculus, with equality iff x=1. Let A= arithmetic mean of $ a_1, a_2....a_k...a_n.$By letting $ x=a_k,$ we get that$ \frac{a_k}{A}\leq e^{\frac{a_k}{A}-1}. $So,$ \frac{a_1}{A}\frac{a_2}{A}.....\frac{a_n}{A}\leq...

If this question was to come up in a HSC 4u exam, then it would usually appear as the later parts of question 14 (paper of average difficulty), or the early parts of question 15 (if the paper is easy that year).

Are the counters returned to the box after each trial? If so, call a number which satisfies the given conditions as a 'good' number. So the required probability is (probability of exactly one 'good' number produced in the first four trials) * (probability of a 'good' number produced in the fifth...

Re: HSC 2015 4U Marathon - Advanced Level sorry looks like I didnt check BOS in time to read your previous post. (this was mainly because I had a major japanese assignment to work on, luckily it's all done now). But yes I do see the error that I previously made now that Ive read your post.

Re: HSC 2015 4U Marathon - Advanced Level If f(1)=2,$ then we get by letting y=1, that$ f(x-1)+2(1-f(x-1))-f(1-x)=0 \rightarrow 2=f(x-1)+f(1-x). $ But from equation A, we know that $f(1-x)=1-f(x-2). So$ f(x-1)-f(x-2)=1, which means that the gradient of the function is always 1, meaning that it...

Re: HSC 2015 4U Marathon - Advanced Level I'll do this in two separate posts, because otherwise the equation is too long f(xy)=f(x)f(y)-f(x+y)+1.$ It's already been established that $f(0)=1, $and letting x=-1, and y=1 (to somehow try make use of f(0)), we get that$ f(-1)=f(-1)f(1),$ so we...

Re: HSC 2015 4U Marathon - Advanced Level by letting each of the expressions equal to u^2 and v^2 respectively, we can deduce that m and n must both be odd for u and v to be integers. Hence, we only consider the case where m and n areboth odd integers. But this is a contradiction as for each...

Re: HSC 2015 4U Marathon Fixing up my notation big time, please wait

Re: HSC 2015 4U Marathon Breaking my Bored of Studies fast to help a fellow Boser in need. \\ i) $Obviously statement holds true for n=1$. \ $Assume true for some n=j. We proceed by induction.$ 1+ \sum_{k=0}^{m}S_{k}(j+1)\binom{m+1}{k+1} =...

Re: HSC 2015 4U Marathon - Advanced Level Yep, this was essentially the same as what I did. You can motivate this by the observation that the only the coefficients of the last 3 terms stay the same, and since the question is asking about the roots, I figured that the relationship between roots...

Re: HSC 2015 4U Marathon - Advanced Level Saw this just then, just a quick note that these days the responses may be slow because year 12s have trials, university is getting busy and some people like me don't log on very often. Regardless, here is the soln. $ai) By definition$...

Re: HSC 2015 4U Marathon - Advanced Level this is a good question, a little easier than the questions which are usually presented here, but harder than the questions on the other thread. $Show that there are no more than 2005 real numbers x so that $ \...

Re: HSC 2015 4U Marathon - Advanced Level 18(n^2+3)=n^3. \ $LHS is even, therefore RHS is even, meaning that n is even(since RHS is n^3).$ $This means that the 2^3 = 8$ $divides the LHS, so 4 divides n^2+3.$ $But, for even n, n^2+3$ $is odd, and so 4 can't divide it $ hence, there are no...

Re: HSC 2015 4U Marathon - Advanced Level ah yep that was a typo, now fixed. (forgot a bracket)

Re: HSC 2015 4U Marathon - Advanced Level $Similarly, the other side length of this slice 'b', is given by $ b= \frac{S(H-j-h)}{H}. $ Summing the thin slices from j= k-h to 0 gives us:$ V_{khP}= \lim_{\delta j\rightarrow 0}\sum_{0}^{k-h} ab\delta j = \int_{0}^{k-h}...