HSC 2014 MX2 Marathon (archive) (4 Viewers)

integral95 · Aug 14, 2014

Re: HSC 2014 4U Marathon

aDimitri said:
Gonna skip lots of lines, especially in the earlier parts, because this is extremely long.
$\\ (i) \quad u=(1-lnx) \quad dv = dx \quad I_{n} = [x(1-lnx)^{n}]_{1}^{e} + n\int_{1}^{e}(1-lnx)^{n-1}dx \\ I_{n} = -1 + nI_{n-1} \\ \\ (ii) \quad I_{3} = 6e - 16 \\ \\ (iii) \quad I_{n} = -1 + nI_{n-1} \\ = -1 -n + n(n-1)I_{n-1} \\ = -1 -n -n(n-1) - n(n-1)(n-1)I_{n-3} \\ ... \\ = n!I_{0} - \frac{n!}{n!} - \frac{n!}{(n-1)!} - ... - \frac{n!}{1!} \\ = n!e - \frac{n!}{n!} - \frac{n!}{(n-1)!} - ... - \frac{n!}{1!} - \frac{n!}{0!} \\ \frac{I_{n}}{n!} = e - \sum_{r=0}^{n} \frac{1}{r!} \\ \\ (iv) \quad \text{Consider the graph of (1-lnx)} \\ \text{In the domain that the integral describes, f(x) is continuous and positive between the points (1,1) and (0,0)} \\ \text{Hence, it is clear that for all values n, the area will be a positive number.} \\ I_{n} \geq0 \\ \text{Since the function lies between 1 & 0, raising it to any positive power will cause a decrease.} \\ \text{Hence, the maximum value occurs at the minimum n, n=0. } \\ I_{n} \leq I_{0} = e - 1 \\ 0 \leq I_{n} \leq e-1$

hahaha got the equation too long message and had to split up part v

$\\ (v) \text{Once again examine the graph. It is clear minimum value for } I_{n} \text{ occurs as } n \to \infty \\ \lim_{n \to \infty} I_{n} = 0. \\ \text{Hence, the desired statement follows quite simply from (iii)} \\ e - \lim_{n \to \infty}\sum_{r=0}^{n} \frac{1}{r!} = 0 \\ \lim_{n \to \infty}\sum_{r=0}^{n} \frac{1}{r!} = e$

Lol for part iv) i think you meant the points (1,1) and (e,0)

aDimitri · Aug 14, 2014

Re: HSC 2014 4U Marathon

integral95 said:
Lol for part iv) i think you meant the points (1,1) and (e,0)

Yes sorry, typo

My hand written version included the diagram

Sy123 · Aug 16, 2014

Re: HSC 2014 4U Marathon

$\\ $i) Show that if the sum of the digits of a number is divisible by 3, then the number is divisible by 3$ \\ \\ $ii) Show that if the obtained by taking the last two digits of the number on its own, is divisible by 4, then the original number is divisible by 4$ \\ \\ $iii) Show that a number is only divisible by 5 if the number ends in 0 or 5$ \\ \\ $iv) Show that if the sum of every second digit of a number is the same as the sum of every other digit, or that their difference is divisible by 11, then that number is divisible by 11$$

Stygian · Aug 16, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $i) Show that if the sum of the digits of a number is divisible by 3, then the number is divisible by 3$ \\ \\ $ii) Show that if the obtained by taking the last two digits of the number on its own, is divisible by 4, then the original number is divisible by 4$ \\ \\ $iii) Show that a number is only divisible by 5 if the number ends in 0 or 5$ \\ \\ $iv) Show that if the sum of every second digit of a number is the same as the sum of every other digit, or that their difference is divisible by 11, then that number is divisible by 11$$

for the first part take any number x

x= a1 + a2.10 + a3.10^2.........an.(10^(n-1))

= a1 + 9.a2 + a2 + 99a3 + a3...........+ 999999(n-2 9s)an + an

= (a1 +a2 +a3 +.......+an) + 3(3a1 + 33a2 +...........33333333(n-2 3s)an)

3(3a1 + 33a2 +...........33333333(n-2 3s)an) is divisible by 3

(a1 +a2 +a3 +.......+an) is the sum of the digits of x

therefore if the sum of the digits of x is divisible by 3 then = (a1 +a2 +a3 +.......+an) + 3(3a1 + 33a2 +...........33333333(n-2 3s)an) is divisible by 3 and thus x is divisible by 3

for part ii

let x be any positive integer

x= a1 + a2.10 + a3.10^2.........an.(10^(n-1))

= a1 +a2.10 + 10^2(a3 + a4.10 + a5.10^2 +.......... + an(10^n-3))

10^2 is divisible by 4 (since 4 times 25 is 100)

so 10^2(a3 + a4.10 + a5.10^2 +.......... + an(10^n-3)) is divisible by 4

the last two digits of the number are given by a1 +a2.10

therefore if the last two digits of the number are divisible by 4, a1 +a2.10 is also divisible by 4 and therefore a1 +a2.10 + 10^2(a3 + a4.10 + a5.10^2 +.......... + an(10^n-3)) is divisible by 4

therefore xi s divisible by 4

Stygian · Aug 16, 2014

Re: HSC 2014 4U Marathon

Stygian said:
for the first part take any number x

x= a1 + a2.10 + a3.10^2.........an.(10^(n-1))

= a1 + 9.a2 + a2 + 99a3 + a3...........+ 999999(n-2 9s)an + an

= (a1 +a2 +a3 +.......+an) + 3(3a1 + 33a2 +...........33333333(n-2 3s)an)

3(3a1 + 33a2 +...........33333333(n-2 3s)an) is divisible by 3

(a1 +a2 +a3 +.......+an) is the sum of the digits of x

therefore if the sum of the digits of x is divisible by 3 then = (a1 +a2 +a3 +.......+an) + 3(3a1 + 33a2 +...........33333333(n-2 3s)an) is divisible by 3 and thus x is divisible by 3

for part ii

let x be any positive integer

x= a1 + a2.10 + a3.10^2.........an.(10^(n-1))

= a1 +a2.10 + 10^2(a3 + a4.10 + a5.10^2 +.......... + an(10^n-3))

10^2 is divisible by 4 (since 4 times 25 is 100)

so 10^2(a3 + a4.10 + a5.10^2 +.......... + an(10^n-3)) is divisible by 4

the last two digits of the number are given by a1 +a2.10

therefore if the last two digits of the number are divisible by 4, a1 +a2.10 is also divisible by 4 and therefore a1 +a2.10 + 10^2(a3 + a4.10 + a5.10^2 +.......... + an(10^n-3)) is divisible by 4

therefore xi s divisible by 4

for part iii it's similar to part ii

let x be any positive integer

x= a1 + a2.10 + a3.10^2.........an.(10^(n-1))

= a1 + 10(a2 + a3.10 + a4.10^2 +.......... + an(10^n-2))

10 is divisible by 5

therefore

10(a2 + a3.10 + a4.10^2 +.......... + an(10^n-2)) is divisible by 5

if a1 is 5 then it is divisible by 5 and a1 + 10(a2 + a3.10 + a4.10^2 +.......... + an(10^n-2)) is divisible by 5

if a1 is 0 then a1 + 10(a2 + a3.10 + a4.10^2 +.......... + an(10^n-2)) is divisible by 5

therefore x is divisible by 5

aDimitri · Aug 16, 2014

Re: HSC 2014 4U Marathon

Stygian said:
for the first part take any number x

x= a1 + a2.10 + a3.10^2.........an.(10^(n-1))

= a1 + 9.a2 + a2 + 99a3 + a3...........+ 999999(n-2 9s)an + an

= (a1 +a2 +a3 +.......+an) + 3(3a1 + 33a2 +...........33333333(n-2 3s)an)

3(3a1 + 33a2 +...........33333333(n-2 3s)an) is divisible by 3

(a1 +a2 +a3 +.......+an) is the sum of the digits of x

therefore if the sum of the digits of x is divisible by 3 then = (a1 +a2 +a3 +.......+an) + 3(3a1 + 33a2 +...........33333333(n-2 3s)an) is divisible by 3 and thus x is divisible by 3

for part ii

let x be any positive integer

x= a1 + a2.10 + a3.10^2.........an.(10^(n-1))

= a1 +a2.10 + 10^2(a3 + a4.10 + a5.10^2 +.......... + an(10^n-3))

10^2 is divisible by 4 (since 4 times 25 is 100)

so 10^2(a3 + a4.10 + a5.10^2 +.......... + an(10^n-3)) is divisible by 4

the last two digits of the number are given by a1 +a2.10

therefore if the last two digits of the number are divisible by 4, a1 +a2.10 is also divisible by 4 and therefore a1 +a2.10 + 10^2(a3 + a4.10 + a5.10^2 +.......... + an(10^n-3)) is divisible by 4

therefore xi s divisible by 4

part 4 is essentially the same as part 1.
separate a1, a3, a5 etc. then assume that a1+ a3+ a5 etc = a2 + a4 + a6 + etc.
you can factorise out an 11 from there.

for the difference bit, you just add a2+a4+a6+... to the series and factor out an 11, in the process you have to subtract a2+a4+a6+...
this means you're left with an 11 factored out of one bit, and as a remainder you have a1 - a2 + a3 - a4 + a5 - a6...
therefore if the difference is divis by 11, the whole thing is divis by 11

Sy123 · Aug 17, 2014

Re: HSC 2014 4U Marathon

$\\ $Let the sequence$ \ a_{n+1} = pa_n - a_{n-1} \ $have$ \ a_0 = 1 \ $and$ \ a_1 = p \\ \\ $Show that the polynomial$ \ x^n - a_nx + a_{n-1} \ $is divisible by$ \ x^2-px+1$

go_cadel · Aug 17, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $Let the sequence$ \ a_{n+1} = pa_n - a_{n-1} \ $have$ \ a_0 = 1 \ $and$ \ a_1 = p \\ \\ $Show that the polynomial$ \ x^n - a_nx + a_{n-1} \ $is divisible by$ \ x^2-px+1$

Screen Shot 2014-08-17 at 9.56.22 AM.png

Sy123 · Aug 17, 2014

Re: HSC 2014 4U Marathon

$\\ $Let$ \ x,y \ $be positive reals so that$ \ (1+x)(1+y) = 2 \ $show that$ \ xy + \frac{1}{xy} \geq 6$

Axio · Aug 17, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $Let$ \ x,y \ $be positive reals so that$ \ (1+x)(1+y) = 2 \ $show that$ \ xy + \frac{1}{xy} \geq 6$

$(1+x)(1+y)=2\\1+x+y+xy=2\\xy=1-x-y\\\therefore xy+\frac{1}{xy}\\=1-x-y+\frac{1}{1-x-y}\\=\frac{x^{2}+2xy+y^{2}+2-2x-2y}{1-x-y}\\=\frac{(x+y)^{2}+2(1-x-y)}{1-x-y}\\=\frac{(x+y)^{2}}{xy}+2\\$

... is this at all close? All I can think of doing is:

$(x+y)^{2}\geq 0\\\frac{(x+y)^{2}}{xy}\geq 0\\\frac{(x+y)^{2}}{xy}+2\geq 2\\\therefore xy+\frac{1}{xy}\geq 2$

go_cadel · Aug 17, 2014

Re: HSC 2014 4U Marathon

Axio said:
$(1+x)(1+y)=2\\1+x+y+xy=2\\xy=1-x-y\\\therefore xy+\frac{1}{xy}\\=1-x-y+\frac{1}{1-x-y}\\=\frac{x^{2}+2xy+y^{2}+2-2x-2y}{1-x-y}\\=\frac{(x+y)^{2}+2(1-x-y)}{1-x-y}\\=\frac{(x+y)^{2}}{xy}+2\\$

... is this at all close? All I can think of doing is:

$(x+y)^{2}\geq 0\\\frac{(x+y)^{2}}{xy}\geq 0\\\frac{(x+y)^{2}}{xy}+2\geq 2\\\therefore xy+\frac{1}{xy}\geq 2$

The first part was on the right track.
Rtp: (x+y)^2/xy + 2 >=6
Rtp: (x+y)^2>=4xy
But (x-y)^2>=0
Hence x^2 + y^2 >= 2xy
And (x+y)^2 >= 4xy as required

Sy123 · Aug 21, 2014

Re: HSC 2014 4U Marathon

$\\ $Three points, A, B and C are randomly chosen on a circle, find the probability that the lengths of AB, BC and AC are all less than the radius of the circle.$$

Sy123 · Aug 23, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $Three points, A, B and C are randomly chosen on a circle, find the probability that the lengths of AB, BC and AC are all less than the radius of the circle.$$

Is there an issue with the wording of this question? I don't think it is too hard for this marathon no?

Kurosaki · Aug 23, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $Three points, A, B and C are randomly chosen on a circle, find the probability that the lengths of AB, BC and AC are all less than the radius of the circle.$$

Hey Sy, is the answer for this 0?

Drawing up a circle centre O and radius r and choosing any 3 random points for A, B, C, connect chords AB, BC and AC. Let angle ACB be $\theta$ .
Applying the sine rule and the fact that the angle at the centre of the circle is twice the angle at the circumference subtended by the same arc:

$\frac{r}{\cos \theta}=\frac{AB}{\sin 2 \theta}$ , which simplifies to give:

$AB=2r \sin \theta$ .
$AB< r $iff $0< \theta < \frac{\pi}{6}$
A simple application of the same procedure for the other chords yields the same inequality as above. but then that means a triangle will not exist as their sum will be less than pi.

Kurosaki · Aug 23, 2014

Re: HSC 2014 4U Marathon

Wait wait wait, I think I screwed up, it's also possible that you can have $\pi > \theta > \frac{5 \pi}{6}$ which guarantees they are all less than the radius...hmmm... about $\frac{1}{6}$ then? Don't know how to calculate it exactly though..
One would have to divide by 3 though as it could be any angle in the triangle, which would make it about $\frac{1}{18}$ ?

Idk I'm probably completely wrong.

glittergal96 · Aug 23, 2014

Re: HSC 2014 4U Marathon

Simple trig shows that two points on the circumference have distance at most r if and only if the angle subtended by their respective radii is at most pi/3.

Because of rotational symmetry, the situation is the same as picking two numbers $-\pi\leq \phi,\theta \leq \pi$ randomly (the signed angles BOA and COA), and asking what the probability is that all three angles AOB, BOC, COA are smaller than pi/3 in size.

For a given $\phi\in [-\pi/3,\pi/3]$ , $\theta$ must lie in an interval of length $\frac{2\pi}{3}-|\phi|$ .

So visualising the full probability space as the rectangle $(\phi,\theta)$ in the plane, we calculate the probability by finding the area of the subregion that obeys the angle condition and dividing by the rectangles area. This calls for integration:

$P=\frac{1}{4\pi^2}\int_{-\pi/3}^{\pi/3} \left(\frac{2\pi}{3}-|\phi|\right) \, d\phi=\frac{1}{4\pi^2}\left(\frac{4\pi^2}{9}-\frac{\pi^2}{9}\right)=\frac{1}{12}.$

(I could probably explain this better but hopefully it's clear what I'm doing.)

Sy123 · Aug 24, 2014

Re: HSC 2014 4U Marathon

I got glittergal96's answer

Originally when I posted this question I simply had a glance and thought it was simple common sense, but the problem is more complex than that and probably should of been in the advanced thread

What I do is fix one point (symmetry), then I let one point be a variable point, let its angle between it and the original fixed point be $\beta$ . In order for a length to be less than the radius, the differences between angles must be within $\frac{\pi}{3}$ . Which yields an integral that is the same as glittergal96

However what I did was:

$P = \frac{$length of interval$}{$length of circle$}$

So when doing integration, I did:

$P = \frac{\int_{-\pi/3}^{\pi/3} 2\pi/3 - \beta d\beta}{\int_{-\pi/3}^{\pi/3} 2\pi d\beta} = \frac{1}{4}$

$Pr(AB, BC, AC \leq 1) = Pr(AB \leq 1) \times (Pr(AC \leq 1) \times Pr(BC \leq 1))$

So lets take A as the fixed point, the probability that C is within range of A and B for some B in the range of A is 1/4 (i.e. 1/4 is equal to the bracket product)

But we still have to compute the probability that B is IN range of A, and that is simply 1/3 (the Pr outside of brackets)

Multiply them and we get 1/12

faisalabdul16 · Aug 24, 2014

Re: HSC 2014 4U Marathon

Need assistance with an inequality question (have not done this with my teacher yet, only looking through past papers and understanding the solutions) and this might be easy to you guys.

You are given - a + 1/a greater than or equal to 2.

Show that if a>0, b>0, c>0 are real numbers, then

i) b+c/a + c+a/b + a+b/c is greater than or equal to 6.

go_cadel · Aug 24, 2014

Re: HSC 2014 4U Marathon

faisalabdul16 said:
Need assistance with an inequality question (have not done this with my teacher yet, only looking through past papers and understanding the solutions) and this might be easy to you guys.

You are given - a + 1/a greater than or equal to 2.

Show that if a>0, b>0, c>0 are real numbers, then

i) b+c/a + c+a/b + a+b/c is greater than or equal to 6.

b/a + a/b>=2 (using the given a+1/a >=2)
now similarly c/a + a/c >=2
and b/c+c/b >=2
Summing the above gives the required inequality.

faisalabdul16 · Aug 24, 2014

Re: HSC 2014 4U Marathon

go_cadel said:
b/a + a/b>=2 (using the given a+1/a >=2)
now similarly c/a + a/c >=2
and b/c+c/b >=2
Summing the above gives the required inequality.

Ohhhhhhhhhhhh how stupid of me, can't believe i didn't see that. b/a + a/b = b^2 + a^2/ab which is greater than two from (a-b)^2>0
And then similarly for the rest

HSC 2014 MX2 Marathon (archive) (4 Viewers)

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