HSC 2013 MX2 Marathon (archive) (1 Viewer)

Sy123 · Sep 5, 2013

Re: HSC 2013 4U Marathon

$A number from the sequence$ \ \ F_n = F_{n-1} + F_{n-2} \ , \ F_0 = 0 \ F_1=1 \ \ $is called a Fibonacci number$

$A formula for the$ \ n^{th} \ $Fibonacci number is$

$F_n = \frac{1}{\sqrt{5}} \left(\left(\frac{1+\sqrt{5}}{2} \right )^n - \left( \frac{1-\sqrt{5}}{2} \right)^n \right )$

$Using this formula or otherwise, prove that$ \ x \ $is a Fibonacci number$

$If and only if$ \qquad 5x^2+4 \qquad $or$ \qquad 5x^2-4 \qquad $is a perfect square$

asianese · Sep 5, 2013

Re: HSC 2013 4U Marathon

Think you meant:

prove that x is a Fibonacci number if and only if 5x^2+4 or 5x^2-4 is a perfect square. (just being picky).

Sy123 · Sep 5, 2013

Re: HSC 2013 4U Marathon

asianese said:
Think you meant:

prove that x is a Fibonacci number if and only if 5x^2+4 or 5x^2-4 is a perfect square. (just being picky).

Yeah that sounds more correct, thank you

asianese · Sep 5, 2013

Re: HSC 2013 4U Marathon

Think of the subsets of $\{ 1,2,\dots,n\}$ as picking $0,1,2,\dots$ elements from the set. That is, if you choose 0 elements, you have 1 choice. If you choose 1 element, you have n choices, for 2, nC2. Ie: $\binom{n}{0} + \binom{n}{1}+\cdots+\binom{n}{n} = 2^n$ which is clearer from your point of view from the expansion of $(1+1)^n.$ On replacement of n by 2n we get the result.

Sy123 · Sep 8, 2013

Re: HSC 2013 4U Marathon

$$The complex numberss$ \ \ a_1, a_2, a_3, \dots, a_n$

$$satsify$ \ \ |a_i| \leq A \ \ $for$ \ i=1,2, \dots, n \ \ $for positive real$ \ A \geq 1$

$For the complex equation$

$a_n z^n + a_{n-1}z^{n-1} + \dots + a_1 z = B \ \ $for positive real$ \ B \leq 1$

$Show that there is no solution in$ \ z \ $for$

$|z| \leq \frac{|B|}{|A| + |B|}$

EDIT 1: One sec, I may have made a mistake
EDIT 2: It should be fine now
EDIT 3: Changed some restrictions, I hope its fine now (B < 1, A > 1)

rural juror · Sep 8, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
We know that:

$a+b \geq 2\sqrt{ab}$

Now let $a=x^2y^6$ and $b=x^6y^2$

$x^2y^6 + x^6y^2 \geq 2\sqrt{x^8y^8}$

$x^2y^6 + x^6y^2 \geq 2x^4y^4$

Multiply both sides by 3 to establish the inequality:

$3x^2y^6 + 3x^6y^2 \geq 6x^4y^4$

Now consider:

$(x^2-y^2)^4 \geq 0$

$x^8 - 4x^6y^2 + 6x^4y^4 - 4x^2y^6 + y^8 \geq 0$

$x^8 + y^8 + 6x^4y^4 \geq 4x^6y^2 + 4x^2y^6$

Split the RHS up into the following:

$x^8 + y^8 + 6x^4y^4 \geq x^6y^2 + x^2y^6 + 3x^6y^2 + 3x^2y^6$

$x^8 + y^8 + 6x^4y^4 - 3x^6y^2 - 3x^2y^6 \geq x^6y^2 + x^2y^6$

$x^8 + y^8 - (3x^6y^2 + 3x^2y^6 - 6x^4y^4) \geq x^6y^2 + x^2y^6$

Now using the inequality we establish before, we get:

$3x^6y^2 + 3x^2y^6 - 6x^4y^4 \geq 0$

Let this be A:

$x^8 + y^8 - A \geq x^6y^2 + x^6y^2$

Hence since $A \geq 0$

$x^8+y^8 \geq x^6y^2 + x^2y^6$

$x^8+y^8 \geq x^2y^2(x^4 + y^4)$

$\frac{x^8+y^8}{x^2y^2} \geq x^4 + y^4$

$\frac{x^6}{y^2} + \frac{y^6}{x^2} \geq x^4 + y^4$

i might have a quicker way and easier to find,

looking at x^6(y^-2 - x^-2) > y^6(y^-2 - x^-2) *just assume the greater than sign is greater than or equal to, this inequality can be easily shown for x>y and y>x to hold true, then all you have to do is move the fractions over and you have the result. max 3 lines or so

rural juror · Sep 8, 2013

Re: HSC 2013 4U Marathon

oh same solution as sys thatll teach me to scroll down

Sy123 · Sep 9, 2013

Re: HSC 2013 4U Marathon

$$Let$ \ \ I_n = \int_0^{\pi /4} \tan^{2n+1} x \ dx$

$Find a recurrence formula for this integral, and use it to prove$

$\ln 2 = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}$

Sy123 · Sep 10, 2013

Re: HSC 2013 4U Marathon

$Take the set of all positive integers$ \ S \ $which contains all positive integers without a zero in their representation$

$$S = \{1,2, \dots , 9, 11, 12, \dots , 19, 21, \dots \}$

$$Let each element in the set$ \ S \ $be$ \ \ a_k \ , \ k=1,2, \dots$

$Consider the sum$

$S_n = \sum_{k=1}^n \frac{1}{a_k}$

$$Show that as$ \ n \to \infty \ \ S_n \to \infty$

Drongoski · Sep 10, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$Let$ \ \ I_n = \int_0^{\pi /4} \tan^{2n+1} x \ dx$

$Find a recurrence formula for this integral, and use it to prove$

$\ln 2 = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}$

$I_n = \int ^{\frac {\pi}{4}} _0 tan^{2n+1}x dx = \int ^a _a tan ^{2n-1}x(sec^2 x -1)dx \\ \\ = \int ^a _0 tan^{2n-1}x sec^2x dx - \int ^a _0 tan ^{2n-1}xdx \\ \\ = \left [\frac {tan ^{2n} }{2n} \right ] ^a _0- I_{n-1} = \frac {1}{2n} - I_{n-1}\\ \\ I_0 = \int ^a _0 tan x dx = \left [ -ln |cos x| \right ] ^a _0 = ln \sqrt 2 \\ \\ I_n = \frac {1}{2n} - I_{n-1} = \frac {1}{2n} - \frac {1}{2(n-1)} + I_{n-2} = \cdots \\ \\ = \frac {1}{2n} - \frac {1}{2(n-1)} + \frac {1}{2(n-2)} - \cdots + \frac {(-1) ^{n-1}}{2} - I_0\\ \\ = \frac {1}{2} \sum ^n _1 \frac {(-1) ^{(n-1)}}{n} - I_0 \\ \\ $Now $ 0 < I_n \leq \int _0 ^a ( \frac {4x}{\pi}) ^n dx = \left [ (\frac {4}{\pi})^ n \frac {x}{n+1} \right ] ^ {\frac {\pi}{4}} _0 = \frac {\pi}{4(n+1)} \to 0 $ as n $ \to \infty.\\ \\ \therefore $ by the Sandwich Theorem $ \\ \\ \lim _{n \to \infty} I_n = 0 \implies \sum ^\infty _{n=1} \frac {(-1) ^{n-1}}{n} = 2 \times I_0 = 2 \times ln \sqrt 2 = ln 2$

Above solution messy and may contain errors. Have been making it up as I went along.

Sy123 · Sep 11, 2013

Re: HSC 2013 4U Marathon

Drongoski said:
$I_n = \int ^{\frac {\pi}{4}} _0 tan^{2n+1}x dx = \int ^a _a tan ^{2n-1}x(sec^2 x -1)dx \\ \\ = \int ^a _0 tan^{2n-1}x sec^2x dx - \int ^a _0 tan ^{2n-1}xdx \\ \\ = \left [\frac {tan ^{2n} }{2n} \right ] ^a _0- I_{n-1} = \frac {1}{2n} - I_{n-1}\\ \\ I_0 = \int ^a _0 tan x dx = \left [ -ln |cos x| \right ] ^a _0 = ln \sqrt 2 \\ \\ I_n = \frac {1}{2n} - I_{n-1} = \frac {1}{2n} - \frac {1}{2(n-1)} + I_{n-2} = \cdots \\ \\ = \frac {1}{2n} - \frac {1}{2(n-1)} + \frac {1}{2(n-2)} - \cdots + \frac {(-1) ^{n-1}}{2} - I_0\\ \\ = \frac {1}{2} \sum ^n _1 \frac {(-1) ^{(n-1)}}{n} - I_0 \\ \\ $Now $ 0 < I_n \leq \int _0 ^a ( \frac {4x}{\pi}) ^n dx = \left [ (\frac {4}{\pi})^ n \frac {x}{n+1} \right ] ^a _0 = \frac {\pi}{4(n+1)} \to 0\\ \\ \therefore $ by the Sandwich Theorem $ \\ \\ \lim _{n \to \infty} I_n = 0 \implies \sum ^\infty _{n=1} \frac {(-1) ^{n-1}}{n} = 2 \times I_0 = 2 \times ln \sqrt 2 = ln 2$

Maybe not quite correct yet. May need to have a look at it again

Yep that is correct.

$What is the largest integer$ \ n \ $for which$ \ n^3+100 \ $is divisible by$ \ n+10 \ $?$

Qb · Sep 11, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$What is the largest integer$ \ n \ $for which$ \ n^3+100 \ $is divisible by$ \ n+10 \ $?$

First post, let's see if I can get the latex right -- I'm studying how you've done it lol

$Using polynomial division: $ \ n^3+100 = (n+10)(n^2-10n+100) - 900 \$

$Therefore $ \ \frac{n^3+100}{n+10} = n^2-10n+100 - \frac{900}{n+10} \$

$Both sides have to be integers, so the fraction must also be an integer. Given we're looking for the largest value of $ \ n\ $, this corresponds to the fraction being equal to 1, or $ \ n+10 = 900 \ $ giving $ \ n = 890 \ $.$

Sy123 said:
$Take the set of all positive integers$ \ S \ $which contains all positive integers without a zero in their representation$

$$S = \{1,2, \dots , 9, 11, 12, \dots , 19, 21, \dots \}$

$$Let each element in the set$ \ S \ $be$ \ \ a_k \ , \ k=1,2, \dots$

$Consider the sum$

$S_n = \sum_{k=1}^n \frac{1}{a_k}$

$$Show that as$ \ n \to \infty \ \ S_n \to \infty$

Are you sure this one is correct? I'm fairly certain I can prove it has a lower bound of 9 and an upper bound of 90 pretty easily...

Sy123 · Sep 11, 2013

Re: HSC 2013 4U Marathon

Qb said:
First post, let's see if I can get the latex right -- I'm studying how you've done it lol

$Using polynomial division: $ \ n^3+100 = (n+10)(n^2-10n+100) - 900 \$

$Therefore $ \ \frac{n^3+100}{n+10} = n^2-10n+100 - \frac{900}{n+10} \$

$Both sides have to be integers, so the fraction must also be an integer. Given we're looking for the largest value of $ \ n\ $, this corresponds to the fraction being equal to 1, or $ \ n+10 = 900 \ $ giving $ \ n = 890 \ $.$

Are you sure this one is correct? I'm fairly certain I can prove it has a lower bound of 9 and an upper bound of 90 pretty easily...

The first answer is correct, as for the second one, let me check my proof again, I got the result from 'Art of Problem SOlving' and it asked whether it diverged and converged.

Feel free to post your proof of the upper and lower bounds though

EDIT: Yeah my proof is wrong, I didn't consider the numbers where a zero is in the middle -.-

Sy123 · Sep 11, 2013

Re: HSC 2013 4U Marathon

Another good Putnam question

$Let$ \ k \ $be a positive integer.$

$The$ \ n$-th derivative of$ \ \ \frac{1}{x^k-1} \ \ $has the form$

$\frac{P_n(x)}{(x^k-1)^{n+1}} \ \ $, where$ \ \ P_n(x) \ \ $is a polynomial.$$

$$Find$ \ \ P_n(1)$

rural juror · Sep 13, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Another good Putnam question

$Let$ \ k \ $be a positive integer.$

$The$ \ n$-th derivative of$ \ \ \frac{1}{x^k-1} \ \ $has the form$

$\frac{P_n(x)}{(x^k-1)^{n+1}} \ \ $, where$ \ \ P_n(x) \ \ $is a polynomial.$$

$$Find$ \ \ P_n(1)$

Is it (-1)^n-1*n!*k^n???

fionarykim · Sep 13, 2013

Re: HSC 2013 4U Marathon

How are u guys typing so much maths here LOOOL im amazed

Also the questions just seem to just confuse my head, too hard ):

Sy123 · Sep 13, 2013

Re: HSC 2013 4U Marathon

rural juror said:
Is it (-1)^n-1*n!*k^n???

Yep =)

rural juror · Sep 13, 2013

Re: HSC 2013 4U Marathon

seemed a bit on the easy side for putnam surely???

Sy123 · Sep 13, 2013

Re: HSC 2013 4U Marathon

rural juror said:
seemed a bit on the easy side for putnam surely???

I thought so too, its problem A1 for 2002 iirc.

HeroicPandas · Sep 14, 2013

Re: HSC 2013 4U Marathon

$$Let a, b, c be positive real numbers with sum 3. \\ \\ Prove that \sqrt{a} + \sqrt{b}+\sqrt{c} \geq ab + bc + ca$

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