HSC 2015 MX1 Marathon (archive) (4 Viewers)

PhysicsMaths · Jan 12, 2015

Re: HSC 2015 3U Marathon

Kaido said:
rofl, pre sure u dont learn that in yr 12 ^

i reckon its 30!!/15!! ?

You could do that, but intended for you to use 3u methods
It is solvable using intuition!

dan964 · Jan 12, 2015

Re: HSC 2015 3U Marathon

@kaido
you do its the last thing you learn **
I originally thought 30!/15!
or 30P15
which is massive

but if you reread it
it is just 2^15 * (28 * 24 * 20 * 16 * 8 * 4)
= 2^22*7!

**edit: only if the question was actually what you thought it was.

dan964 · Jan 12, 2015

Re: HSC 2015 3U Marathon

(a) Prove by mathematical induction that
2^(11n) - 1 is divisible by 23

(b) Hence deduce that
2^2047 -2 is divisible by 2047

leehuan · Jan 12, 2015

Re: HSC 2015 3U Marathon

PhysicsMaths said:
Next question!

Express (30 x 28 x 26 x ... x 2) / (15 x 13 x 11 x ... x 1)
in factorial notation

(i'm running out of ideas too)

Answer: (2^22)(7!) checked WolframAlpha
Working: http://i1164.photobucket.com/albums/q566/Rui_Tong/IMG_1234_zps72e72f45.jpg

leehuan · Jan 12, 2015

Re: HSC 2015 3U Marathon

dan964 said:
@kaido
you do its the last thing you learn **
I originally thought 30!/15!
or 30P15
which is massive

but if you reread it
it is just 2^15

**edit: only if the question was actually what you though it was.

At first, I thought it was 2^15 as well. But it's a product, not a sum.

leehuan · Jan 12, 2015

Re: HSC 2015 3U Marathon

dan964 said:
(a) Prove by mathematical induction that
2^(11n) - 1 is divisible by 23

(b) Hence deduce that
2^2047 -2 is divisible by 2047

a) http://i1164.photobucket.com/albums/q566/Rui_Tong/IMG_1236_zpsa25e12d7.jpg
b) Still working on, and somewhat stuck.

Kaido · Jan 12, 2015

Re: HSC 2015 3U Marathon

part b) seems surprisingly simple

Carrotsticks · Jan 12, 2015

Re: HSC 2015 3U Marathon

leehuan said:
a) http://i1164.photobucket.com/albums/q566/Rui_Tong/IMG_1236_zpsa25e12d7.jpg
b) Still working on, and somewhat stuck.

Hint: Factorise out 2 and then use (a).

LostInHSC · Jan 12, 2015

Re: HSC 2015 3U Marathon

Carrotsticks said:
Hint: Factorise out 2 and then use (a).

$\begin{align*} &= 2^{2047}-2 \\ &= 2\left ( 2^{2046} -1\right ) & \\ &= 2\left ( 2^{11 \times 186} -1 \right ) \end{align*} \\ =2\left ( 2^{11n}-1 \right )\\=2\times 23n \\ =46n$

i've been here for a while

FrankXie · Jan 12, 2015

Re: HSC 2015 3U Marathon

LostInHSC said:
$\begin{align*} &= 2^{2047}-2 \\ &= 2\left ( 2^{2046} -1\right ) & \\ &= 2\left ( 2^{11 \times 186} -1 \right ) \end{align*} \\ =2\left ( 2^{11n}-1 \right )\\=2\times 23n \\ =46n$

i've been here for a while

$2^{11 \times 186} -1=2048^{186}-1, $ and try to factorise $ a^n-b^n. $ But this is not the hence method.$$

$$ The hence deduce method: from the proof of part (a), the conclusion can be actually strengthened to be $ 2^{11n}-1 $ is divisible by $ 2047(=2^{11}-1).$

LostInHSC · Jan 13, 2015

Re: HSC 2015 3U Marathon

LostInHSC said:
$\begin{align*} &= 2^{2047}-2 \\ &= 2\left ( 2^{2046} -1\right ) & \\ &= 2\left ( 2^{11 \times 186} -1 \right ) \end{align*} \\ =2\left ( 2^{11n}-1 \right )\\=2\times 23n \\ =46n$

i've been here for a while

FrankXie said:
$2^{11 \times 186} -1=2048^{186}-1, $ and try to factorise $ a^n-b^n. $ But this is not the hence method.$$

$$ The hence deduce method: from the proof of part (a), the conclusion can be actually strengthened to be $ 2^{11n}-1 $ is divisible by $ 2047(=2^{11}-1).$

So, knowing $2^{11}-1 = 2047$ , should my solution be:

$\begin{align*} &= 2^{2047}-2 \\ &= 2\left ( 2^{2046} -1\right ) & \\ &= 2\left ( 2^{11 \times 186} -1 \right ) \, \,\,\textit{which is in the form}\,\,\,\, (2^{11n}-1) \\ &= 2 \times 2047n \\ &= 4094n \\ &= 2047(2n) \\ &= 2047M\, \,\textit{\textit{Thus divisible by 2047.}} \end{align*}$

Kaido · Jan 13, 2015

Re: HSC 2015 3U Marathon

LostInHSC said:
So, knowing $2^{11}-1 = 2047$ , should my solution be:

$\begin{align*} &= 2^{2047}-2 \\ &= 2\left ( 2^{2046} -1\right ) & \\ &= 2\left ( 2^{11 \times 186} -1 \right ) \, \,\,\textit{which is in the form}\,\,\,\, (2^{11n}-1) \\ &= 2 \times 2047n \\ &= 4094n \\ &= 2047(2n) \\ &= 2047M\, \,\textit{\textit{Thus divisible by 2047.}} \end{align*}$

Fourth line assumes something that isn't yet true.

LostInHSC · Jan 13, 2015

Re: HSC 2015 3U Marathon

Kaido said:
Fourth line assumes something that isn't yet true.

So we must prove by induction that $2^{11n} -1$ is divisible by $2047$ first?

$P(1)\rightarrow = \, \, 2^{11\times1} -1 =2047 \,\,\,\,\,\,\,\,\,\,\, $Therefore P(1) is true.$ \\ $Assuming P(k) is true \rightarrow 2^{11k} -1 = 2047M \\$

$\begin{align*} \mathrm{Now}, P(k+1)&\rightarrow 2^{11k+11} -1 \\ &= 2^{11k} \times 2^{11} -1 \\ \text{Now from our assumption, it is known that }2^{11k} = 2047M +1 \\ \therefore \,&= 2^{11}(2047M +1)-1 & \\ &= 4192256M +2047 \\ &=2047(2048M+1) \end{align*} \\ \therefore P(k+1) \, \, $is true. \\ ...... True by mathematical induction.$

Kaido · Jan 13, 2015

Re: HSC 2015 3U Marathon

Yup, seems legit

(since no allocated marks, safest bet)

FrankXie · Jan 22, 2015

Re: HSC 2015 3U Marathon

NEXT QUESTION -- a mc on perm and comb

Four shoes are selected from five distinct pairs of shoes. In how many ways can these four shoes include exactly one pair?
A. 240 B. 140 C. 120 D. 60

integral95 · Jan 29, 2015

Re: HSC 2015 3U Marathon

FrankXie said:
NEXT QUESTION -- a mc on perm and comb

Four shoes are selected from five distinct pairs of shoes. In how many ways can these four shoes include exactly one pair?
A. 240 B. 140 C. 120 D. 60

I got B, with 8C2 x 5 .....not too confident with this one.

InteGrand · Jan 29, 2015

Re: HSC 2015 3U Marathon

FrankXie said:
NEXT QUESTION -- a mc on perm and comb

Four shoes are selected from five distinct pairs of shoes. In how many ways can these four shoes include exactly one pair?
A. 240 B. 140 C. 120 D. 60

I'm getting C) 120.

5 ways to select the pair.

From the remaining 8 shoes (note that all shoes are different), we need to pick 2, but avoid a pair, and there are ⁸C₂ – 4 ways to do this (subtracting 4 since we don't want to make a pair from the last 2 we pick, and there are 4 ways to make a pair).

So answer is 5(⁸C₂ – 4) = 120.

FrankXie · Jan 29, 2015

Re: HSC 2015 3U Marathon

NEXT QUESTION

$$ For what values of positive constant $ a, $ the coefficient of $ x^{13} $ in the expansion of $ (3+ax)^{20} $ is the greatest.$

Drsoccerball · Jan 29, 2015

Re: HSC 2015 3U Marathon

FrankXie said:
NEXT QUESTION

$$ For what values of positive constant $ a, $ the coefficient of $ x^{13} $ in the expansion of $ (3+ax)^{20} $ is the greatest.$

sorry guys havnt learnt Latex yet but:
(20C13)(3^7)(ax)^13 is the greatest term.
Therefore for it to be the greatest term 13th term/12th term has to be less than 1
so you get ((20C13)(3)^7(a)^13)/((20C12)(3^8)(a^12)) <1
When you re arrange you get a< 39/8

braintic · Jan 30, 2015

Re: HSC 2015 3U Marathon

Drsoccerball said:
sorry guys havnt learnt Latex yet but:
(20C13)(3^7)(ax)^13 is the greatest term.
Therefore for it to be the greatest term 13th term/12th term has to be less than 1
so you get ((20C13)(3)^7(a)^13)/((20C12)(3^8)(a^12)) <1
When you re arrange you get a< 39/8

Assuming you are using the Cambridge formula for Tn:

For T13 to be the greatest term, you need T13/T12 to be GREATER than 1, giving a>39/8

BUT you also need T13/T14 >1, giving a<6

So the solution is 39/8 < a < 6 (or 'less than or equal' if you allow for equal greatest terms)

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