HSC 2015 MX2 Marathon (archive) (4 Viewers)

kawaiipotato · Sep 25, 2015

Re: HSC 2015 4U Marathon

$Let a number be in the form: $ a_n a_{n-1} ... a_1 a_0 $ where $ a_n $ is the 'n'th digit$

$a_n a_{n-1} ... a_1 a_0 = a_0 + 10*a_1 + 100*a_2 + ... + 10^{n-1}a_{n-1} + 10^{n}a_{n}$

$= a_0 + (9+1)a_2 + ... + (9+1)^{n-1}a_{n-1} + (9+1)^{n}a_{n}$

$= a_0 + (9+1)a_2 + ... + ( ^{n-1}C_0 9^{n-1} + ... + ^{n-1}C_{n-1} )a_{n-1} + ( ^{n}C_0 9^n + ... + ^{n}C_{n} )a_{n}$

$= a_0 + a_1 + a_2 + ... + a_n + 9[ $ the rest $ ] $ since everything else will have a '9' in front $$

$Dividing by 3,$

$= \frac{a_0 + a_1 + a_2 + ... + a_n}{3} + 3[ $ the rest $ ]$

$So for it to be divisible by 3, $ a_0 + a_1 + a_2 + ... + a_n = 3k $ where k is an integer$

$\therefore a_0 + a_1 + a_2 + ... + a_n $ must be divisible by $ 3$

Is this correct method? My notation and method seems weird

Paradoxica · Sep 25, 2015

Re: HSC 2015 4U Marathon

kawaiipotato said:
$Let a number be in the form: $ a_n a_{n-1} ... a_1 a_0 $ where $ a_n $ is the 'n'th digit$

$a_n a_{n-1} ... a_1 a_0 = a_0 + 10*a_1 + 100*a_2 + ... + 10^{n-1}a_{n-1} + 10^{n}a_{n}$

$= a_0 + (9+1)a_2 + ... + (9+1)^{n-1}a_{n-1} + (9+1)^{n}a_{n}$

$= a_0 + (9+1)a_2 + ... + ( ^{n-1}C_0 9^{n-1} + ... + ^{n-1}C_{n-1} )a_{n-1} + ( ^{n}C_0 9^n + ... + ^{n}C_{n} )a_{n}$

$= a_0 + a_1 + a_2 + ... + a_n + 9[ $ the rest $ ] $ since everything else will have a '9' in front $$

$Dividing by 3,$

$= \frac{a_0 + a_1 + a_2 + ... + a_n}{3} + 3[ $ the rest $ ]$

$So for it to be divisible by 3, $ a_0 + a_1 + a_2 + ... + a_n = 3k $ where k is an integer$

$\therefore a_0 + a_1 + a_2 + ... + a_n $ must be divisible by $ 3$

Is this correct method? My notation and method seems weird

That is correct. I believe the standard notation to express a string of digits is to place a horizontal bar above the entire string.

Next question.

Given $a+b \geq 2\sqrt{ab}$ , find the minimum value of $e^{2x} + e^x + e^{-x} + e^{-2x}$

Hence, determine the minimum value of $\sum_{k=1}^{n}{\left(e^{kx} + e^{-kx}\right )}$ by induction or other means.

Drsoccerball · Sep 25, 2015

Re: HSC 2015 4U Marathon

Paradoxica said:
That is correct. I believe the standard notation to express a string of digits is to place a horizontal bar above the entire string.

Next question.

Given $a+b \geq 2\sqrt{ab}$ , find the minimum value of $e^{2x} + e^x + e^{-x} + e^{-2x}$

Hence, determine the minimum value of $\sum_{k=1}^{n}{\left(e^{kx} + e^{-kx}\right )}$ by induction or other means.

$A) e^{2x}+e^{-2x} \geq 2$

$e^x+e^{x} \geq 2$

$e^{2x} +e^x +e^{-x} +e^{-2x} \geq 4$

$\therefore $ Minimum value $ =4$

$B) \sum{e^{kx}+e^{-kx}} \geq 2(1+1+1+1+...+1)$

$\sum_{k=1}^n{e^{kx} +e^{-kx}} \geq 2n$

$\therefore $ minimum value $ =2n$

Note: I don't know why you would suggest induction...

kawaiipotato · Sep 25, 2015

Re: HSC 2015 4U Marathon

Paradoxica said:
That is correct. I believe the standard notation to express a string of digits is to place a horizontal bar above the entire string.

So like a conjugate bar for complex numbers?

Paradoxica · Sep 25, 2015

Re: HSC 2015 4U Marathon

kawaiipotato said:
So like a conjugate bar for complex numbers?

Code:

\overline{a_0a_1a_2a_3\dots a_{n-2}a_{n-1}a_n}

Yes. e.g. $\overline{a_0a_1a_2a_3\dots a_{n-3}a_{n-2}a_{n-1}a_n}$ is an "n" digit number in a given base

Drsoccerball said:
Note: I don't know why you would suggest induction...

I was a bit rushed for time to leave cabra library as I was writing this question, induction seemed like a way to do it at the time. You could say I made an...

inductive

guess

asdfghjklqwerty · Sep 25, 2015

Re: HSC 2015 4U Marathon

https://gyazo.com/b24c32ca76662a2a96aaae5c7ada144a

can someone tell me how and why this holds true without actually integrating? thanks

kawaiipotato · Sep 25, 2015

Re: HSC 2015 4U Marathon

Let u = pi/2 - x for both integrals
LHS integral = int from -pi/2 to pi/2 of cos^8 x dx
> 0 as it is even.
RHS integral = int from -pi/2 to pi/2 of sin(4pi - 8x) = -sin8x dx
= 0 as it is odd
Holds true

dan964 · Sep 25, 2015

Re: HSC 2015 4U Marathon

asdfghjklqwerty said:
https://gyazo.com/b24c32ca76662a2a96aaae5c7ada144a

can someone tell me how and why this holds true without actually integrating? thanks

graph the thing?
Express one in terms of the other, and prove that the integral (by graphing) is either positive or negative
and split the integral up.

RealiseNothing · Sep 26, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
Prove that if $n$ is a positive integer such that $n>3$ , then at least one of $n$ , $n+2$ or $n+4$ is composite.

n has to be odd, and since 2 and 3 are co-prime three consecutive odd numbers must take all possible values mod 3. This means there is exactly one of the values equal to 0 mod 3 and hence composite.

porcupinetree · Sep 26, 2015

Re: HSC 2015 4U Marathon

asdfghjklqwerty said:
http://www.sciweavers.org/upload/Tex2Img_1443239914/render.png

use the substitution t=tan (x/2) to show this equals pi/4
can someone do this question please? thanks

The link you posted doesn't work

asdfghjklqwerty · Sep 26, 2015

Re: HSC 2015 4U Marathon

oh ok this should work

porcupinetree · Sep 26, 2015

Re: HSC 2015 4U Marathon

asdfghjklqwerty said:
oh ok this should work View attachment 32460

Don't know why you'd use t formulae.

asdfghjklqwerty · Sep 26, 2015

Re: HSC 2015 4U Marathon

That's exactly what I did but the q asked to show that the answers Pi/4 then verify it using t formula

porcupinetree · Sep 26, 2015

Re: HSC 2015 4U Marathon

asdfghjklqwerty said:
That's exactly what I did but the q asked to show that the answers Pi/4 then verify it using t formula

I'm struggling to see a way to solve it via t-method, at least not without another substitution (I don't know what additional sub would work anyway)

Drsoccerball · Sep 26, 2015

Re: HSC 2015 4U Marathon

porcupinetree said:
I'm struggling to see a way to solve it via t-method, at least not without another substitution (I don't know what additional sub would work anyway)

When you reach the step $I= \frac{1}{2}\int_0^{\frac{\pi}{2}}{1}$ use the t method

But instead of $u=tan\frac{x}{2} let u=tanx$

kawaiipotato · Sep 27, 2015

Re: HSC 2015 4U Marathon

Don't think this would fall under any topic in 4U but I found a nice math/logic puzzle

$Five couples go to a party and start shaking hands. You don't shake your spouse's hand or (of course) your own. One of the men, Jake, surveys everyone and asks how many they shook hands. Every person gives a different answer. Jake's wife is the last to answer. How many handshakes did she have?$

Drsoccerball · Sep 27, 2015

Re: HSC 2015 4U Marathon

kawaiipotato said:
Don't think this would fall under any topic in 4U but I found a nice math/logic puzzle

$Five couples go to a party and start shaking hands. You don't shake your spouse's hand or (of course) your own. One of the men, Jake, surveys everyone and asks how many they shook hands. Every person gives a different answer. Jake's wife is the last to answer. How many handshakes did she have?$

Does everyone have to shake at least one persons hand?

kawaiipotato · Sep 27, 2015

Re: HSC 2015 4U Marathon

No since everyone has a different answer we can deduce that the remaining 9 people must match up to 0,1,..., 8 handshakes each

asdfghjklqwerty · Sep 27, 2015

Re: HSC 2015 4U Marathon

https://gyazo.com/4b06663b490112e35fca6bb7110488ae

for 1)c i got the first half but couldnt get the 2nd half of the inequality. i didnt know how to prove k!e^(k-1)<(1/4)(k+1)^(k+1)
can someone please try? thanks

kawaiipotato · Sep 27, 2015

Re: HSC 2015 4U Marathon

asdfghjklqwerty said:
https://gyazo.com/4b06663b490112e35fca6bb7110488ae

for 1)c i got the first half but couldnt get the 2nd half of the inequality. i didnt know how to prove k!e^(k-1)<(1/4)(k+1)^(k+1)
can someone please try? thanks

$\int_2^{k+1} \ln(x) dx > $ part b $ $ will give you the answer $ ($ as it is above the y = ln(x-1) curve $)$

mr mai

HSC 2015 MX2 Marathon (archive) (4 Viewers)

Well-Known Member

-insert title here-

Well-Known Member

Well-Known Member

-insert title here-

Member

Attachments

Well-Known Member

what

what is that?It is Cowpea

not actually a porcupine

Member

not actually a porcupine

Member

not actually a porcupine

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 4)