HSC 2017 MX1 Marathon (1 Viewer)

Paradoxica · Oct 8, 2016

si2136 said:
Could you show your working out? I read the question wrong with 10 steps but with 15 steps, this is my working out:

15 Steps

3 Step // 1 Step

5 // 0 = 1 Way
4 // 3 = 35 Ways
3 // 6 = 84 Ways
2 // 9 = 55 Ways
1 // 12 = 13 Ways
0 // 15 = 1 Way

Therefore 189 ways.

You're right, I was counting the wrong diagonal set.

oops

si2136 · Oct 8, 2016

Paradoxica said:
You're right, I was counting the wrong diagonal set.

oops

It's okay

leehuan · Oct 8, 2016

This is going well beyond the scope of MX1 now but...

$\text{A recurrence relation is defined by each term being dependent on the previous terms. E.g. the Fibonacci sequence follows }\\F_n=F_{n-1}+F_{n-2},F_0=0, F_1=1$

$\text{Suppose, as in Drsoccerball's example, you may only climb 1 or 3 steps at once. Form a recurrence for }a_n\\\text{, where }a_n\text{ is the number of ways of reaching the }nth\text{ step. Include as many initial conditions as you believe are necessary.}$

Drsoccerball and Paradoxica are not allowed to post their solutions

Drsoccerball · Oct 8, 2016

leehuan said:
This is going well beyond the scope of MX1 now but...

$\text{A recurrence relation is defined by each term being dependent on the previous terms. E.g. the Fibonacci sequence follows }\\F_n=F_{n-1}+F_{n-2},F_0=0, F_1=1$

$\text{Suppose, as in Drsoccerball's example, you may only climb 1 or 3 steps at once. Form a recurrence for }a_n\\\text{, where }a_n\text{ is the number of ways of reaching the }nth\text{ step. Include as many initial conditions as you believe are necessary.}$

Drsoccerball and Paradoxica are not allowed to post their solutions

This is even going beyond the course of MX2... It's evil to leave them wandering...

Hint: Instead of counting up count the ways you can finish.

pikachu975 · Oct 8, 2016

leehuan said:
This is going well beyond the scope of MX1 now but...

$\text{A recurrence relation is defined by each term being dependent on the previous terms. E.g. the Fibonacci sequence follows }\\F_n=F_{n-1}+F_{n-2},F_0=0, F_1=1$

$\text{Suppose, as in Drsoccerball's example, you may only climb 1 or 3 steps at once. Form a recurrence for }a_n\\\text{, where }a_n\text{ is the number of ways of reaching the }nth\text{ step. Include as many initial conditions as you believe are necessary.}$

Drsoccerball and Paradoxica are not allowed to post their solutions

What about Integrand

leehuan · Oct 8, 2016

Drsoccerball said:
This is even going beyond the course of MX2... It's evil to leave them wandering...

Hint: Instead of counting up count the ways you can finish.

Eh, true I guess.

You can answer after 7 days if you still want to.

pikachu975 said:
What about Integrand

InteGrand probably will leave it alone haha

jathu123 · Oct 8, 2016

leehuan said:
This is going well beyond the scope of MX1 now but...

$\text{A recurrence relation is defined by each term being dependent on the previous terms. E.g. the Fibonacci sequence follows }\\F_n=F_{n-1}+F_{n-2},F_0=0, F_1=1$

$\text{Suppose, as in Drsoccerball's example, you may only climb 1 or 3 steps at once. Form a recurrence for }a_n\\\text{, where }a_n\text{ is the number of ways of reaching the }nth\text{ step. Include as many initial conditions as you believe are necessary.}$

Drsoccerball and Paradoxica are not allowed to post their solutions

$$\noindent$a_{n}=a_{n-1}+a_{n-3},\\ $ $a_{1}=1\\a_{2}=1\\ a_{3}=2$

not 100% sure though

Drsoccerball · Oct 8, 2016

jathu123 said:
$$\noindent$a_{n}=a_{n-1}+a_{n-3},\\ $ $a_{1}=1\\a_{2}=1\\ a_{3}=2$

not 100% sure though

Yep. So the basic idea is which steps can I be on as I take my last step:

$a_n = a_{n-1} ($ If my last step is 1 $) + _{n-3} ($ If my last step is 3 $ )$

Extra question: Write a recurrence for the amount of ways I can climb n stairs using 1,2,3... or n step(s).

Should be easy now...

Paradoxica · Oct 8, 2016

leehuan said:
This is going well beyond the scope of MX1 now but...

$\text{A recurrence relation is defined by each term being dependent on the previous terms. E.g. the Fibonacci sequence follows }\\F_n=F_{n-1}+F_{n-2},F_0=0, F_1=1$

$\text{Suppose, as in Drsoccerball's example, you may only climb 1 or 3 steps at once. Form a recurrence for }a_n\\\text{, where }a_n\text{ is the number of ways of reaching the }nth\text{ step. Include as many initial conditions as you believe are necessary.}$

Drsoccerball and Paradoxica are not allowed to post their solutions

I just constructed a diagonal line along the rows of Pascal's Triangle and proceeded to count which cells represent a set of possible paths, then sum them all together. Which is basically what si1236 did.

No complicated enumeration required.

Paradoxica · Oct 8, 2016

Drsoccerball said:
Yep. So the basic idea is which steps can I be on as I take my last step:

$a_n = a_{n-1} ($ If my last step is 1 $) + _{n-3} ($ If my last step is 3 $ )$

Extra question: Write a recurrence for the amount of ways I can climb n stairs using 1,2,3... or n step(s).

Should be easy now...

Hence, express the general solution as the powers of the roots of the recurrence's characteristic polynomial.

si2136 · Oct 12, 2016

General question. Are you allowed to start on the RHS for Trig Identities? I never have done that before.

Green Yoda · Oct 12, 2016

si2136 said:
General question. Are you allowed to start on the RHS for Trig Identities? I never have done that before.

yes

frog1944 · Oct 12, 2016

Are you allowed to use MX2 techniques in MX1?

si2136 · Oct 12, 2016

frog1944 said:
Are you allowed to use MX2 techniques in MX1?

No.

pikachu975 · Oct 12, 2016

frog1944 said:
Are you allowed to use MX2 techniques in MX1?

Yes

eyeseeyou · Oct 12, 2016

si2136 said:
No.

pikachu975 said:
Yes

What?

pikachu975 · Oct 12, 2016

eyeseeyou said:
What?

There's like barely any 4 unit techniques applicable to 3 unit anyway so doesn't matter

Drsoccerball · Oct 12, 2016

si2136 said:
No.

Yes you can... But if you do generally the questions become harder.

si2136 · Oct 12, 2016

Drsoccerball said:
Yes you can... But if you do generally the questions become harder.

I actually didn't know that

Blitz_N7 · Oct 13, 2016

pikachu975 said:
$\\ \text{i)} \\ \text{This is legit just differentiating to find gradient then using point gradient formula...} \\ \\ \frac{dy}{dx} = \frac{x}{6} \\ \\ \text{Subbing in x = 6p} \\ \\ \frac{dy}{dx} = p \\ \\ \text{Sub into point gradient formula} \\ y - 3p^2 = p(x-6p) \\ y - 3p^2 = px - 6p^2 \\ y = px - 3p^2 \\ \\ \text{ii)} \\ \text{Find where it hits the y axis by subbing in x = 0 to the tangent equation} \\ \text{When x} = 0, y = -3p^2 \\ \text{Then use the internal division formula to find point A} \\ (\frac{1 \times 6p + 2 \times 0}{1 + 2}, \frac{1 \times 3p^2 + 2 \times -3p^2}{1+2}) \\ \\ = (2p, - p^2) \\ \text{Find point A in terms of the parameters then do the usual simultaneous equation to solve it} \\ x = 2p \\ \\ p = \frac{2}{x} \\ \\ \text{Sub into y} \\ y = -(\frac{2}{x})^2 \\ \\ y = \frac{-x^2}{4}$

This took years to type and the answer is probably wrong lol

I think you used internal division in the ratio 2:1 instead of 1:2.

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