HSC 2015 MX1 Marathon (archive) (1 Viewer)

[integral95]

Re: HSC 2015 3U Marathon

Ok, now where is my actual pie?

pi*e =

 

[rand_althor]

Re: HSC 2015 3U Marathon

A one-player card game is played by placing 13 cards (Ace through King in that order) in a circle. Initially all the cards are face-up and the objective of the game is to flip them face-down. However, a card can only be flipped face-down if another card that is '3 cards away' is face-up. For example, one can only flip the Queen face-down if either the 9 or the 2 (or both) are face-up. A player wins the game if they can flip all but one of the cards face-down. Given that the cards are distinguishable, find the number of ways it is possible to win the game.

 

[kawaiipotato]

Re: HSC 2015 3U Marathon

A one-player card game is played by placing 13 cards (Ace through King in that order) in a circle. Initially all the cards are face-up and the objective of the game is to flip them face-down. However, a card can only be flipped face-down if another card that is '3 cards away' is face-up. For example, one can only flip the Queen face-down if either the 9 or the 2 (or both) are face-up. A player wins the game if they can flip all but one of the cards face-down. Given that the cards are distinguishable, find the number of ways it is possible to win the game.

On the first turn, do we choose a specific card to turn face down? (hence there are different ways to win)

 

[rand_althor]

Re: HSC 2015 3U Marathon

On the first turn, do we choose a specific card to turn face down? (hence there are different ways to win)

I think you can choose any card as long as it meets the conditions to be flipped.

 

[braintic]

Re: HSC 2015 3U Marathon

A one-player card game is played by placing 13 cards (Ace through King in that order) in a circle. Initially all the cards are face-up and the objective of the game is to flip them face-down. However, a card can only be flipped face-down if another card that is '3 cards away' is face-up. For example, one can only flip the Queen face-down if either the 9 or the 2 (or both) are face-up. A player wins the game if they can flip all but one of the cards face-down. Given that the cards are distinguishable, find the number of ways it is possible to win the game.

I've found the source for your question, and I don't understand their solution. Do you?

 

[rand_althor]

Re: HSC 2015 3U Marathon

I've found the source for your question, and I don't understand their solution. Do you?

Yea I think I do. I wouldn't be able to come up with that in an exam though. Which part didn't you understand?

 

[InteGrand]

Re: HSC 2015 3U Marathon

Yea I think I do. I wouldn't be able to come up with that in an exam though. Which part didn't you understand?

Doubt this question would be in the exam, and if it were, there'd be guiding steps as sub-parts of the question.

 

[rand_althor]

Re: HSC 2015 3U Marathon

Doubt this question would be in the exam, and if it were, there'd be guiding steps as sub-parts of the question.

True. It was from a math tournament.

 

[braintic]

Re: HSC 2015 3U Marathon

Yea I think I do. I wouldn't be able to come up with that in an exam though. Which part didn't you understand?

If I blindly accept the following two statements, then the calculation itself is fine. But they are not obvious to me.

(1) Similarly, the game cannot be won if an isolated card is created on the 3rd, 4th, 5th, ..., 11th turn.

(2) This is because any subsequent card must be one chosen from the two cards on either side of the 'face-down block' (otherwise more than one isolated card is created).

 

[rand_althor]

Re: HSC 2015 3U Marathon

If I blindly accept the following two statements, then the calculation itself is fine. But they are not obvious to me.

(1) Similarly, the game cannot be won if an isolated card is created on the 3rd, 4th, 5th, ..., 11th turn.

(2) This is because any subsequent card must be one chosen from the two cards on either side of the 'face-down block' (otherwise more than one isolated card is created).

(1) I understood this part by actually testing it out. I'll refer to flipping as crossing out on the circle. I drew up the circle, crossed out the top card (the Ace on my circle) and then the one on the right (4). I then made an isolated card by skipping the third card (7) and crossing out the fourth (10). If you continue crossing out all cards clockwise, you eventually reach the Jack which can't be crossed out as it becomes isolated after the 8 and Ace are crossed out. So we end up with two isolated cards, i.e. we've lost the game. I'd imagine there would be similar endings if the isolated card was made on the 4th, 5th, ..., 11th turns. Not sure how to prove this non-empirically though.

(2) The isolated card must be made last to win the game. To do this, we have to pick the cards such that the flipped over cards are all next to each other. If we don't, what happened in (1) will happen, meaning we don't win the game.

 

[braintic]

Re: HSC 2015 3U Marathon

(1) I understood this part by actually testing it out. I'll refer to flipping as crossing out on the circle. I drew up the circle, crossed out the top card (the Ace on my circle) and then the one on the right (4). I then made an isolated card by skipping the third card (7) and crossing out the fourth (10). If you continue crossing out all cards clockwise, you eventually reach the Jack which can't be crossed out as it becomes isolated after the 8 and Ace are crossed out. So we end up with two isolated cards, i.e. we've lost the game. I'd imagine there would be similar endings if the isolated card was made on the 4th, 5th, ..., 11th turns. Not sure how to prove this non-empirically though.

(2) The isolated card must be made last to win the game. To do this, we have to pick the cards such that the flipped over cards are all next to each other. If we don't, what happened in (1) will happen, meaning we don't win the game.

But in (1), you have only jumped three at a time (or multipled of three at a time). How do you test out a more random selection?

 

[rand_althor]

Re: HSC 2015 3U Marathon

But in (1), you have only jumped three at a time (or multipled of three at a time). How do you test out a more random selection?

If you draw the circle as the question suggests ("13 cards (Ace through King in that order) in a circle."), you could try a random selection. For example, starting at the top (Ace) and skipping every second card (2, 4, 6, ...) so that we have every first flipped over (A, 3, 5, ...). You can go around until you reach the 4, which is isolated as the Ace and 7 have been flipped over. At this point, all the remaining cards are isolated (correct me if I'm wrong here, may have made a mistake in my working).

 

[InteGrand]

Re: HSC 2015 3U Marathon

Well in the solutions' equivalent game, if we form an isolated card I before the 12th move, at this stage, there must be at least one that is face up apart from I. This is because we have flipped over fewer than 12 cards, so there must be more than 13 – 12 = 1 face-up cards. Since we have an isolated card, there must be at least two face-down cards (the cards adjacent to I). Consider a set of adjacent face-up cards (other than the set {I}) that lie between two face-down cards (such a set clearly exists (because we are before the 12^th move). Say that there are N cards in this set (clearly N is an integer between 1 and 10 for this game; we can easily generalise it to a game where there are X cards to start off with (here X = 13)). The objective in order to win will be to eventually flip all these cards over to face-down. We show this will be impossible (i.e. an isolated card must be created from within this set).

If N is 1, then already isolated.

If N is 2, then whichever of these two we flip, the other card becomes isolated.

If N is 3, then flipping any of these cards reduces it to one of the cases above.

etc., i.e. assume an isolated card will eventually be created for all lengths N = 1,2,3,...,k, for some specific integer k (assumption for strong induction).

Consider if the length of the ''adjacent-face-ups chain'' is k + 1 cards now. When we turn over one of these cards, the chain is split and gets reduced to a case of a set of 1 or 2 or ... or k face-up cards adjacent. This will result in the formation of an isolated card by the induction assumption.

Hence we will always obtain a new isolated card.

Thus, it is impossible to win if an isolated card is created before the 12^th move (in general before the (X – 1)^th move, where X is the number of cards in total).

 

[InteGrand]

Re: HSC 2015 3U Marathon

If I blindly accept the following two statements, then the calculation itself is fine. But they are not obvious to me.

(1) Similarly, the game cannot be won if an isolated card is created on the 3rd, 4th, 5th, ..., 11th turn.

(2) This is because any subsequent card must be one chosen from the two cards on either side of the 'face-down block' (otherwise more than one isolated card is created).

Oh, and for claim (2) here, it follows from the induction proof above; if we choose a card that is not one of the ones adjacent to the 'face-down block', then we create two chains of adjacent face-up cards (each with length > 0) in between two face-down ones, each of which we proved always results in the creation of an isolated card, and so two isolated cards will be created, which we don't want.

I.e. the second card we flip to face-down must be one of the ones either side of our chosen first card (otherwise we create those chains and make two isolated cards).

Then the third card must be one on the ends of the two adjacent face-down cards.

Etc. (as you can see, we just create a chain of adjacent face-down cards until we reach a final card, which can no longer be flipped, and the game is won; this is the only process by which a game is won, and always results in a won game.)

 

[davidgoes4wce]

Re: HSC 2015 3U Marathon

The number of different arrangements of the letters of the word REGISTER which begin and end with letter R is:

 

[Drsoccerball]

Re: HSC 2015 3U Marathon

The number of different arrangements of the letters of the word REGISTER which begin and end with letter R is:

360.

 

[kawaiipotato]

Re: HSC 2015 3U Marathon

360.

 

[davidgoes4wce]

Re: HSC 2015 3U Marathon

I thought about it the wrong way. This was a MCQ from Girraween High School ( I have never heard of their school) trials but I would have treated the R as times 2 since the R could be in either positions (front or back or in reverse order)

So the lesson from this is they have to be in 'fixed positions'

 

[InteGrand]

Re: HSC 2015 3U Marathon

I thought about it the wrong way. This was a MCQ from Girraween High School ( I have never heard of their school) trials but I would have treated the R as times 2 since the R could be in either positions (front or back or in reverse order)

So the lesson from this is they have to be in 'fixed positions'

The R's are identical, so having reverse order results in the same word. If the letters to go on the end were non-identical (e.g. A and B), then we would need to account for both orders if asked for how many words are possible with A and B at the endpoints.

 

[rand_althor]

Re: HSC 2015 3U Marathon