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HSC 2013 MX2 Marathon (archive) (6 Viewers)

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GoldyOrNugget

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Re: HSC 2013 4U Marathon

Nup. It's a circle, and also I think you have to use cases.

The answer is 105/7 = 15. You don't have to use cases -- you can prove that for any of these bead strings, none of its rotations will be identical to itself, so there's no risk of double counting: 7!/(4!*2!*1!) will count every rotation precisely 7 times.

For the record: ['ggrggyr', 'gggrgyr', 'ggrggry', 'ggggyrr', 'ggrgygr', 'gggrygr', 'gggrrgy', 'ggggrry', 'ggrrggy', 'ggggryr', 'gggrgry', 'ggrgrgy', 'gggyrgr', 'gggygrr', 'ggygrgr']
 

RealiseNothing

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Re: HSC 2013 4U Marathon

The answer is 105/7 = 15. You don't have to use cases -- you can prove that for any of these bead strings, none of its rotations will be identical to itself, so there's no risk of double counting: 7!/(4!*2!*1!) will count every rotation precisely 7 times.

For the record: ['ggrggyr', 'gggrgyr', 'ggrggry', 'ggggyrr', 'ggrgygr', 'gggrygr', 'gggrrgy', 'ggggrry', 'ggrrggy', 'ggggryr', 'gggrgry', 'ggrgrgy', 'gggyrgr', 'gggygrr', 'ggygrgr']
Couldn't you just treat it as a circle:

 

GoldyOrNugget

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Re: HSC 2013 4U Marathon

Couldn't you just treat it as a circle:

Only once you've made the observation that all rotations are different to each other. With your method, if the beads were RGRG (in which case, you can rotate it by two beads and get RGRG again), you'd get 3!/(2!2!) = 6/4 (uh oh...)
 

study1234

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Re: HSC 2013 4U Marathon

The answer is 105/7 = 15. You don't have to use cases -- you can prove that for any of these bead strings, none of its rotations will be identical to itself, so there's no risk of double counting: 7!/(4!*2!*1!) will count every rotation precisely 7 times.

For the record: ['ggrggyr', 'gggrgyr', 'ggrggry', 'ggggyrr', 'ggrgygr', 'gggrygr', 'gggrrgy', 'ggggrry', 'ggrrggy', 'ggggryr', 'gggrgry', 'ggrgrgy', 'gggyrgr', 'gggygrr', 'ggygrgr']
The answer is actually 9. You have to consider when the bracelet if flipped (i.e. anticlockwise is the same as clockwise). This makes combinations such as ggrggry the same as ggrggyr. These are the combinations: (ggrggyr, gggrgyr, ggggyrr, ggrgygr, gggrrgy, ggrrggy, ggggryr, gggrgry, ggrgrgy).

However, does anyone know to get the answer as 9 by using permutations and combinations, rather than trial and error?
 

Sy123

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Re: HSC 2013 4U Marathon





 
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Sy123

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Re: HSC 2013 4U Marathon



I got different!

You may be correct, I did a the substitution x=sin^2 u which got me to a trig recurrence. So maybe I made a mistake.
But I'm not sure I don't have the answer with me.
 

hayabusaboston

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Re: HSC 2013 4U Marathon

Was about to ask if you come up with all these questions yourself Sy, but then I think I saw this question in cambridge somewhere. Hm.
Anyway, im here to request your participation in the "Accurate IQ test" thread started by Hypem. That is all.
 

hayabusaboston

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Re: HSC 2013 4U Marathon



I got different!
Do you even know how much u confused me with ur polynomials question, where u made it a^2 not a, that had me going for hours and hours trying to figure out how u got a^2 and it turns out it was a all along lol. Btw If U have a cool conics question can u plz give to me.
 

Sy123

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Re: HSC 2013 4U Marathon

totally not giving away the answer here ;)
Hmm is there perhaps a really elegant solution here? It is an early IMO question and I was able to solve it with a bit of algebra and substitution.
 

RealiseNothing

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Re: HSC 2013 4U Marathon

Hmm is there perhaps a really elegant solution here? It is an early IMO question and I was able to solve it with a bit of algebra and substitution.
It's just backwards multiplication of two numbers. The original number ends in 6, and when multiplied by 4 gets the same number but with a 6 out the front instead of the end:



Multiply by the 4 and carry the 2 (since you multiply by the units and carry the tens)



Multiply by the 8 and carry the 1:



Multiply by the 3 and carry the 3:



Multiply by the 5 and carry the 1:



Multiply by the 1 and carry the 2:



Since we have just 6, we stop here, and so our number is all the numbers we multiplied by (if you read the 1st number of each step going from top to bottom) - 648351 - but we reverse it as we did the multiplcation in the reverse order to get 153846.
 
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