Riddles/Brainteasers

[EverettKings]

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Specifically, where does [ QUOTE ]
In group B there are 50-x coins, (x-y) of which are heads .

[/ QUOTE ] this assumption come from?

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In group A we had y heads, and we have x total heads, so that leaves (x-y) for group B.

[Hornacek]

tuq, me > you

[Requin]

I thought X represents the total number of coins in group A?
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In group A there are x coins

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It can't be both that and the number of heads in total.

OOOOOOOOOOOooooooooh I missed where they told us how many heads there are.

[Terrabon98]

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Great start. In group A there are x coins, y of which are heads and (x-y) of which are tails. In group B there are 50-x coins, (x-y) of which are heads. If we flip all coins in group A, then will will now have y tails and (x-y) heads (as all heads become tails and all tails become heads), exactly the same as group B.

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Yeah so that ended up being totally solvable and a lot easier than I thought it would be. I suck.

[/ QUOTE ]Don't feel too bad, I still don't get it. Say we split it into 2 groups, Group A having 20 coins 10 of which are HU and 10 of which are TU, Group B having 30 coins 15 HU 15 TU. Flipping all of group A gives us 10 HU coins, which isn't equal to the 15 HU coins in group B. What am I missing?

Specifically, where does [ QUOTE ]
In group B there are 50-x coins, (x-y) of which are heads .

[/ QUOTE ] this assumption come from?

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Just remember that the two piles are complements. An example might make it easier.

EXAMPLE: You have 50 coins, 20 heads and 30 tails. You split them into two piles, and the pile you are manipulating has 20 coins, 5 heads and 15 tails. Therefore if you have 5 heads in this pile, there will be 15 in the other pile. Now take our pile, and flip every coin. The 5 heads become tails, and the 15 tails become heads. You now have equal amounts in each pile

[Josh W]

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Thats not too hard right? I cant do it while multitabling cause it'll need pythagoras' formula, but it shouldnt be too hard.

Edit: or, some sinus cosinus stuff. Anyway you have 2 edges of a triangle, you just need to calculate the third.

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what are the two edges?

[/ QUOTE ]You can't just make it a triangle with 42 and 10 edges, and find the hypotenuse?

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The hypotenuse is longer than either side of a right triangle, right? So, your answer is definitely longer than 42.

You could acheive 42 just by walking straight down (11), straight along the bottom (30), and straight up (1). So your pythagorean way is definitely not optimal.

Josh

[GuyOnTilt]

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"Is either one of the following true?
A) The sky is blue and 'da' means yes
B) The sky is not blue and 'ja' means yes"


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So the True God will always answer "da" and the False God "ja", right. Gotcha so far.


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So we have forced 'da' to mean yes and 'ja' to mean no by applying this logic.

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I don´t think you have. You´ve just forced the True God to always answer with "ja", regardless of whether ja means Yes or No (and similarly with the False God). That DOES NOT mean that "ja" means "yes".

[Yeti]

GoT,

In the riddle you gave, x+y < 100 is supposed to be given in the question. But I guess it was pretty obvious anyway that the solutions wouldn't be that large.

[GuyOnTilt]

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In the riddle you gave, x+y < 100 is supposed to be given in the question. But I guess it was pretty obvious anyway that the solutions wouldn't be that large.

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No, x+y can be > 100. I ended up not working on this at all today. I´ll do it on my plane/bus ride tomorrow if I´m not sleeping.

[Yeti]

OK, I googled the puzzle and that's what I read.

[srjunkacct]

Re: the product and sum riddle:

Goldbach's conjecture is useful.

If the Sum guy had been given an even number, he would not be able to make his statement, as the sum could conceivably be a sum of two primes, whereby Product guy would have been able to deduce the two primes.