no one wrote:miyano_shiho wrote:I don't know if you already know this but here goes...
You have 20 cards, 6 with faces up and 14 with faces down. You are in a dark room and you can't see them. How will you divide the cards into two groups so that the number of cards with faces up in each group are equal? Flipping the cards is allowed.
has anyone solved this riddle?
Oh, sorry I forgot about my riddle.

[spoiler=Solution]Suppose you divide the cards into two groups: one containing x cards and the other having (20-x) cards.
In one group there are x cards, containing h cards with faces up and t cards with faces down.
In the other group there are (20-x) cards, containing (6-h) cards with faces up and (14-t) cards with faces down.
Then, you flip all those x cards, resulting in t cards with faces up and h cards with faces down. (Flipping none wouldn't work since it gives us [h=6-h] and it is only possible iff h=3. Flipping only some of the cards wouldn't work either, since we wouldn't be sure of the outcome. Flipping all the cards in the other group as well would just reverse the orientations of the cards, so it wouldn't work too. So we are left with flipping all the cards in only one group...)
In equation form (after flipping):
x = h + t
where h = cards with faces down and t = cards with faces up
The two groups should have the same number of cards with faces up so we get this equation:
t = 6 - h, or
h + t = 6
It then follows that x=6 and (20-x)=14.
So to solve that riddle, you should divide the cards into two groups containing 6 and 14 cards, and flip all those 6 cards.
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I actually thought of that but I said to myself, "It's probably wrong. It doesn't even sound like a word." XD
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