(Heard this from a friend today)
There are a 100 coins laid out on a table in a dark room. 20 of them are showing heads, and 80 are tails. Your job: When you leave the dark room, the coins should be split into two groups, both of which contain exactly the same number of Heads. You cannot see the coins and you cannot tell heads from tails just by feeling them.
This can be done deterministically. Your time starts now!
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10 comments:
i like i like... i like puzzles...
ok my best guess (based on the way you've worded it) is you simply count 50 coins and put them on one side and leave the other 50 on the other side. You DID ask that the coins be split up into two groups with an equal number of heads... well each coin DOES have a head and a tail...
SO, as long as you split them into two equal groups both would have an equal number of heads...
so do u intend on having a new puzzle everyday? cuz that'd be cool... if you do i'll send you some...
ciao
Well that's a smart answer =) but there's a proper solution to this, even if you consider only the exposed faces..
I wasn't planning on making this a regular feature, but whenever I do come across one I like (and have solved), I'll post it here just to see you guys struggle and prove myself better =).
$%*#! I HATE I HATE I HATE ... you!
ok this one's eating me up and I know you're enjoying it...
Ok once again... your wording... dark room... are we talking about a room that is pitch black or a dark room in a photographer's lingo... cuz if its the later... well the red light's on and you can see the coins... but i think this is rubbish cuz, you go on to say that you cannot see the coins... BAH! ok fine will have to respond later...
i'm enjoying this more and more..
No, this is not a play on words... here's a hint for you, dimwit =)
HINT ALERT!
Hint 1: You need to have the same number of heads, but I didn't say anything about the number of tails.
HINT ALERT!
Hint 2: What if, instead of 100 coins with 20 heads + 80 tails, I said 2 coins, 1 head + 1 tail?
if there were just two coins, you would just flip the other one over and have one head in each group. But that still doesnt answer the question, cuz how do you know which coin is heads and which tails prior to flipping one over to make them both heads?
I could have just as easily said you'd have to eliminate all the heads and thus you would still have an 'equal' number of heads, i.e., zero...
Look obviously I'm too stupid for this puzzle, so you've gotta help me out... cuz its been driving me crazy... mail me the answer :)
Zero heads on both sides *is* an equal number of heads, that's valid.
Lemme know if i'm close :P...
* Treat the prob like recursion. At the simplest form, we have two coins. Doesn't matter which is what. Flip one of them and the answer's good.
* Now treat each pair as a 'unit'. When combining two units, flip one of them. (all coins in that unit get flipped )
* So the 'units' keep getting larger .. first 2 coins, then 4, then 8 and so on.
.
.
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I keep getting stuck with what to do with the remaining coins at the end (odd number).. :( .. which is why i'm not sure if this is anywhere near the solution..
@Randomizer
It's hard to give you a hint that's useful yet not practically giving away the answer... ok, think about this same game, but with only 3 coins.
--SPOILER ALERT--
Hey I got it I think.
flip 20 coins and put them in one group. The rest of the 80 will automatically have the same number of heads.
For eg. if out of the 20 coins we picked, 5 were originally heads, then the new dynamics are 20-5=15 heads in the bigger group.. and 15 tails were converted to heads in our smaller group... so 15=15 heads
neat question :)
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