Once again.. Another coin puzzle after this and this (I wonder whether puzzles have anything else other than hats, kings and coins :D)
I heard this from Rushabh Sheth (Mech Sophie) who got it from Vivek Jha (Elec Thirdie).
There are 100 coins on the table out of which 50 are tailface up and 50 are head face up. You are blind folded and there is no way to determine which side is up by rubbing, etc. You have to divide the 100 coins in two equal halves such that both have equal number of coins with tails face up. (This obviously implies that the two have equal number of coins with heads face up)
Second part: There are 100 coins on the table out of which 10 are tailface up and 90 are head face up. You are blind folded and there is no way to determine which side is up by rubbing, etc. You have to divide the 100 coins in two halves (not necessarily equal) such that both have equal number of coins with tails face up.
Update (02/01/10): Sorry for inducing confusion in the system. :P Solution given by Vivek Jha, posted in comments by me!!
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Source: Sent to me by Gaurav Sinha Problem: Siddhant writes a Maths test and correctly answers 5 out of 6 Arithmetic questions and 20...

This is not a puzzle. So, for those of you who follow this puzzle blog, please bear with me for just one post. Interesting Math in this art...

Source: Sent to me by Gaurav Sinha Problem: Siddhant writes a Maths test and correctly answers 5 out of 6 Arithmetic questions and 20...

Let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process; B keep tos...
For the first part, divide 100 coins in two sets of 50 coins each.
ReplyDeleteThe first set has x heads up coins and 50x tails up coins.
The first set has 50x heads up coins and x tails up coins.
Not flip the coins in the second set. So, both have x coins heads up and 50x coins tails up. :D
For the second part,
Doing the same operation again and again, we will approach towards the solution.
2nd part...you gave incorrect solution....Solution is "deterministic" and simple.....few operations
ReplyDeletehit me with the solution!! :D
ReplyDeleteDivide into two regions of 10 coins and 90 coins...then flip a;; coins of the heap of 10 coins.
ReplyDeleteIf originally there were x heads among those 10 then there were (10x) heads among 90. Flipping the 10 coin set will give you x tails and (10x) heads in that set.
So,you get (10x) heads in both.
Second part of the question was wrong. Hence removed.
ReplyDelete@viki
The question said that we have to divide the 100 coins into two sets of 50 coins each such that both have equal number of heads up. Your method divided it into two unequal parts.
"This obviously implies that the two have equal number of coins with heads face up".
ReplyDeleteThat was just for 50 heads and 50 tails in the beginning.
Sorry for my mistake. Reposted the second part and now the correct solution for both parts. Thanx Vivek.
ReplyDeleteFirst part:
Divide 100 coins in two sets of 50 coins each.
The first set has x heads up coins and 50x tails up coins.
The second set has 50x heads up coins and x tails up coins.
Not flip the coins in the second set. So, both have x coins heads up and 50x coins tails up. :)
Second part:
Divide 100 coins in two sets, one of 10 coins and other of 90 coins.
The first set has x heads up coins and 10x tails up coins.
The second set has 90x heads up coins and x tails up coins.
Not flip the coins in the first set. So, both have x coins tails up.
Hence, done. :)
Thanx again vivek (@viki).
Excellent, thanks for this. It was winding me up not knowing the answer.
ReplyDeleteNB: "Not flip the coins" should read "Now flip the coins".