Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

you have a pile of 100 coins with 90 tails up and 10 heads up.You have to divide this pile of 100 coins into two piles(may be of unequal size).You are blind and you have to divide it such that both the piles have equal number of heads up.

share|improve this question
1  
You forgot to mention you're also wearing gloves, so you can't feel each coin's face either. –  Joe Z. Sep 23 '13 at 1:52
    
ohkk..but that was understood i guess! –  TSP1993 Sep 25 '13 at 5:21

1 Answer 1

Instead of tackling this problem with 100 coins, try a simpler version of it:

Let's say I give you a pile of 3 coins, 2 are heads up and the 3rd is tails up. You're blind-folded, but since there are only three coins, its easy to picture how they might be stacked when I give them to you.

From top to bottom:

THH

HTH

HHT

If it were the first case, THH, what should we do? Let's first move the top coin, T, into a second pile, so you have HH in one and T in the other. Now how do I make sure they both have the same number of heads up/tails up coins? I can flip that one coin I moved over, so now I have HH and H. DONE!

If it were the second case, let's try the same thing. Move the top coin, H, into a second pile, so now we have TH in one pile and H in the other. What should we do? Flip the H we just moved over!

Lastly, if we have HHT, again, move the top H over so we have HT and H. I'm going to guess you get the trick now.

Now, this problem was set up with 3 coins where only 1 coin was flipped. Can you imagine what might happen when we go back to your problem of 100 coins with 10 flipped?

share|improve this answer
    
yeah i got it now!! –  TSP1993 Sep 25 '13 at 5:20

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.