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You are in an absolutely dark room, with (a large number of) coins strewn all over the floor. You know that exactly 20 coins are tails up. You need to split the coins into two groups, such that both have the same number of tails. How would you achieve this?

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    $\begingroup$ You forgot to mention you're wearing gloves, and can't take them off. $\endgroup$ – Joe Z. May 7 '15 at 12:56
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Assuming the two groups don't need to be of equal size:

Take a group of $20$ coins, and flip them

This should work because:

You take $x$ (with $0\le x\le20$) tails-up coins, and change them to heads, thus leaving $20-x$ tails-up coins on the floor.
You've also taken $20-x$ heads-up coins, and changed them to tails.

You now have a group of $20$ coins and the group of remaining coins, both containing $20-x$ tails-up coins

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  • $\begingroup$ This includes my initial reaction of an answer: "Two groups of 0 coins each" $\endgroup$ – Rj Geraci Oct 2 '15 at 17:22
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Some lateral thinking answers:

  • Turn the light on.
  • Put on your night vision goggles.
  • Use your smartphone as a flashlight.
  • Start a fire.
  • Use your well trained cat to find the coins.
  • Open the door and use the sunlight.
  • Melt all the coins and reforge two coins with tails on both sides.
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Make two groups of coins, and in one group take any 20 coins out of large number of coins and change the faces of each of these coins. Now both groups will have an exactly equal number of "tails" faces.

Example:

Suppose we have 6 coins which are "tails" in one group, then other group will have 14... i.e, (20-6) "tails" coins (because it is given that only 20 coins are "tails")

After changing each face of 20 coins in the group, now it will have 6 "heads" faces & 14 "tails" coins.

In this way we have made the number of tails faces the same in both groups. (You can assume any number of tale face in any group, and then you can apply the concept)

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