There are 100 coins placed on the table. 90 with heads facing up and 10 with tails facing up. There is a blind man who obivously cannot see which coins have heads on top. Nor can he make out by feeling the coins. He has been asked to divide the coins into two parts with each part having equal number of heads. How will he do it?

Presumably he knows there are 10 coins that are tails up, in which case he splits the coins into two parts, one with 90 coins and the other with 10. Then he flips every coin in the larger part over.

We know there are 10 tails all together, so let's say there are exactly n tails in the larger part and 10 - n tails in the smaller part. But that means there are n HEADS in the smaller part, as it has 10 coins in total. When the blind man then turns all the coins in the larger part over, he turns heads into tails and tails into heads. Therefore there are now exactly n heads in the larger part, the same as in the smaller part. ðŸ˜€

same as with 90, just the opposite effect. 100 coins, 10 are heads. u take 10 random coins, say 6 heads and 4 tails. that means that there are 4 heads and 86 tails in the other bundle, flip the first bundle and u have 4 heads and 6 tails-that's 4 heads in each bundle!

Originally posted by genius same as with 90, just the opposite effect. 100 coins, 10 are heads. u take 10 random coins, say 6 heads and 4 tails. that means that there are 4 heads and 86 tails in the other bundle, flip the first bundle and u have 4 heads and 6 tails-that's 4 heads in each bundle!

G

Of course. But the original problem had 90 heads and 10 tails. I couldn't think of any solutions other than the one I used.