There are 100 5 ruble coins on a table. 10 of them are tails-up. Your zodiac sign is Taurus. While blindfolded, can you separate the coins into 2 piles, each with the same number of tails-up coins?

Any takers? Hints: type of coin is irrelevant, as long as they have 2 distinct faces (heads/tails), the coins don't care when your birthday is, and you can flip coins if you so choose.

Figured it out. Take ten coins out and flip them over into a second pile. Say you picked up 4 tails and 6 heads, turn them over and you have 4 heads and 6 tails in your small pile and by moving 4 tails from the big pile to the small pile your left with 6 tails in the big pile. It works any way you slice it. I learned this one with 16 coins in high school, but I guess it works with any number of coins that has its square root as the number of tails up coins. Edit: just confirmed this in real life with sixteen pennies on a table, four of them tails up. Closing my eyes and flipping over four coins at random into a seperate pile always leaves the same number of tails up coins in each pile. 1 tail in each pile 2 tails each pile 3 tails each pile 0 tails each pile, and obviously if this were inversed it would be 4 tails each pile. Try it for yourself! It's kinda cool! Fun stuff charmadillo!!

Initial pile = 90 H (heads) 10 T (tails) Remove at random ten coins, say one 1T 9H, big pile now equals 81H 9T. Flip the removed coins in your new small pile and 1T 9H turns into 1H 9T, giving you nine tails up coins in each pile. Try that with any combo of heads and tails coins (2T 8H, 7T 3H, 10T 0H, 0T 10H etc.), as long as you are only removing and flipping ten coins, or an amount equal to the square root of the total number of coins you will always end up with the same number of tails up coins in each pile.