you have a 100 coins laying flat on a table, each with a head side and a tail side. 10 of them are heads up, 90 are tails up. you can't feel, see or in any other way find out which side is up. split the coins into two piles such that there are the same number of heads in each pile.
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Pick any 10 coins and make a small pile. You now have 90 coins in a pile (call it pile A), and 10 coins in a new pile (call it pile B). You don't know where the heads are, but you do know that there are 10 of them in total, and therefore however many heads there are in pile A, there is 10 minus this many in pile B.
Let's do an inventory:
Pile A contains:
- n heads
- 90-n tails
Pile B contains:
- 10-n heads
- the rest tails. How many tails exactly? Well out of 10 coins, 10-n are heads, which means the number of tails is 10-(10-n) = n tails.
Wait...what? pile A has n heads, and pile B has n tails? So we know that there are always the same number of heads in pile A as there are tails in pile B, even if we don't know exactly what number that is? So flip over all the coins in pile B. We now have the same number of heads in both piles.
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Let's do an inventory:
Pile A contains:
- n heads
- 90-n tails
Pile B contains:
- 10-n heads
- the rest tails. How many tails exactly? Well out of 10 coins, 10-n are heads, which means the number of tails is 10-(10-n) = n tails.
Wait...what? pile A has n heads, and pile B has n tails? So we know that there are always the same number of heads in pile A as there are tails in pile B, even if we don't know exactly what number that is? So flip over all the coins in pile B. We now have the same number of heads in both piles.
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Answered by
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Nice question :)
First we pick up any 10 coins and place them in a pile.
100 - 10 = 90 coins in another pile.( Pile 1)
10 coins form Pile 2
But we don't have any idea how many heads or tails are there.
So using little mathematics,
In 1st pile --> 90 coins in total
Number of heads be x.. ...........(i)
Number of tails will be 90 - x
In 2nd pile ,
Number of heads will be 10 - x
Number of tails = 10 - (10 - x) ...................since total number of coins = 10
Number of tails = 10 - 10 + x = x ..........(ii)
Now let's compare (i) and (ii)
We see that number of heads in 1st pile = number of tails in 2nd pile.
So, even if we don't know the exact number coins we will be able to say with confidence that there will be equal number of heads in 1st pile and tails in 2nd pile.There will be same number of heads in both the piles,flip all the coins in 2nd pile!
First we pick up any 10 coins and place them in a pile.
100 - 10 = 90 coins in another pile.( Pile 1)
10 coins form Pile 2
But we don't have any idea how many heads or tails are there.
So using little mathematics,
In 1st pile --> 90 coins in total
Number of heads be x.. ...........(i)
Number of tails will be 90 - x
In 2nd pile ,
Number of heads will be 10 - x
Number of tails = 10 - (10 - x) ...................since total number of coins = 10
Number of tails = 10 - 10 + x = x ..........(ii)
Now let's compare (i) and (ii)
We see that number of heads in 1st pile = number of tails in 2nd pile.
So, even if we don't know the exact number coins we will be able to say with confidence that there will be equal number of heads in 1st pile and tails in 2nd pile.There will be same number of heads in both the piles,flip all the coins in 2nd pile!
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