Question

I will give brainliest to a satisfactory answer:
2016 coins are placed on table with 50 heads up and remaining tails up.Suppose you are blindfolded and the only thing you can do is flip the coins.How can you separate 2016 coins into two groups such that each group has equal number of head?

Answer 1

Here is the solution:

●Divide the coins into 2 parts -1st part with 50 coins and the 2nd part containing the remaining coins(i.e the rest 1966 coins)

●Let the no.of heads in the 2nd part be x.
So no.of heads in 1st part =50-x [since total no.of heads=50]
So the no.of tails in 1st part=50-(50-x)=x

●We now flip the first 50 coins .So the heads become tails and tails become heads.
So now the no.of heads in 1st set becomes =x
which is equal to the number of heads in 2nd part.

SO THE NUMBER OF HEADS ON BOTH SIDES BECOMES EQUAL.

Thank You for the Wonderful Question.
Hope this helps you. : )

Answer 2

2016 coins are placed on table with 50 heads up and remaining tails up.Suppose you are blindfolded and the only thing you can do is flip the coins.How can you separate 2016 coins into two groups such that each group has equal number of ...