You are blindfolded and taken into a room with two tables. There are coins scattered on one table. You are told the number of coins which are heads up on this table. The second table is empty. You are allowed to move coins from one table to another or flip them. Before you leave the room, there must be an equal number of coins heads up on each table. How can you do it?

Sunday, April 1, 2012

Please email your solutions to puzzles@cotpi.com.

### 7 comments

### Credit

This puzzle is based on a puzzle from folklore:

- Haspel, Aaron. "How to solve it." God of the Machine. 22 Nov. 2006. University of Cambridge. 5 Apr. 2012 <http://www.godofthemachine.com/?p=596>. Problem 2.

## Abhishek Sethi solved this puzzle:

Let the number of coins with heads up be \(x\). Now I transfer \(x\) coins to the second table. Then I flip all the coins on the second table. This ensures that both tables have an equal number of coins with heads up. Initially, I have \(x\) coins which are heads up. I transfer \(x\) coins to the second table. If the second table has \(y\) coins with heads up, then we know that the first table has \(x - y\) coins with heads up, so in order to have \(x - y\) coins with heads up on the second table, I flip all coins on the second table. In this way, I can ensure that both tables have an equal number of coins with heads up.

## Siddhesh Chaubal solved this puzzle:

Let the number of coins with heads up be \(x\). Shift \(x\) coins arbitrarily to the second table and flip all the coins on the second table. After shifting \(x\) coins to the second table, if there were \(y\) coins with heads up on it, then after flipping them, it has \(x - y\) coins with heads up. Since there were \(x\) coins with heads up initially, the first table has \(x - y\) coins with heads up now. Thus, both tables have an equal number of coins with heads up.

## Gabriel Elpers solved this puzzle:

Given a table with \(m\) heads, select any \(m\) coins from the first table, move them to the second table, and flip all the coins just moved. Supposing that the number of tails in our selection was \(k\), and hence the number of heads removed from the first table was \(m - k\), there are now \(k\) heads on the second table and \(m - (m - k) = k\) heads on the first table, which solves the problem.

## Atul S Vasu solved this puzzle:

Let there be \(n\) coins, \(k\) heads up. Take \(m\) coins and move them to the other table and flip them. Now, let there be \(x\) coins heads up on the first table. Thus, there are \(m - (k - x)\) coins heads up on the second table. We want \(x = m - (k - x)\), so \(m = k\).

## Gnanasenthil G solved this puzzle:

Let the number of coins heads up be \(h\). Retain \(h\) coins on the first table and transfer the remaining ones to the second table without flipping them. Then flip all coins on the first table. Now, there are \(h\) coins on the first table. Let the number of heads on the first table be \(x\). Then the number of tails on it is \(h - x\) and the number of heads in the second table is also \(h - x\). If all the coins in the first table are flipped, the number of heads in the first table becomes \(h - x\). Now both tables have an equal number of coins heads up.

## Srijit Dutt solved this puzzle:

Let the number of coins heads up on the first table be \(n\). Pick \(n\) coins from the first table and place them on the second table. Flip all \(n\) coins on the second table. We will be left with an equal number of coins heads up on both tables.

The following table shows the number of coins heads up on both tables:

## Adam Callanan solved this puzzle:

Let us assume that \(x\) of an unknown number of coins on the first table are heads up. The first step would be to take \(x\) coins off the first table and place them on the second table, being careful to keep their facing. Now let us assume that there are \(y\) coins heads up on the second table, such that \(y\) is less than or equal to \(x\). That would mean on the first table, there remain \(x - y\) coins heads up.

Now if we have \(x\) coins on the second table, \(y\) of which are heads up, there are \(x - y\) coins tails up on the second table. If we flip over every single coin on the second table, we end up with \(x - y\) coins heads up on both tables.