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Merging by a strange rule

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The numbers $2$ , $3$ , $4$ , $6$ and $12$ are written on a board. A move erases any $2$ numbers $a$ and $b$ and writes $\frac{ab}{a + b}$ in their place. After $4$ moves a single number remains, whatever order the moves take. What is that number? Give the answer as a fraction in lowest terms.

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Hint

Look at what the move does to the reciprocals of the numbers.

Solution

If $c = \frac{ab}{a + b}$ , then $\frac{1}{c} = \frac{a + b}{ab} = \frac{1}{a} + \frac{1}{b}$ .
So each move keeps the sum of the reciprocals of all the numbers on the board unchanged.
At the start that sum is $\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} + \frac{1}{12} = \frac{4}{3}$ .
The last number $x$ therefore has $\frac{1}{x} = \frac{4}{3}$ , so $x = 1 \div (\frac{4}{3}) = \frac{3}{4}$ .

Answer: $\frac{3}{4}$

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