# Triangle of numbers on a blackboard

Ten rows of numbers are written on a blackboard. The first row has one $1$, the second has two $2$'s, and so on up through the tenth row with ten $10$'s: $$\begin{gather*} 1\\ 2,\;2\\ 3,\;3,\;3\\ \vdots\\ 10,\;10,\ldots,\;10 \end{gather*}$$ Choose two of the numbers on the board, erase them, and write their product divided by their sum (which will likely be a fraction). Repeat the process until only one number remains.

What is the largest value that the remaining number could be? Also, what is the smallest value?

## 1 Answer

Notice that

$$\frac{1}{\frac{xy}{x+y}}=\frac{x+y}{xy}=\frac{1}{x}+\frac{1}{y}$$

So the sum of the reciprocals of the numbers on the board is always preserved. The sum of the reciprocals of the numbers in each row is $1$, so the sum of all the reciprocals is $10$. Therefore, the last number left must be $\frac{1}{10}$.

• I want to point out the general idea here that also solves this problem: apply a transformation to the numbers so that the combining operation on the transformed numbers is commutative and associative. Here, inverting makes the operation be addition, and there, adding 1 makes the operation be multiplication. – xnor Oct 15 '15 at 4:23