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?


Notice that


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 $\frac{1}{10}$.

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    $\begingroup$ 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. $\endgroup$ – xnor Oct 15 '15 at 4:23

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