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 1


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 $\frac{1}{10}$.

  • 5
    $\begingroup$ 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, 2015 at 4:23

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