Estonian Mathematical Olympiad

At all points throughout the process, every number on the board can be written as $2^{\alpha }{3}^{\beta }$ for some $0\le \alpha ,\beta \le 100$. Moreover, the list of exponents of both $2$ and $3$ does not change throughout the process. Indeed, in a step of the first kind, two numbers $2^{{\alpha }_1}{3}^{{\beta }_1}$ and $2^{{\alpha }_2}{3}^{{\beta }_2}$ are turned into $2^{\min ({\alpha }_1,{\alpha }_2)}{3}^{\min ({\beta }_1,{\beta }_2)}$ and $2^{\max ({\alpha }_1,{\alpha }_2)}{3}^{\max ({\beta }_1,{\beta }_2)}$, preserving both the exponents ${\alpha }_1,{\alpha }_2$ of $2$ and the exponents ${\beta }_1,{\beta }_2$ of $3$. A step of the second kind is just the reverse of a step of the first kind, so the exponents are preserved again. Thus the lists of both the exponents of $2$ and $3$ are $$\underset{103\text{ copies}}{\underbrace{0,\dots ,0}};1,1,2,2,\dots ,100,100.$$

By the rearrangement inequality, we obtain that the sum of the numbers on the board is the greatest when they are $$\underset{103\text{ copies}}{\underbrace{1,\dots ,1}};6,6,6^2,6^2,\dots ,6^{100},6^{100}.$$ The sum of these numbers equals $101+2\cdot \frac{6^{101}-1}{5}$. These numbers can be achieved by performing a step of the first kind on the numbers $2^{\gamma }$ and $3^{\gamma }$ for every $\gamma =1,\dots ,100$.

By the rearrangement inequality, we obtain that the sum of the numbers on the board is the least when the lists of the exponents of $2$ and $3$ are sorted in opposite directions. This means that the numbers would be $$\begin{array}{cccccccccc}{2}^{100}& 2^{100}& 2^{99}& 2^{99}& 2^{98}& 2^{98}& \dots & 2^{51}& 2^{51}& 2^{50}\\ 2^{50}& 2^{49}& 2^{49}\cdot 3& 2^{48}\cdot 3& 2^{48}\cdot 3^2& 2^{47}\cdot 3^2& \dots & 2\cdot 3^{49}& 3^{49}& 3^{50}\\ 3^{50}& 3^{51}& 3^{51}& 3^{52}& 3^{52}& 3^{53}& \dots & 3^{99}& 3^{100}& 3^{100}.\end{array}$$ Using the formula for the sum of a geometric series for each row, separately for numbers at odd positions and even positions, yields: for the first row, $2^{101}-2^{50}$ and $2^{101}-2^{51}$; for the second row, $3^{51}-2^{51}$ and $3^{50}-2^{50}$; * for the third row, $\frac{1}{2}(3^{101}-3^{50})$ and $\frac{1}{2}(3^{101}-3^{51})$.

Thus the sum of all the numbers is $2^{102}-3\cdot 2^{51}+2\cdot 3^{50}+3^{101}$. This is indeed achievable. For this, we will pair the first desired number with the last one, the second first number with the second last one, etc. The least common multiples of the pairs are $6^{100},6^{100},\dots ,6^{26},6^{26},6^{25}$ and the greatest common divisors are $\underset{103\text{ copies}}{\underbrace{1,1,\dots ,1}};6,6,6^2,6^2,\dots ,6^{24},6^{24}$. Finally, $6^{25}$ is left over. This means that we can use steps of the second kind to achieve the desired numbers from the situation with the maximal sum.