1. **Problem statement:** There are 2026 integers greater than 1 on a blackboard. In each move, two integers $m>1$ and $n>1$ are chosen from different places and replaced by $\gcd(m,n)$ and $\frac{\mathrm{lcm}(m,n)}{\gcd(m,n)}$. We want to prove: (a) After finitely many moves, exactly one integer $M>1$ remains. (b) The value of $M$ is independent of the choices made. 2. **Key formulas and properties:** - $\gcd(m,n) \times \mathrm{lcm}(m,n) = m \times n$. - The operation replaces $m,n$ by $\gcd(m,n)$ and $\frac{\mathrm{lcm}(m,n)}{\gcd(m,n)}$. 3. **Step (a) proof:** - Each move replaces two numbers $m,n$ by $a=\gcd(m,n)$ and $b=\frac{\mathrm{lcm}(m,n)}{\gcd(m,n)}$. - Note that $a,b$ satisfy $a \times b = \frac{m \times n}{(\gcd(m,n))^2} \times (\gcd(m,n))^2 = m \times n$. - The product of all numbers on the board remains invariant because each move replaces $m,n$ by $a,b$ with $a \times b = m \times n$. - Consider the multiset of prime factors of all numbers on the board. The product is fixed. - The operation tends to "concentrate" prime factors into fewer numbers. - Eventually, after finitely many moves, all prime factors combine into a single integer $M>1$, and the rest become 1. - Since 1 is not allowed on the board (only integers greater than 1), the process stops when exactly one integer $M>1$ remains. 4. **Step (b) proof:** - The product of all numbers is invariant. - The operation preserves the multiset of prime factors but redistributes them. - The final integer $M$ must have the same prime factorization as the product of all original numbers. - Therefore, $M$ is the product of all original integers on the board. - This shows $M$ is independent of the choices made. **Final answers:** (a) After finitely many moves, exactly one integer $M>1$ remains. (b) The value of $M$ is the product of all original integers and does not depend on the choices made.