Solution to Problem 1

Let the initial integers be

[ a_1,a_2,\ldots,a_{2026}>1. ]

Part (a): termination and the final number of nontrivial entries

For a positive integer (x), define

[ \Omega(x)=\sum_{p\text{ prime}}v_p(x), ]

where (v_p(x)) is the exponent of the prime (p) in (x). Thus (\Omega(x)) is the number of prime factors of (x), counted with multiplicity.

For a blackboard state (X), define

[ S(X)=\sum_{x\in X}\Omega(x), \qquad R(X)=#{x\in X:x>1}. ]

Suppose a move is made on (m,n>1). Write

[ g=\gcd(m,n), \qquad h=\frac{\operatorname{lcm}(m,n)}{\gcd(m,n)}. ]

For every prime (p), if (a=v_p(m)) and (b=v_p(n)), then

[ v_p(g)=\min(a,b), \qquad v_p(h)=\max(a,b)-\min(a,b)=|a-b|. ]

Consequently,

[ \Omega(g)+\Omega(h) =\Omega(m)+\Omega(n)-\Omega(g). ]

There are two cases.

If (g>1), then (\Omega(g)>0), so (S(X)) strictly decreases.
If (g=1), then the move is [ (m,n)\longmapsto(1,mn). ] In this case (S(X)) is unchanged, but (R(X)) decreases by one.

Therefore the lexicographically ordered pair

[ \Phi(X)=(S(X),R(X))\in\mathbb N\times\mathbb N ]

strictly decreases after every move. The lexicographic order on (\mathbb N\times\mathbb N) is well founded, so only finitely many moves are possible.

At every move the two new integers cannot both be 1, because

[ gh=\operatorname{lcm}(m,n)>1. ]

Thus at least one integer greater than 1 always remains. When the process stops, two integers greater than 1 cannot remain, since they would constitute another legal move. Hence exactly one integer (M>1) remains.

Part (b): the final value is invariant

Fix a prime (p). Consider the multiset of exponents

[ v_p(x_1),v_p(x_2),\ldots,v_p(x_{2026}) ]

of the numbers currently on the blackboard.

A move replaces two exponents (a,b) by

[ \min(a,b),\quad |a-b|. ]

The Euclidean algorithm identity

[ \gcd\bigl(\min(a,b),|a-b|\bigr)=\gcd(a,b) ]

shows that the greatest common divisor of all the (p)-adic exponents is unchanged by every move. Therefore

[ g_p:=\gcd\bigl(v_p(a_1),v_p(a_2),\ldots,v_p(a_{2026})\bigr) ]

is an invariant.

At the terminal state, all entries except (M) are 1. Their (p)-adic exponents are therefore

[ v_p(M),0,0,\ldots,0. ]

The gcd of these exponents is (v_p(M)), so invariance gives

[ v_p(M)=g_p. ]

This holds for every prime (p). Hence

[ \boxed{ M=\prod_{p\text{ prime}} p^{\gcd_{1\le i\le 2026}v_p(a_i)} }. ]

Only finitely many primes occur in this product. The right-hand side depends solely on the initial blackboard, so (M) is independent of all choices made during the process.