2026 IMO Problem 1 — Natural-Language Solution

Problem

There are $2026$ integers greater than $1$ written on a blackboard, not necessarily distinct. In a move, Confucius chooses two integers $m>1$ and $n>1$ from different places on the blackboard and replaces them with

\gcd(m,n) \quad\text{and}\quad \frac{\mathrm{lcm}(m,n)}{\gcd(m,n)}.

He continues to make moves while possible.

Prove that, regardless of Confucius's choices, after finitely many moves exactly one integer $M$ on the blackboard is greater than $1$ .
Prove that $M$ does not depend on Confucius's choices.

Here $\gcd$ denotes the greatest common divisor and $\mathrm{lcm}$ denotes the least common multiple.

Solution

Let the current entries be $a_1,\ldots,a_{2026}$ . Call an entry active if it is greater than $1$ .

Termination

Every entry remains positive after a move. Moreover, at least one of the two new entries is active: if $g=\gcd(m,n)>1$ , then the new gcd entry is active; if $g=1$ , then the other new entry is

\frac{\mathrm{lcm}(m,n)}{g}=\mathrm{lcm}(m,n)>1.

Consider the integer-valued rank

R(a_1,\ldots,a_{2026}) =2027\prod_{i=1}^{2026}a_i +\left|\{i:a_i>1\}\right|.

Suppose a move is made using $m$ and $n$ , and put $g=\gcd(m,n)$ . The product of the two old entries is

mn=g\mathrm{lcm}(m,n),

whereas the product of the two new entries is

g\cdot\frac{\mathrm{lcm}(m,n)}g =\mathrm{lcm}(m,n).

Thus the old total product is $g$ times the new total product. If $g>1$ , the product drops by at least $1$ , so the first term in $R$ drops by at least $2027$ ; meanwhile the active count can change by at most $2026$ . Hence $R$ decreases. If $g=1$ , the product is unchanged, but the gcd entry becomes $1$ , while the lcm entry stays active. The two selected active entries are therefore replaced by only one active entry, so $R$ again decreases.

The rank is a natural number, so an infinite sequence of moves is impossible. At a terminal position there cannot be two active entries, since any two such entries would permit another move. On the other hand, the positivity argument above shows that an active entry always remains. Therefore every play terminates with exactly one entry greater than $1$ .

Independence of the final value

Fix a prime $p$ , and write $v_p(x)$ for the exponent of $p$ in $x$ . For the two selected entries, set

x=v_p(m),\qquad y=v_p(n).

After the move, their two exponents become

\min(x,y) \quad\text{and}\quad \max(x,y)-\min(x,y).

Consequently,

\gcd\bigl(\min(x,y),\max(x,y)-\min(x,y)\bigr)=\gcd(x,y).

All other entries are unchanged. It follows that

d_p=\gcd_{1\le i\le 2026}v_p(a_i)

is invariant under every move.

At a terminal position all entries except the unique active entry $M$ equal $1$ , so all their $p$ -adic exponents are zero. Hence

v_p(M)=d_p

for every prime $p$ . Therefore the final value is forced to be

\boxed{M=\prod_p p^{\gcd_{1\le i\le 2026}v_p(a_i)}}

where only primes dividing the product of the initial entries contribute. This expression depends only on the initial board, so $M$ is independent of every choice made during the process.