HongKong 2022-23 IMO Selection Tests

Answer: $14$

First note that there are exactly three even numbers and one odd number. (If all four numbers are even then all H.C.F.'s would be even, whereas if there are at most two even numbers then at most one H.C.F. can be even.) Suppose $a$ is the odd number and $b,c,d$ are the even numbers. We use $(m,n)$ to denote the H.C.F. of $m$ and $n$. Then $(a,b)$, $(a,c)$ and $(a,d)$ must be $1$, $3$ and $5$ in some order. In particular, we note that $a$ is divisible by $3$ and $5$, whereas exactly one of $b,c,d$ is divisible by $3$, and exactly one of $b,c,d$ is divisible by $5$.

WLOG suppose $(b,c)=2$, $(b,d)=4$ and $(c,d)=k>5$. Then $b$ and $d$ are both divisible by $4$, and $c$ is divisible by $2$ but not $4$. It follows that $k$ is also divisible by $2$ but not $4$. Furthermore, $k$ is not divisible by $3$ since $c$ and $d$ cannot be both divisible by $3$, and similarly $k$ is not divisible by $5$.

Now we know that $k$ is greater than $5$, is even, and is not divisible by $3$, $4$, $5$. The smallest such $k$ is thus $14$. Indeed this works; for instance if the four numbers are $(a,b,c,d)=(15,4,70,84)$, then the H.C.F.'s involving $a$ will be equal to $1$, $3$, $5$, while those not involving $a$ will be equal to $2$, $4$, $14$. It follows that the answer is $14$.