Problem 8 · Six almost-chained numbers
A $2,3$ fork that merges at $6$ saves a doubling.
Integer answer, at most 4 digitsWe want to choose six positive integers so that in every pair from the set one divides the other, except for exactly one pair in which neither divides the other. If $n$ is the largest number of the set, what is the smallest possible value of $n$?
Reasoned solution
Key idea: in a divisibility chain each step multiplies by at least $2$; the single incomparable pair allows one short fork.
If the six numbers formed a full chain ($a_1 \mid a_2 \mid \cdots \mid a_6$), the maximum would be at least $2^5 = 32$. The incomparable pair lets us fork one step: take
where the only incomparable pair is $(2,3)$ and everything else chains up ($1 \mid 2 \mid 6$, $1 \mid 3 \mid 6 \mid 12 \mid 24$). The maximum is $24$, and an exhaustive search confirms no set with maximum $\le 23$ works.