The set of $\{1,2,3,...,63\}$ was divided into three non-empty disjoint sets $A,B$. Let $a,b,c$ be the product of all numbers in each set $A,B,C$ respectively and finally we have determined the greatest common divisor of these three products. What was the biggest result we could get?
Let $Q=\prod ^{63}_{k=1} k$ and $m=\gcd( a,b,c)$.
Note that for any prime $p$, we have $\nu _{p}( m) \leq \nu _{p}( Q) /3$.
For any prime $p >19$, $\nu _{p}( Q) < 3$, which implise $\nu _{p}( m) =0$.
For any prime $p\in [ 11,19]$, $3\leq \nu _{p}( Q) < 6$, which implise $\nu _{p}( m) \leq 1$
We have $\nu _{7}( Q) =10$, which implise $\nu _{p}( m) \leq 3$.
We have $\nu _{5}( Q) =14$, which implise $\nu _{p}( m) \leq 4$.
We have $\nu _{3}( Q) =30$, which implise $\nu _{p}( m) \leq 10$.
We have $\nu _{2}( Q) =57$, which implise $\nu _{p}( m) \leq 19$.
Therefore, $m\leq 2^{19} \cdot 3^{10} \cdot 5^{4} \cdot 3^{3} \cdot 11\cdot 13\cdot 17\cdot 19$
It is not difficult to construct $A,B,C$ such that $m=2^{19} \cdot 3^{10} \cdot 5^{4} \cdot 3^{3} \cdot 11\cdot 13\cdot 17\cdot 19$.
For example,
$
A=\{5,7,8,10,11,13,14,15,16,17,19,20,21,23,24,27,43,48,53,54,61\}\\
B=\{1,3,6,9,12,18,22,25,26,28,30,32,34,35,37,38,41,42,47,59,60\}\\
C=\{2,4,29,31,33,36,39,40,44,45,46,49,50,51,52,55,56,57,58,62,63\}\
$
Hence the answer is $\boxed{2^{19} \cdot 3^{10} \cdot 5^{4} \cdot 3^{3} \cdot 11\cdot 13\cdot 17\cdot 19}$