Partition of numbers : $1, 2, ..., 20$ Integers $1, 2, ..., 20$ are partitioned into $2$ groups. Sum of all integers in one group is equal to $n$ and the product of all integers in another group is also equal to $n$. Find the maximal $n$. 
Since $1 + 2 + ... + 20 = 210$, so the product of all integers in another group is less than $210$. 
Please suggest how to proceed.
 A: Here's a answer by trial and error.
Notice that the product of the smallest $5$ integers is $5!=5\cdot 4\cdot 3\cdot 2\cdot 1=120$ and the product of $6$ integers is bigger then $210$.
Now the sum will be bigger if we choose smaller numbers,lets try to make the sum the max possible value $<210$ by changing the digit $5$,testing $8$ is the biggest digit so the sum doesn't exceed $210$.We have that $1\cdot 2\cdot 3\cdot 4\cdot 8=192$ and the sum is $210-1-2-3-4-8=192$ so we have that $192$ can be $n$.
Now lets try to prove that $n\leq 192$ first $193$ is a prime $194=97\cdot 2$ where $97$ is prime so those are not solutions (because of the product).Now lets try numbers $\geq 195$ then sum of those numbers is $\leq 15$ checking products of $2$ numbers such that their sum is $\leq 15$ the greatest such number is $8\cdot 7=56$, now checking product of $3$ numbers the biggest is $5\cdot 4\cdot 6=120$ since the closer the $3$ numbers the bigger the product is,checking $4$ numbers we get that the biggest product is $2\cdot 3\cdot 4\cdot 5=120$ and at last checking five numbers it's $1\cdot 2\cdot 3\cdot 4\cdot 5=120$ this isn't the answer either so we concluded that $n=192$.
