Divide the set of integers from 1 to 34 into two subsets A & B where the sum of the numbers in set A is equal to the product of the numbers in set B.

Enter this sum = product.

The answer is 544.

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If n is even then B can consist of the numbers $1$ , $(\frac{n}{2})-1$ , $n$ . The product of these is $\frac{n^{2}}{2}-n$

The sum of the numbers in A is then $\frac{n^{2}+n}{2}-1-(\frac{n}{2}-1)-n=\frac{n^{2}}{2}-n$

So B can consist of $1$ , $16$ , $34$ with a product of $1*16*34=\boxed{544}$

Note: for most n, there is more than one solution, but for n=34 the solution is unique. As far as I can tell it is the largest number with a unique solution.