# Mock AIME II 2012 Problems/Problem 10

## Problem

Call a set of positive integers if it can be split into two nonempty disjoint subsets and with such that the product of the elements in and the product of the elements in sum up to the cardinality of . Find the number of sets such that the largest element is less than . (Disjoint subsets have no elements in common, and the cardinality of a set is the number of elements in the set.)

## Solution

Let the cardinality of be . We therefore want the sum of the products of the elements of and to be , and this is constant no matter what the elements of are. The set with the smallest such sum of products is the set , since if one of the elements is not in this set, then we can replace it with one of the elements in this set and get a smaller product. However, notice that whichever subset or contains the element , then the product of the elements of that set will be greater than or equal to , and the product of the elements of the other set will be positive, and so the sum of the products must be greater than . Therefore, no lucky set of positive integers is possible and the answer is .