> [!NOTE] Lemma > No set of nine consecutive integers can be partitioned into two sets with the product of the elements of the first equal to the product of the elements of the second set. ###### Proof 1 BWOC, suppose there exists a set $S$ of nine consecutive integers that can be partitioned into two sets with the product of the elements of the first equal to the product of the elements of the second set. Consider the prime factors of the product $p$ of the elements of $S.$ None of the elements of $S$ is divisible by primes at least $11$ since the number of elements of $S$ are divisible by such primes is at most one and it can't be one. $p$ is a square so its congruent to either $1,4,9,3,12,10$ modulo $13.$ But also $p$ must be congruent to the product of a sublist of length $19$ of $1,2,3,\dots,12,$ that is, $p$ is either congruent to $11,6, 7$ or $2$ modulo $13$ - a contradiction. ###### Proof 2: