The following is a proof in my textbook that no set of nine consecutive integers can be partitioned into two sets with the product of the elements of the first set equal to the product of the elements of the second set.
proof. Assume that such numbers do exist and let us look at their prime factorizations. For primes $p$ greater than $7$, at most one of the numbers can be divisible by $p$, and the partition cannot exist. The prime factors of the given numbers can be only $2,3,5,$ and $7$.
We now look at repeated prime factors. Because the difference between two numbers divisible by $4$ is at least $4$, at most three of the nine numbers are divisible by $4$. Also, at most one is divisible by $9$, at most one by $25$, and at most one by $49$. Eliminating these at most $3+1+1+1 = 6$ numbers, we are left with at least three numbers among the nine that do not contain repeated prime factors. They are among the divisors of $2\cdot3\cdot5\cdot7$, and so among the numbers $$2,3,5,6,7,10,14,15,21,30,35,42,70,105,210$$
Because the difference between the largest and the smallest of these three numbers is at most $9$, none of them can be greater than $21$. We have to look at the sequence $1,2,3, \cdots,29$. Any subsequence of consecutive integers of length $9$ that has a term greater than $10$ contains a prime number greater than or equal to $11$, which is impossible. And from $1,2, \cdots,10$ we cannot select nine consecutive numbers with the required property. This contradicts our assumption, and the problem is solved.
In the argument, why do we say that the difference between the largest and the smallest of these three numbers is at most $9$? I thought that the difference would be at most $8$.
