I tried to solve this question, but I had no idea how to approach this question. I checked the solution in Putnam and Beyond, but the solution is also very hard to understand. Can you explain the solution step-by-step?
The solution:
Assume that such numbers 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. Thus the prime factors of the given numbers can be only 2, 3, 5, 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 x 3 x 5 x 7, 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,…, 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,…, 10 we cannot select nine consecutive numbers with the required property. This contradicts our assumption, and the problem is solved.
Note: I know there is exactly the same question asked 8 years ago, but it’s also very hard to understand, and I am interested in understanding this particular solution.
Best Answer
Let's try to solve it without the solution in the book.
The numbers are $n+1$, $n+2$, $\ldots$, $n+9$, with $n\ge 0$ a natural number. Assume that we can divide them in two groups with equal products. First group contains at least $5$ numbers, the other at most $4$. The product of the numbers in the first group is $$\ge \prod_{k=1}^5(n+k)$$ the product of numbers in the second group is $$\le \prod_{k=6}^9 (n+k)$$
Now, for $n\ge 5$ we have $$\prod_{k=1}^5(n+k) > \prod_{k=6}^9 (n+k)$$
So we necessarily have $0 \le n \le 5$. Now we eliminate all these cases for $n$, since there are some primes appearing here.