Show that no set of nine consecutive integers can be partitioned into two sets where the products of all elements of two sets are equal.

Question

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 $2 \times 3 \times 5 \times 7$, 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 $1, 2, 3,\dots, 29$. 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 $1, 2,\dots , 10$ 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.

For example, ${2^8,2^7,2^3}$ and ${1,2,2^2,2^4,2^5,2^6}$ partitions a set of nine integers into two sets whose products are equal. — Thomas Andrews, Nov 24 '23 at 02:49
@ThomsAndrews Sorry, a typo. It should be 9 consecutive integers. — , Nov 24 '23 at 02:50
Likely to be a duplicate. Yup, discussed here. Not voting to close as a duplicate right away, because my vote would take immediate effect, and the focus is arguably different. If somebody agrees to use that (or this) as a duplicate target, I will then act. — Jyrki Lahtonen, Nov 24 '23 at 03:16
CH Park, could you please tell us which old question you referred to? The one I linked to (now in the right margin under Linked)? Is the solution you posted from Putnam and beyond or where? — Jyrki Lahtonen, Nov 24 '23 at 03:31
Yes, Dr.Lahtonen the one you linked here. The solutions are confusing and there are only hints there. I think it is useful to discuss again here. — , Nov 24 '23 at 03:32
Anyway, it is ok to ask about a specific step (or steps) in an argument. Which step in particular you have trouble with? — Jyrki Lahtonen, Nov 24 '23 at 03:33
@JyrkiLahtonen Why do we look at the repeated prime factors in the first place? — , Nov 24 '23 at 03:35

orangeskid · Accepted Answer · 2023-11-24T03:15:44.990

3

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.

edited Nov 24 '23 at 03:15

answered Nov 24 '23 at 03:03

orangeskid

56,630

2

This is how I would have approached it, too. – Jyrki Lahtonen Nov 24 '23 at 03:13
But the question is a rerun :-( – Jyrki Lahtonen Nov 24 '23 at 03:16

score 0 · Answer 2 · answered Jul 07 '24 at 16:27

OP observes that the sequence must be composed of integers that have no prime factors greater than $7$. We also know that if the two subsets have an equal product, then the product of all members of the set must be a perfect square. If we work $\bmod 13$ we find that sets running from $(1,\dots 9),(2,\dots 10),(3,\dots 11),(4,\dots 12)$ have products $\equiv (11,6,7,2)$ respectively. But these values are not quadratic residues $\bmod 13$, so the product of any set of nine consecutive integers which contain no factors of $13$ is not a perfect square, and therefore there are no two subsets with an equal product of its members.

Show that no set of nine consecutive integers can be partitioned into two sets where the products of all elements of two sets are equal.

2 Answers2