Prove that a set of nine consecutive integers cannot be partitioned into two subsets with same product
Problem: Prove that a set of nine consecutive integers cannot be partitioned into two subsets with same product
My attempt:
This problem would be solved if it could be proven that 9 consecutive numbers cannot exist such that all the numbers are of the form $2^a 3^b 5^c 7^d$ where $a,b,c,d$ are integers.
A suitable hint would be appreciated, I don not want a complete solution.
Edit:
If there existed a prime factor greater than $7$ of any number present in the set, then its multiple cannot exist b/c $22-11>9$
Edit 2: As mentioned by Erick, my reasoning is not valid for numbers $1-9$ and $2-9$, let this be an exception.
Hint
Show that none of the 9 consecutive numbers can be a multiple of $13$.
Denote the product of the 9 consecutive numbers by $a$. Show that $a$ must be square.
Look at the remainder of $a$ modulo $13$. Use 1. to see that there are only $4$ possibilities. Use 2. to find a contradiction in all 4 cases.
(This is inspired by this answer. Note that taking the residues modulo $11$ won't work but, quite surprisingly, $13$ does.)
Another idea is to write your numbers $a,a+1,...,a+8$. Let $A$ and $B$ your equal products. Then one of them (say $A$) has at most $4$ factors, and the second (B) has $\geq 5$ factors. Then $A\leq (a+8)^4$ and $B\geq a^5$, hence $(a+8)^4\geq a^5$, this show $a\leq 15$. Now you can finish using the point 2) of the answer of @azimut, using for $a=1$ the fact that the exponent of the prime $5$ is $1$ in the product $a(a+1)...(a+8)$, for $a=2$, the prime $7$, etc.