Partitioning integers into sets

Question

Try to find out if it is possible to partition 6 consecutive positive integers into two sets, $A_1$ and $A_2$ such that the product of the elements in $A_1$ is equal to the product of the elements in $A_2$?

Also, check if this is possible for 20 consecutive positive integers?

I said:

Let $K=$ positive integer, then we need to partition:

$$K, K+1, K+2, K+3, K+4, K+5 $$

This wasnt helping me so i worked with numbers.

I started out with:

$$ 1,2,3,4,5,6 $$

and i can't seem to partition it.

So i worked with:

$$ 2,3,4,5,6,7 $$

I tried:

$A_1=4,7,3=84$ $A_2=2,5,6=60$

and I tried more cases but trial and error does not seem to be effective. Any strategies or patterns anybody else notice?

Answer 1

You checked that $\{1,2,3,4,5,6\}$ does not work. So we look for solutions with smallest entry greater than $1$. It will turn out that there are none.

If there are such integers, the only prime factors of each are $2$, $3$, or $5$. For any prime $p$ greater than $5$ can divide at most one of our numbers, say $a$. And then if $a$ is put into say $A_1$, the product of the elements of $A_1$ is divisible by $p$ but the product of the elements of $A_2$ is not.

Exactly $3$ of our numbers are odd. Can their prime factors be only $3$ and $5$? If so, two of them have to be powers of $3$, which cannot happen with numbers all greater than $1$.