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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?

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  • $\begingroup$ I am not saying there is or isn't a solution but the sets don't have to be the same size. $\endgroup$ – John Douma Oct 27 '15 at 1:45
  • $\begingroup$ Small observation: At least one of those six numbers will be divisible by 5. So at least one of the two products will be divisible by 5. If they're equal, then both must be divisible by 5. Only if the divisible-by-five number is the first or last will another be divisible by 5. So you only need to look at sequences like 0,1,2,3,4,5 or 5,6,7,8,9,10. Analyzing divisibility might help you figure out a little more . $\endgroup$ – John Hughes Oct 27 '15 at 1:48
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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 $1$.

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