Question from Putnam and Beyond # 2 I just started working with Putnam and Beyond.


*Show that no set of nine consecutive integers can be partitioned into two sets with the product of the elements of the first set equal to the product of the elements of the second set.


I've only read the first paragraph of the solution from the book and it is drastically different from my approach. I have not read the entire solution from the book, hoping to still work on this problem. 
I thought I had a solution, but seeing that the book tried to approach it in a very different way, I am suspicious of it. I only ask for a hint why my proof is not a correct. 
My solution:
Suppose such integers exists. Let $\mathcal{O}$ be the set of such nine integers. Let the partitions be $A$ and $B$. Let $P(A)$ denote the product of integers of set $A$.
Thus, assume $P(A)=P(B)$. Now divide $P(A),$ and $P(B)$ with the highest power of $3$ that divides, $P(A)$. Denote the result with $Q(A)=Q(B)$. Observe that each integer of $\mathcal{O}$ belongs to exactly one of the equivalence classes in $\mod 9$. Thus, $Q(A)$ and $Q(B)$ are the products of integers in $\mathcal{O}$ in the following equivalence classes $\mod 9$:
$$\bar{1},\bar{2},\bar{4},\bar{5},\bar{7},\bar{8}$$
For integer $x$, $x\mod9=d\mod9$, where $d$ is the sum of digits of $x$. The sum of the digits of $Q(A)Q(B)$ is $29\mod9=2\mod9$. Therefore, $Q(A)=1\mod9=Q(B)$. But the product of the equivalence classes listed above is $8\mod9$, thus, if $Q(A)=1\mod9$ then $Q(B)=8\mod9$. Contradiction.
 A: I don't see how you are getting that the sum of the digits of $Q(A)Q(B)$ to be 29. You won't be able to prove that two partitions must be different mod 9; for example, consider the partition $A=\{4\}$ and $B=\{1,2,3,5,6,7,8,9\}$. The product of all elements in B stripped of all powers of 3 is 1120, which is also $4\mod 9$. This means we have to work with a different type of argument.
When approaching a problem like this, it might be easier to consider a bounding argument. Since we're working with 9 consecutive integers, call them $n,n+1,\dots,n+8$. 
We know that one of the groups will have at least 5 elements. We can show that for $n\ge11$, $n^5>(n+8)^4$, so the group with 5 elements will clearly have a greater product.
For $n<11$, we can always check manually that there is a prime between $n$ and $n+8$ that appears just once, which means that it is either in group A or group B. This means that exactly one of the products of the two groups will be divisible by that prime, meaning that the products cannot be equal.
A: Your method involves "dividing out powers of $3$". However, consider a number such as $6$ ; dividing by $3$ still leaves the factor $2$ which will make a contribution to $Q(A)$ or $Q(B)$. Therefore the assertion that $Q(A)Q(B)$ is $2$ mod $9$ is false.
Example
Let $A=\{ 4,8,10,11\},B=\{ 5,6,7,9,12\}$. Then 
$$Q(A)Q(B)=4\times8\times10\times11\times5\times2\times7\times1\times4\equiv 1 \mod 9.$$
