If positive integer $x$ is a multiple of $6$ and positive integer $y$ is a multiple of $14$, is $xy$ a multiple of $105$? Yes or no?
My answer key says that "the product $xy$ must carry at least two $2$s, one $3$, and one $7$." referring to the prime factorization of $6$ and $14$.
That last part is false right? $xy$ only needs one $2$, one $3$, and one $7$ right? For example: the prime factorization of $42$ is $7 \cdot 3 \cdot 2$ and $42$ is divisible by both $14$ and $6$. You don't need two $2$s right?
2) What if $x$ is now also a multiple of $25$? Now is $xy$ a multiple of $105$?
I think now $xy$ is now divisible by $105$ since $105$'s prime factorization is : $5 \cdot 7 \cdot 3$. We need a $5$ in the factorization of a number that is a multiple of $105$ somewhere and so having $x$ be a multiple of $25$ solves that problem.
Am I somewhere in the realm of correct thinking?
Your presentation of the question is a bit confusing, but I gather that you’re asked to determine whether either of the options $(1)$ and $(2)$ would provide enough additional information to let you conclude that $xy$ is a multiple of $105$. If that interpretation is correct, then your conclusion that $(2)$ is sufficient is correct.
Looking at the prime factorizations of $6,14$, and $105$ is the right thing to do. Since $x=6m$ for some integer $m$, and $y=14n$ for some integer $n$, we know that $$xy=6\cdot14mn=84mn=2^2\cdot3\cdot7mn\;.$$ We also know that $105=3\cdot5\cdot7$, so $xy$ is missing only a factor of $5$ in order to be a multiple of $105$. The additional information that $x$ is a multiple of $25=5^2$ would indeed ensure that $xy$ has that needed factor of $5$, while the information that $x$ is a multiple of $9=3^2$ would not.
Note that $xy$ really does have two factors of $2$. It’s true that $42$, which has only one, is a multiple of both $6$ and $14$, but that’s not relevant here: $x$ is a multiple of $6$ and $y$ is a multiple of $14$, so $xy$ is a multiple of $6\cdot14$, which has two factors of $2$.