If $x$ is the remainder when a multiple of $4$ is divided by $6$, and $y$ is the remainder when a multiple of $2$ is divided by $3$, maximise $x+y$. The question is: if $x$ is the remainder when a multiple of $4$ is divided by $6$, and $y$ is the remainder when a multiple of $2$ is divided by $3$, what is the greatest possible value of $x+y$?
The book says "the greatest value of $4$ is divided by $6$, which produces a remainder of $4$.  The greatest value of $y$ is when $2$ is divided by, which produces a remainder of $2$.  Therefore, the greatest value of $x+y$ is $6$."
I think what's throwing me off is the phrase "multiple of $4$" bc it makes me think that any multiple of $4$ can be divisible by $6$ (i.e. $24/6 = 4$).  The books answer doesn't use multiples, just the $4$ and $2$, respectively.  I don't understand how this works.  Can someone please clarify?

In order to get to the book's conclusion, you can test out a few numbers:
$4$ leaves remainder $4$ when divided by $6$.
$8$ leaves remainder $2$ when divided by $6$.
$12$ leaves remainder $0$ when divided by $6$.
$16$ leaves remainder $4$ when divided by $6$.
Then you can observe the possible remainders are $0, 2$ and $4$. This is because $16$ is $12$ more than $4$, a multiple of $6$, so the remainders will follow the same pattern after $16$.
There is another way to think about this pattern. When you add $4$ to a number, the remainder will be the same as if you subtract $2$ from the number, which explains the 'decreasing by $2$ pattern'.
Now try this with small multiples of $2$ and find the remainder when they are divided by $3$. This gives the book's answer of $4 + 2 = 6$.
