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?


The greatest value of $\color{red}{x \text{ is when}}$ 4 is divided by 6, which produces a remainder of 4.

Note that $4$ divided by $6$ produces the remainder $4$. In general, division by $6$ can produce remainders $0,1,2,3,4,5$. A multiple of $4$ divided by $6$ can not produce the remainder $1,3,5$, but only $0,2,4$ (why?), hence $4$ is the greatest possible remainder.

The greatest value of $y$ is when 2 is divided by $\color{red}3$, which produces a remainder of 2.

Note that $2$ divided by $3$ produces the remainder $2$. In general, division by $3$ can produce remainders $0,1,2$. A multiple of $2$ divided by $3$ can produce $0,1,2$, hence $2$ is the greatest possible remainder.
