<s>[INST] 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?
 [/INST] So let $N = 4M$ be a multiple of $4$.  If you divide by $6$ and take the remainder $x$ so that $N=4M = 6k + x$ what are the possible values of $x$ if $x$ is not negative and $x < 6$.
Well, It's possible that $6$ divides into $N=4M$.  If so $x = 0$.
And it's possible that $6$ doesn't divide into $N=4M$.  What are the possible values of $x$.
Well, $4M = 6k+x$ so $2M =3k + \frac x2$.  $2M$ and $k$ are  is whole numbers, so $\frac x2$ is a whole number.  And $x < 6$ so $x$ can be: $0, 2,4$.
Can $x =4$.  Well of course.  $4 = 0*4 + 4$ has a remainder of $4$.  So does $28 = 6*4 + 4$. 
So $x \le 4$.
....
Now let $N = 2K$ but a multiple of $2$.  If you divide by $3$ and take the remainder so that $N =2k +y$ what are the possible values of $y$ if $y$ is not negative and $y < 3$.
Well, $y$ can only be $0, 1,$ or $2$.  Which are possible.
All of them are  $2=0*3 + 2$ so $y$ can be $2$. And $4=1*3 +1$ so $y$ can be $1$.  And $6= 3*2 + 0$ so $y$ can be $0$.
So $y \le 2$.  So $x + y \le 4 + 2 =6$.  
That's it.
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