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?
 A: 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 $4 + 2 = 6$.
A: 
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.
A: Take any integer $m$ and divide it by $6$. The possible remainders are $0,1,2,3,4,5$. Therefore, if we take $4m$ and divide it by $6$ then the possible remainders are $0,4,2,0,4,2$. The maximum remainder when a multiple of $4$ is divided by $6$ is $4$. 
Similarly, if you divide an integer $n$ by $3$ the possible remainders are $0,1,2$. Therefore, if we take $2n$ and divide it by $3$ then the possible remainders are $0,2,1$. The maximum remainder when a multiple of $2$ is divided by $3$ is $2$. 
Hence the maximum of the sum of the two remainders is $4+2=6$.
A: 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.
