The possible remainders when a multiple of 4 is divided by 6, and when a multiple of 2 is divided by 3 I'm currently studying for the GRE and a specific type of question has me stumped. I have a two part question, but first here is one problem and my work:
"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"
Since it says greatest I thought GCM and wrote(with arbitrary variables)
$$N = 24a + x$$
$$M = 6b + y$$
I thought that since 2 & 3 are factors of 4 & 6, respectively, maybe I could equate them:
$$24a+x=6b+y$$
Or maybe N is twice a large as M:
$$24a+x=2(6b+y)$$
1) How do I solve this? Am I even on the right track?
This kind of question has had me stumped for a while but the books explanation has me even more lost. 
"The greatest possible remainder for a multiple of 4 being divided by 6, happens when 4 is divided by 6. When 4 is divided by 6 the result is 0 with a remainder of 6."
.....Don't remainders by nature have to be smaller than the divisor......
2)How is that possible? Shouldn't the result be 0 with a remainder of 4? 
How can there be a remainder of 6 when the divisor is 6? I found a similar problem and solution here: Finding a number given its remainder when divided by other numbers and here: How to find the greatest remainder of a number that is a multiple of another number But the more I read the more lost I was. 
At this point I feel like a child throwing spaghetti about trying to make art. I have no idea how to even start on this problem any help is appreciated.
 A: When any integer $a$ is divided by $6$ then the possible remainders 
are $0,1,2,3,4,5$. So if $4a$ is divided by $6$, then the possible remainders will be $0,4,2,0,4,2$ respectively (if you are familiar with modular arithmetic then this can be written more elegantly). So the maximum possible remainder when a multiple of $4$ is divided by $6$ is $4$. 
Likewise when you divide any integer $b$ by $3$, the possible remainders are $0,1,2$. In which case the remainders when $2b$ is divided by $3$ will be $0,2,1$. Here the maximum is $2$. Thus the maximum of the sum will be $4+2=6$.
A: The pattern which is followed when divided by 6 a multiple of 4 is 0,4,2 starting from 0 and pattern followed by remainder when a multiple of 2 is divided is 0,2,1 when started from 0 so greatest value if $x+y=4+2=6$. Hope it helped you.
A: Multiples of 4 are: 4,8,12,16,20,24 etc.. and multiples of 2: 2,4,6,8,10,12,14 etc..
Divide the listed multiples of 4 by 6 to check the remainder by finding the remainder in each multiple. It is obvious that 6 goes into 16 with a remainder of 4 giving the value for X. Remember the question said the greatest possible value and in this case is remainder 4. Apply the same method to multiples of 2 divided by 3. 3 goes into 8, 2 times with a remainder of 2 giving the value for Y. Therefore X+Y=6
