# 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.

• Think in $\pmod 6$. Then the multiples of 4 are either 4,2,0. Think in $\pmod 3$. Then the multiples of 2 are 2,1,0. Thus, the greatest posible value is $2+4=6$ – EA304GT Nov 15 '15 at 6:18
• If the book said what you say it said, the book is wrong, and you are right. The remainder when you divide 4 by 6 is 4. – Gerry Myerson Nov 15 '15 at 6:20
• Thank you for replying. @EA304GT Is the "mod" mentioned the same as the modular arithmetic Anurag A mentioned? – Broguen Whetstone Nov 16 '15 at 20:32
• Thank you so much! @Gerry Myerson I feel so much better that someone else says the remainder is 4! – Broguen Whetstone Nov 16 '15 at 20:34
• @BroguenWhetstone Yes, I meant modular arithmetic – EA304GT Nov 16 '15 at 23:18

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 $4+2=6$.
• When you divide an even number (like $4a$) by an even number (like 6), the remainder has to be an even number. Think about it, Broguen! – Gerry Myerson Nov 16 '15 at 22:17
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.