My question is related to the following
(#5 from the online Number Properties question bank)

"When the positive integer x is divided by 11, the quotient is y and the remainder 3. When x is divided by 19, the remainder is also 3. What is the remainder when y is divided by 19?


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I was able to establish two equation: x=11y +3, and x=19z+3. - From there I was not sure what to do and so I had guessed the answer (incorrectly).


The given solution equated both of the equations I had come up with to obtain 11y = 19z

The soltion then goes on to state the following:

"The question asks us what the remainder is when y is divided by 19. From the equation we see that 11y is a multiple of 19 because z is an integer. y itself must be a multiple of 19 since 11, the coefficient of y, is not a multiple of 19.

If y is a multiple of 19, the remainder must be zero."

I do not understand this - why does y itself have to be a multiple of 19?

Thanks

Hi Carla,

Well let me try....

From the stem Question we get that (X-remainder) i.e (x-3) is multiple of both 11 and 19 rt?

Now both 11 and 19 are prime numbers hence..the only way for a sumber to be multiple of both 11 and 19 is n*11*19 where n is >0 for example numbers will be

11*19 = 209
2*11*19 = 418
3*11*19 = 627

So now when you divide 209 by 11 what's the quotient is 19, 418 divide by 11 the quotient is 38=2*19 rt?

Back to the question so what all can be the values of x ..212, 421, 630 rt?

say you divide 212 by 11 the quotient y is 19 and remiander is 3 rt?

say you divide 421 by 11 the quotient y is 38 and remiander is 3 rt?

So in any above the case when y is divided by 19 the remainder is always going to zero.

Well it's a very conceptual question for which I don't think equation needs to be formed or something. I mean may be these things comes with practice ...or something. The only TRICKY part in the problem was to un-wrap the wordings of the Problem and keep track of what the question is throwing on you.

I hope it was help if not let me know may be I can try again.

Saurabh Malpani

Hi,


That was very helpful - thank you. I just wanted to send a few notes to make sure that I get it and I think I am almost there...


From the two equations we had

x = 11y + 3
x = 19z + 3

You re-arranged these as follows:

(x-3) = 11y
(x-3) = 19z

From this you were able to conclude that the number (x-3) is a multiple of both 11 and 19. Taking note that both 11 and 19 are prime numbers. (Something that had actually not jumped out at me...)

The only way for some number, in this case (x-3) to be a multiple of both 11 and 19 there would have to be some product such that
(This was actually an important step that I am not sure I would have come to on my own... )

(x-3) = n*11*19
where is some integer n>0



So if we use some examples:


n=1 --> (x-3) = 1*11*19 = 209 ==> x=212
n=2 --> (x-3) = 2*11*19 = 418 ==> x= 421 etc....

I actually just realized where I was not making the connection.. the key is that Y (which now I realize why you bolded) is what ties the whole thing together....

SO if we are working with the first statement in the problem that was x/11 gives quotient Y and remainder 3 and we use our example of x = 212
we get 212/11 = 19 with remainder 3. Where Y = 19 = quotient...(which we will later have to divide by 19 to check for the value of that new remainder....)

Also if we use x = 421 : 421/11 = 38 with remainder 3. So Y in this case = 38 = 2*19 = quotient...

So any Y is actually going to be a multiple of 19.. and from the last part of the question: Y/19 = some unknown quotient and a (possible) remainder.. and the question wants to know that remainder.. since all the quotients were divisible by 19.. then this new remainder would have to be zero...

Wow... I think I just figured that out as I was writing. Thanks so much.. I think I will post this anyways in case anyone else finds this helpful.

Thanks!!

Carla

I am glad it helped!!!

Saurabh Malpani

Nice job guys! I love it when you figure it out for yourselves - that's the best kind of learning! :)

_________________
Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT

there is one VERY fast solution to this problem, which no one figured out back when this was originally posted.

namely:

we need a remainder of 3 if x is divided by 11, and also a remainder of 3 if x is divided by 19.
since the answer choices are NUMBERS (i.e., they don't depend on x), it follows that we can get a solution to this problem if we can find ANY valid value for x.

the hard way:
start with 11 + 3 = 14, and then list all the subsequent numbers that give remainders of 3 upon division by 11: 25, 36, 47, etc.
start with 19 + 3 = 22, and then list all the subsequent numbers that give remainders of 3 upon division by 19: 41, 60, 79, etc.
if you continue these lists far enough, you'll find that x = 212 is in both lists. therefore, you can let x = 212.
since dividing 212 by 11 gives 19 (remainder 3), we have y = 19. therefore the answer to the problem is 0.

the easy way:
don't forget about x = 3.
if we divide 3 by 11, the remainder is 3. if we divide 3 by 19, the remainder is also 3.
therefore, x = 3 works!
if we divide 3 by 11, the quotient is 0 (11 doesn't go into 3 at all) and the remainder is 3. therefore, y = 0, so the answer to the problem is 0.

Hi Ron,

the easy way:
don't forget about x = 3.
if we divide 3 by 11, the remainder is 3. if we divide 3 by 19, the remainder is also 3.
therefore, x = 3 works!
if we divide 3 by 11, the quotient is 0 (11 doesn't go into 3 at all) and the remainder is 3. therefore, y = 0, so the answer to the problem is 0.

Can you please explain this easy way in more detail. I dont quite understand how you have remainder of 3 when you divide 3 by 11 and 3 by 19. Also how does x = 3, please explain..

11 "goes into" 3 zero times (since 11>3), after which 3 are "left over." In algebraic form, this is: 3 = 11*0 + 3 = 11y + 3, where y = 0.

19 "goes into" 3 zero times (since 19>3), after which 3 are "left over." In algebraic form, this is: 3 = 19*0 + 3 = 19z + 3, where z = 0.

I agree with Ron; number listing is better than algebra for remainder problems because
(1) There is ALWAYS a pattern in remainder problems.
(2) The list lets you SEE the pattern, whereas algebra hides the pattern.

So you asked "why is x = 3?" We didn't know that at first! We list numbers that x could be, then see which possibilities work for both constraints:

If x divided by 11 gives a remainder of 3, x could be 3, 14, 25, 36, 47, 58, 67,...
[See the pattern? x is 3 more than each multiple of 11. The terms on the list are spaced 11 apart.]

If x divided by 19 gives a remainder of 3, x could be 3, 22, 41, 60, ...
[again, the pattern is "start with 3 and add 19 to get each next term."]

Since x has to be a value that is on both lists (i.e. a number that meets both constraints) x = 3 works. There are other numbers that work (x = 212, 421, etc.), but since they all work, it doesn't matter which one we use. Use the easy one, x = 3.

_________________
Emily Sledge
Instructor
ManhattanGMAT