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! :)