When the positive integer x is divided by 11, the quotient is y and th

If you decide to go with quotient/remainder formula approach, then I'd suggest to express the info in the stem with it. And then look whether we can somehow manipulate with the expressions at hand to answer the question.

(1) When the positive integer x is divided by 11, the quotient is y and the remainder 3 --> \(x=11y+3\);
(2) When x is divided by 19, the remainder is also 3 --> \(x=19q+3\).

Easy to spot that \(19q+3=11y+3\) --> \(19q=11y\) --> \(y=\frac{19q}{11}\). Now as \(y\) and \(q\) are integers and 19 is prime then \(\frac{q}{11}\) must be an integer --> \(y=19*integer\) --> \(y\) is a multiple of 19, hence when divide by 19 remainder is 0.

Answer: A.

Hope its clear.
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Take LCM of 19 & 11 = 209
Adding 3 = 212
Say x = 212

212/11: Quotient = 19 (=y) & remainder = 3

212/19: Quotient = 11 & remainder = 3

19/19: Quotient = 1 & remainder = 0 = Answer = A

Find some numbers that fit the question.

We are told that x and y both give a remainder of 3 when divided by 19 or 11.

We can easily construct a number that fits this.

x = 19*11 + 3

x / 11 = 19 + 3 / 19

so y = 19

remainder when y / 19 is 0

A

Sol: Given x=11y+3 or y= (x-3)/11
Also x=19a+3 Substitute for x in the above equation we get

y= (19a+3-3)/11 or y=19a/11. Note that "a" is an integer and multiple of 11

y/19 =a/11 where a/11 is an integer and hence remainder is 0

Ans is A
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Please find the solution as attached,

19z + 3 = x = 11y + 3

Or, x = 19*11 + 3 { z = 11 and y = 19 }

or, x = 202

y/19 = Quotient 1 and remainder 0

Hence , answer will be (A) 0

Hope this helps !!

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Bunuel, what is the relationship between prime and divider in this case?
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We have \(y=\frac{19q}{11}\). 19 is not a multiple of 11, thus for 19q/11 to be an integer q must be a multiple of 11.
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Hi All,

We're told that...
X/11 = Y remainder 3
X/19 = something remainder 3

With the first piece of information, we know that X is 3 greater than a multiple of 11; with the second piece of information, we know that X is 3 greater than a multiple of 19. To have a remainder of 3 when you divide X by BOTH 11 and 19, X must be a number that is 3 greater than a MULTIPLE of BOTH 11 and 19.

We're asked what the remainder would be when Y is divided by 19. At this point, you might recognize that you could choose Y=19 and solve from there. If you don't recognize why that that relationship exists, then here's a more step-heavy way to get to the correct answer:

X = (11)(19) + 3 = 209 + 3 = 212

212/11 = Y remainder 3
212 = 11Y + 3
209 = 11Y
209/11 = Y
19 = Y

We're ultimately asked what the remainder would be when 19 is divided by 19. The remainder is 0.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

A fast approach is to find a value of x that meets the given conditions.

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....
Notice that x = 3 meets the above conditions.
3 divided by 11 = 0 with remainder 3. In this case, y = 0
Likewise, 3 divided by 11 also leaves a remainder of 3

What is the remainder when y is divided by 19?
In the above example, y = 0
So, when we divide 0 by 19, the remainder is 0

Answer: A

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We can create the following expressions:

x = 11y + 3

and

x = 19Q + 3

Thus:

11y + 3 = 19Q + 3

11y = 19Q

11y/19 = Q

Since Q is an integer and 11 is not divisible by 19, then y must be divisible by 19, and hence the remainder when y is divided by 19 is zero.

Answer: A
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Hi Bunuel,

I understand the solutions. I just had a doubt regarding y and q. Might be a stupid question. But how do we know that they are integers for sure?

Hi uc26,

The concept of "remainders" ONLY involves integers. For example, if you're dealing with 6/4, then you know that that is 1.5; with remainders though, 6/4 = 1 remainder 2.

When you're dealing with remainders on the GMAT, neither the numerator nor the denominator can involve any decimals (re: you cannot ask for a remainder of 4.7/1.3).

GMAT assassins aren't born, they're made,
Rich

As per the given question stem,

y should be a multiple of 19

And, x will be equal to 19*11*a + 3 (a being a random integer)

Hence, answer is 0.

OA: A
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Given: When the positive integer x is divided by 11, the quotient is y and the remainder 3
therefore: x=11y + 3
Keep in note, both x and y here are integers

Given: When x is divided by 19, the remainder is also 3 (lets call the quotient q)
x = 19q+3

Manipulating this expression
(x-3)/19 = q
We know x = 11y+3. Substituting for x in the above equation
11y+3-3 /19 = q
11y/19 = q
We know that y is an integer and 11 is not divisible by 19
So since it can be any integer, it could be 19
Hence 11x19/19 = 11 with remainder 0
Hence A