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When the positive integer x is divided by 11, the quotient [#permalink]
13 Apr 2010, 04:49

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E

Difficulty:

35% (medium)

Question Stats:

74% (02:44) correct
26% (02:18) wrong based on 218 sessions

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?

Re: Remainder Problem [#permalink]
13 Apr 2010, 05:07

7

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Expert's post

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Hussain15 wrote:

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?

A.4 B.3 C.2 D.1 E.0

(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.

Subtract (2) from (1) --> 19q=11y --> y=\frac{19q}{11}. Now as y and q are integers and 19 is prime \frac{q}{11} must be an integer --> y=19*integer --> y is a multiple of 19, hence when divide by 19 remainder is 0.

Re: Remainder Problem [#permalink]
13 Apr 2010, 05:40

2

This post received KUDOS

Hussain15 wrote:

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?

A.4 B.3 C.2 D.1 E.0

IMHO E

We have two cases.. x = 11*y + 3 and x = 19*m + 3... where just like y..its an integer..

equating both the equations.. 11*y + 3 = 19*m + 3 y = 19*m / 11. now both 11 and 19 are prime...so m has to be a multiple of 11.. Only then we can get y as integer.. so let say m= 11*p, where p is a positive integer..

so y = (19 * 11 * p)/11 and when y is divide by 19..we get remainder as zero.

Re: Remainder Problem [#permalink]
10 Sep 2010, 02:07

3

This post received KUDOS

Any Number which when divided by divisor d1,d2, etc. leaving same remainder "r" takes the form of "K+r" where k = LCM (d1,d2)

In this case the divisors are 11 & 19 and remainder is 3. so LCM (11,19) = 209 So N= 209+3 = 212 Also X=d1q+3 ; which means d1q=209 & d1=11 therefore q=19

And ( y divided by 19)19/19 leaves remainder 0.

Answer is E

This approach took me less than 50 secs. hope it helps. _________________

Consider giving Kudos if my post helps in some way

Re: Remainder Problem [#permalink]
10 Sep 2010, 13:39

Hey guys,

Looks like this question is under control - I liked seeing that subject line since I just threw up a blog post specifically on remainders today. If you're interested, you can see it at: http://blog.veritasprep.com/2010/09/gmat-tip-of-week-remainder.html _________________

Help w/ Remainder Question (MGMAT Online Question Bank) [#permalink]
29 Nov 2010, 21:15

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?

A) 0 B) 1 C) 2 D) 3 E) 4

So I got this far:

If x divided by 11 has a quotient of y and a remainder of 3, x can be expressed as x = 11y + 3, where y is an integer (by definition, a quotient is an integer). If x divided by 19 also has a remainder of 3, we can also express x as x = 19z + 3, where z is an integer.

We can set the two equations equal to each other: 11y + 3 = 19z + 3 11y = 19z

This is where I get lost

How does 11y = 19z help me determine what the remainder is when y is divided by 19???? Is there another step somewhere?

Here's the rest of the explanation:

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.

Answer is A) 0

Please tell me there's an easier way to do this then to "assume" y is a multiple of 19 so then there's no remainder. Thoughts??

Re: Help w/ Remainder Question (MGMAT Online Question Bank) [#permalink]
29 Nov 2010, 21:34

both 11 and 19 are prime numbers therefore z and y have to be multiples of these prime numbers in order for 11y=19Z to be true. Since there prime there is no other way possible. And if y is a multiple of Y, then Y/19 will result in a 0 remainder.

Dividing By 11 and 19 [#permalink]
27 Dec 2010, 15:42

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?

A) 0 B) 1 C) 2 D) 3 E) 4

Could you please explain the answer?? I am confused by the following explanation:

If x divided by 11 has a quotient of y and a remainder of 3, x can be expressed as x = 11y + 3, where y is an integer (by definition, a quotient is an integer). If x divided by 19 also has a remainder of 3, we can also express x as x = 19z + 3, where z is an integer.

We can set the two equations equal to each other: 11y + 3 = 19z + 3 11y = 19z

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.

Re: When the positive integer x is divided by 11, the quotient [#permalink]
16 Oct 2013, 13:23

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Re: Remainder Problem [#permalink]
16 Dec 2013, 22:50

Bunuel wrote:

Hussain15 wrote:

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?

A.4 B.3 C.2 D.1 E.0

(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.

Subtract (2) from (1) --> 19q=11y --> y=\frac{19q}{11}. Now as y and q are integers and 19 is prime \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: E.

Thanks or the explanation. Hw you figured out q/11 is an integer. I know that q is an integer and 11 also; but cannot guarantee that q/ 11 - will be an integer. Please clarify.

Re: Remainder Problem [#permalink]
17 Dec 2013, 00:55

Expert's post

rango wrote:

Bunuel wrote:

Hussain15 wrote:

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?

A.4 B.3 C.2 D.1 E.0

(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.

Subtract (2) from (1) --> 19q=11y --> y=\frac{19q}{11}. Now as y and q are integers and 19 is prime \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: E.

Thanks or the explanation. Hw you figured out q/11 is an integer. I know that q is an integer and 11 also; but cannot guarantee that q/ 11 - will be an integer. Please clarify.

Let me ask you a question: how can y be an integer if \frac{q}{11} is not? _________________

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