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When the positive integer x is divided by 11, the quotient

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

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

Answer: E.
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Re: Remainder Problem [#permalink] New post 13 Apr 2010, 05:40
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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



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.

OA plz..
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Re: Remainder Problem [#permalink] New post 13 Apr 2010, 06:03
x=11*k + 3, where 11*k = y
x=19*m + 3

=> 11*k +3 = 19*m+3
=> y+3 = 19*m + 3
=> y = 19*m
So y is divisible by 19 with 0 remainder.

Answer is E.
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Re: Remainder Problem [#permalink] New post 10 Sep 2010, 02:07
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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.
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Re: Remainder Problem [#permalink] New post 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
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Re: Remainder Problem [#permalink] New post 10 Sep 2010, 14:28
great explanation Bunuel... thanks!
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Help w/ Remainder Question (MGMAT Online Question Bank) [#permalink] New post 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 :evil:

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??

Thanks in advance!
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Re: Help w/ Remainder Question (MGMAT Online Question Bank) [#permalink] New post 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.
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Re: Help w/ Remainder Question (MGMAT Online Question Bank) [#permalink] New post 29 Nov 2010, 22:17
From given conditions:

11y + 3 = 19q +3

11y = 19q

This implies y = q

y = q = 0 is the only possibility.

Therefore when Y = 0 is divided by 19, the remainder is 0
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Re: Help w/ Remainder Question (MGMAT Online Question Bank) [#permalink] New post 29 Nov 2010, 22:37
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continued from 11y=19z

z = 11y/19

we know that z is a quotient and hence a whole number

For z to be a whole number 11y has to be divisible by 19. now since 11 is not divisible by 19, Y has to be..

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Dividing By 11 and 19 [#permalink] New post 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.

The correct answer is A.
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Re: Dividing By 11 and 19 [#permalink] New post 27 Dec 2010, 15:47
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Re: Remainder Problem [#permalink] New post 27 Dec 2010, 18:42
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We know by now..
11y + 3 = 19z + 3
=> 11y = 19z

11 and 19 are prime numbers, so to hold the above equation correct, y has to be multiple of 19 and z has to be multiple of 11

11 * 19 = 19 * 11
11 * 38 = 19 * 22
11 * 57 = 19 * 33
and so on..

Basically, y can be 19, 38, 57... which is divisible by 19.
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Re: Remainder Problem [#permalink] New post 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.
:-D
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Re: Remainder Problem [#permalink] New post 17 Dec 2013, 00:55
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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.
:-D


Let me ask you a question: how can y be an integer if \frac{q}{11} is not?
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Re: Remainder Problem [#permalink] New post 17 Dec 2013, 03:18
Let me ask you a question: how can y be an integer if \frac{q}{11} is not?[/quote]

Ok BUT Since x= 11(y) + 3 and x= 19 (q) + 3 ; it means that 11 is greater than q. therefor the result q/ 11 will be fraction not an integer.

Let me what you think of
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Re: Remainder Problem [#permalink] New post 17 Dec 2013, 06:13
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rango wrote:
Let me ask you a question: how can y be an integer if \frac{q}{11} is not?


Ok BUT Since x= 11(y) + 3 and x= 19 (q) + 3 ; it means that 11 is greater than q. therefor the result q/ 11 will be fraction not an integer.

Let me what you think of[/quote]

That's not correct.
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Re: Remainder Problem   [#permalink] 17 Dec 2013, 06:13
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