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When the positive integer x is divided by 11, the quotient [#permalink]
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13 Apr 2010, 05:49
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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
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Re: Remainder Problem [#permalink]
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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]
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13 Apr 2010, 06:40
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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]
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13 Apr 2010, 07:03
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]
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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]
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10 Sep 2010, 14:39
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/gmattipofweekremainder.html
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Re: Remainder Problem [#permalink]
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10 Sep 2010, 15:28
great explanation Bunuel... thanks!



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Help w/ Remainder Question (MGMAT Online Question Bank) [#permalink]
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29 Nov 2010, 22:15
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?? Thanks in advance!



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Re: Help w/ Remainder Question (MGMAT Online Question Bank) [#permalink]
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29 Nov 2010, 22:34
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]
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29 Nov 2010, 23:17
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]
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29 Nov 2010, 23:37
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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.. PS: If you like my post click on +1Kudos
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Dividing By 11 and 19 [#permalink]
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27 Dec 2010, 16:42
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]
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27 Dec 2010, 16:47



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Re: Remainder Problem [#permalink]
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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: When the positive integer x is divided by 11, the quotient [#permalink]
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Re: Remainder Problem [#permalink]
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16 Dec 2013, 23:50
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.
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Re: Remainder Problem [#permalink]
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17 Dec 2013, 01:55
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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Re: Remainder Problem [#permalink]
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17 Dec 2013, 04:18
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: When the positive integer x is divided by 11, the quotient [#permalink]
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Re: When the positive integer x is divided by 11, the quotient
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