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Re: M05-29 [#permalink]
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Bunuel wrote:
Official Solution:


(1) \(p\) is a prime number. If \(p=2\) then the answer is NO but if \(p=11\) then the answer is YES. Not sufficient.

(2) \(2p\) is divisible by 11. Given: \(\frac{2p}{11}=\text{integer}\). Multiply by 2: \(2*\frac{2p}{11}=\frac{4p}{11}=2*\text{integer}=\text{integer}\), but we don't know whether this integer is positive or not: consider \(p=0\) and \(p=11\). Not sufficient.

(1)+(2) Since \(p\) is a prime number and \(2p\) is divisible by 11, then \(p\) must be equal to 11 (no other prime but 11 will yield integer result for \(\frac{2p}{11}\) ), therefore \(\frac{4p}{11}=4\). Sufficient.


Answer: C


Hello,

But we still dont know if it is positive, shouldnt the answer be E?
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Re: M05-29 [#permalink]
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aimtoteach wrote:
Bunuel wrote:
Official Solution:


(1) \(p\) is a prime number. If \(p=2\) then the answer is NO but if \(p=11\) then the answer is YES. Not sufficient.

(2) \(2p\) is divisible by 11. Given: \(\frac{2p}{11}=\text{integer}\). Multiply by 2: \(2*\frac{2p}{11}=\frac{4p}{11}=2*\text{integer}=\text{integer}\), but we don't know whether this integer is positive or not: consider \(p=0\) and \(p=11\). Not sufficient.

(1)+(2) Since \(p\) is a prime number and \(2p\) is divisible by 11, then \(p\) must be equal to 11 (no other prime but 11 will yield integer result for \(\frac{2p}{11}\) ), therefore \(\frac{4p}{11}=4\). Sufficient.


Answer: C


Hello,

But we still dont know if it is positive, shouldnt the answer be E?


Only positive numbers can be primes.
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Re: M05-29 [#permalink]
Bunuel wrote:
Official Solution:


(1) \(p\) is a prime number. If \(p=2\) then the answer is NO but if \(p=11\) then the answer is YES. Not sufficient.

(2) \(2p\) is divisible by 11. Given: \(\frac{2p}{11}=\text{integer}\). Multiply by 2: \(2*\frac{2p}{11}=\frac{4p}{11}=2*\text{integer}=\text{integer}\), but we don't know whether this integer is positive or not: consider \(p=0\) and \(p=11\). Not sufficient.

(1)+(2) Since \(p\) is a prime number and \(2p\) is divisible by 11, then \(p\) must be equal to 11 (no other prime but 11 will yield integer result for \(\frac{2p}{11}\) ), therefore \(\frac{4p}{11}=4\). Sufficient.


Answer: C

Bunuel can you please tell us all the properties of zero which will be useful in GMAT exam
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Re: M05-29 [#permalink]
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I got the answer correct and agree with you Bunuel. I just want to ask that in case of (2) 2p is divisible by 11, can p be a fraction like 11/7, this will still be divisible by 11.
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Re: M05-29 [#permalink]
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knitgroove04 wrote:
I got the answer correct and agree with you Bunuel. I just want to ask that in case of (2) 2p is divisible by 11, can p be a fraction like 11/7, this will still be divisible by 11.


2p is divisible by 11 implies that 2p must be an integer. p there can be a fraction, say 11/2, but not 11/7 because in this case 2p is not an integer.
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Re: M05-29 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re: M05-29 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re: M05-29 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re: M05-29 [#permalink]
I think this is a high-quality question and I agree with explanation. Excellent question! I missed the positive word in the question stem.
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Re: M05-29 [#permalink]
Bunuel wrote:
Official Solution:


(1) \(p\) is a prime number. If \(p=2\) then the answer is NO but if \(p=11\) then the answer is YES. Not sufficient.

(2) \(2p\) is divisible by 11. Given: \(\frac{2p}{11}=\text{integer}\). Multiply by 2: \(2*\frac{2p}{11}=\frac{4p}{11}=2*\text{integer}=\text{integer}\), but we don't know whether this integer is positive or not: consider \(p=0\) and \(p=11\). Not sufficient.

(1)+(2) Since \(p\) is a prime number and \(2p\) is divisible by 11, then \(p\) must be equal to 11 (no other prime but 11 will yield integer result for \(\frac{2p}{11}\) ), therefore \(\frac{4p}{11}=4\). Sufficient.


Answer: C



Bunuel, In second statement when P=0 then 2P will be 0, how can we consider then 0 is divisible by 11.
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Re: M05-29 [#permalink]
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mzaid wrote:
Bunuel wrote:
Official Solution:


(1) \(p\) is a prime number. If \(p=2\) then the answer is NO but if \(p=11\) then the answer is YES. Not sufficient.

(2) \(2p\) is divisible by 11. Given: \(\frac{2p}{11}=\text{integer}\). Multiply by 2: \(2*\frac{2p}{11}=\frac{4p}{11}=2*\text{integer}=\text{integer}\), but we don't know whether this integer is positive or not: consider \(p=0\) and \(p=11\). Not sufficient.

(1)+(2) Since \(p\) is a prime number and \(2p\) is divisible by 11, then \(p\) must be equal to 11 (no other prime but 11 will yield integer result for \(\frac{2p}{11}\) ), therefore \(\frac{4p}{11}=4\). Sufficient.


Answer: C



Bunuel, In second statement when P=0 then 2P will be 0, how can we consider then 0 is divisible by 11.


0 is divisible by all integers (except 0 itself). Divisible means divisible without a remainder, so x is divisible by y, means that x/y = integer. Since 0/11 = 0 = integer, then 0 is divisible by 11.

Hope it's clear.
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Re: M05-29 [#permalink]
I think this is a high-quality question and I agree with explanation. This is a great question that makes sure you pay attention to the details.
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Re: M05-29 [#permalink]
Bunuel wrote:
Is \(\frac{4p}{11}\) a positive integer?


(1) \(p\) is a prime number

(2) \(2p\) is divisible by 11


Hi Bunuel,

During my mock exam I saw very clearly that there was 0, but the 2nd sentence says 2p is divisible by 11. And during the exam I was thinking: does GMAT mean 0 is divisible by 11 in these types of questions? I was stuck on a technicality. Then I thought that "divisible by 11" would mean the same as 2p is a multiple of 11. And I have seen plenty of problems where multiples of 11 are numbers more than or equal to 11 (so not 0). According to the logic given in your explanation, 0 is divisible by any number (I am also aware that GMAT does not test 0/0 cases). So, If I want to avoid any any doubts on the exam, should I think that whenever the problem says divisible by some number 0 is always an alternative?

And also, I would be glad to hear about the multiples of numbers. Am I correct at least about this matter (or is 0 multiple of any number)?
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Re: M05-29 [#permalink]
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Foreheadson wrote:
Bunuel wrote:
Is \(\frac{4p}{11}\) a positive integer?


(1) \(p\) is a prime number

(2) \(2p\) is divisible by 11


Hi Bunuel,

During my mock exam I saw very clearly that there was 0, but the 2nd sentence says 2p is divisible by 11. And during the exam I was thinking: does GMAT mean 0 is divisible by 11 in these types of questions? I was stuck on a technicality. Then I thought that "divisible by 11" would mean the same as 2p is a multiple of 11. And I have seen plenty of problems where multiples of 11 are numbers more than or equal to 11 (so not 0). According to the logic given in your explanation, 0 is divisible by any number (I am also aware that GMAT does not test 0/0 cases). So, If I want to avoid any any doubts on the exam, should I think that whenever the problem says divisible by some number 0 is always an alternative?

And also, I would be glad to hear about the multiples of numbers. Am I correct at least about this matter (or is 0 multiple of any number)?


0 is divisible by all integers (except 0 itself). Divisible means divisible without a remainder, so integer x is divisible by integer y, means that x/y = integer. Since 0/11 = 0 = integer, then 0 is divisible by 11 (0 is a multiple of 11).

Next, multiples of 11 are ..., -22, -11, 0, 11, 22, 33, ... Good news is that on the GMAT all divisibility/remainder questions are limited to positive integers only. So, a question will tell you in advance that all unknowns are positive integers. Though you still should know properties of 0 and that negative integers also can be multiples of a number just in case.

Hope it's clear.
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Re: M05-29 [#permalink]
As per statement 2, can p be equal to a negative integer also?
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Re: M05-29 [#permalink]
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8427791377 wrote:
As per statement 2, can p be equal to a negative integer also?


Yes, for example, if p = -11.
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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