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M14-30

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New post 16 Sep 2014, 00:54
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  35% (medium)

Question Stats:

71% (01:53) correct 29% (02:01) wrong based on 58 sessions

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New post 16 Sep 2014, 00:54
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New post 21 Nov 2016, 09:27
Isnt statement 1 sufficient?
If p=2 then 2p+1=5 prime
if p=3 then 2p+1=7 prime

Hence if p is prime then 2p+1 is prime??
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M14-30  [#permalink]

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New post 16 Dec 2016, 12:22
AmritaSarkar89 wrote:
Isnt statement 1 sufficient?
If p=2 then 2p+1=5 prime
if p=3 then 2p+1=7 prime

Hence if p is prime then 2p+1 is prime??



Since there are so many primes in the early numbers, try testing a larger prime.

if p=17 then 2p+1= 35 Not Prime

Also notice that for any value of p, 2p+1 will be odd. And then you can ask yourself, are all odds prime numbers? No. But that number pattern doesn't emerge until you getting to larger numbers
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Re: M14-30  [#permalink]

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New post 16 Dec 2016, 12:42
This problem took me 8 minutes. Although I got it correct I feel like I must be missing something. Can anyone explain a fast way to arrive at the answer?
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New post 16 Dec 2016, 14:20
Bunuel wrote:
If \(p\) is a positive integer, is \(2p + 1\) prime?


(1) \(p\) is prime.

(2) The units digit of \(p\) is not prime.


from 1

if p = 2 the 2p+1 = 5 ... prime and if p = 7 thus 2p+1 = 15 not prime ... insuff

from 2

if p = 14 then 2p+1 = 29 prime if p = 12 thus 2p+1 = 25 not prime ... insuff

both
p is a prime >= 11 that ends ( units digit ) in a non prime and since all primes after 5 ends in 1 , 3 , 7,9 then we are looking for primes that end in 9 or 1

11 and 19

if p= 11 thus 2p+1 = 23 prime ... if p = 19 thus 2p+1 = 39 not prime ... insuff

E
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New post 17 Dec 2016, 02:50
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meshackb wrote:
This problem took me 8 minutes. Although I got it correct I feel like I must be missing something. Can anyone explain a fast way to arrive at the answer?


I guess that you want to solve this one by using algebric way? No need to do that. In actual GMAT, you could simply find 2 cases satisfied statements but lead to 2 different results. This won't cost much time.

For example:

(1) p is prime.

p=2 so 2p+1=5 is prime
p=3 so 2p+1=7 is also a prime
p=5 so 2p+1=11 is also a prime

It seems good.

Now if p=7 so 2p+1=15 is not a prime. Hence (1) is out.

(2) The units digit of p is not prime.
So units digit of p could be 0, 1, 4, 6, 8, 9.

If p=1 then 2p+1=3 is a prime
If p=4 then 2p+1=9 is not a prime
Hence (2) is also out.

Now, combine (1) and (2).

The smallest positive number which is a prime and its units digit is not a prime is 11.
If p=11 then 2p+1=23 is a prime

The next prime is 13, 17, 19
13, 17 are out because these numbers don't satisfy (2).
Now 19 left. If p=19 then 2p+1=2*19+1=39 is not a prime

Clearly (1) + (2) is insufficient.
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New post 30 Dec 2016, 01:58
yezz wrote:
Bunuel wrote:
If \(p\) is a positive integer, is \(2p + 1\) prime?


(1) \(p\) is prime.

(2) The units digit of \(p\) is not prime.


from 1

if p = 2 the 2p+1 = 5 ... prime and if p = 7 thus 2p+1 = 15 not prime ... insuff

from 2

if p = 14 then 2p+1 = 29 prime if p = 12 thus 2p+1 = 25 not prime ... insuff

both
p is a prime >= 11 that ends ( units digit ) in a non prime and since all primes after 5 ends in 1 , 3 , 7,9 then we are looking for primes that end in 9 or 1

11 and 19

if p= 11 thus 2p+1 = 23 prime ... if p = 19 thus 2p+1 = 39 not prime ... insuff

E


In the second example, you cannot take p=12. Since the last digit is not prime, 12 is not an option. Take 17 and it gives us the answer.
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New post 30 Dec 2016, 13:02
theincredible wrote:
yezz wrote:
Bunuel wrote:
If \(p\) is a positive integer, is \(2p + 1\) prime?


(1) \(p\) is prime.

(2) The units digit of \(p\) is not prime.


from 1

if p = 2 the 2p+1 = 5 ... prime and if p = 7 thus 2p+1 = 15 not prime ... insuff

from 2

if p = 14 then 2p+1 = 29 prime if p = 12 thus 2p+1 = 25 not prime ... insuff

both
p is a prime >= 11 that ends ( units digit ) in a non prime and since all primes after 5 ends in 1 , 3 , 7,9 then we are looking for primes that end in 9 or 1

11 and 19

if p= 11 thus 2p+1 = 23 prime ... if p = 19 thus 2p+1 = 39 not prime ... insuff

E


In the second example, you cannot take p=12. Since the last digit is not prime, 12 is not an option. Take 17 and it gives us the answer.




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New post 30 Dec 2016, 13:02
U r right


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New post 30 Jun 2017, 11:51
Prime numbers can't be represented through linear equations. If that would have been the case, we wouldn't need super computers.
So, E is the answer.
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New post 10 Feb 2019, 03:03
I think this is a poor-quality question and I don't agree with the explanation. The question clearly says - if P is a positive integer. So in option 1 if P is prime, then P can only be equal to 2 (since that is the only even prime integer). Hence, 2*2+1 = 5 is a prime number.

However, for 2 if the unit digit of p is not prime and P is even (from the question) - so let's say 14 - 14*2+1 = 29 which is prime, however, 16 gives us 33 which is not prime. Hence, only solution 1 is sufficient.
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New post 10 Feb 2019, 03:10
NilayanDasGupta wrote:
I think this is a poor-quality question and I don't agree with the explanation. The question clearly says - if P is a positive integer. So in option 1 if P is prime, then P can only be equal to 2 (since that is the only even prime integer). Hence, 2*2+1 = 5 is a prime number.

However, for 2 if the unit digit of p is not prime and P is even (from the question) - so let's say 14 - 14*2+1 = 29 which is prime, however, 16 gives us 33 which is not prime. Hence, only solution 1 is sufficient.


Why it's necessary in (1) p to equal to 2? Why it cannot be 11? Or 19? Both are prime integers and if p=11, then 2p+1 IS primes (23) but if p=19, then 2p+1 is NOT a prime (39).
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