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Prime Number

Expert replies
by Aman verma » Sun Jun 06, 2010 9:22 am
Q: P, Q, R and S are four distinct natural numbers and three of them are prime numbers. Is P an even prime ?

I. Among PQ , QR , RS , and QS only one is even.

II. S is a two digit number which when divided by 6 leaves a remainder of 3.
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Source: — Data Sufficiency |

by akhpad » Sun Jun 06, 2010 10:01 am
Statement 1:
Q, R, and S occur at 2 places. So these cannot be even

But P can be even prime or just even number.
P =2, Q and R prime, S = 15
P = 4; Q, R, and S are prime

Insufficient

Statement 2:
S = 6x + 3 = 3(2x+1) = ODD number multiple of 3
P, Q, and R must be prime but any one of these can be even prime.
Insufficient

Statement 1 and 2
P, Q, and R ==> Prime
S ==> ODD number multiple of 3

P =2, Q and R prime, S = 15

Sufficient

Answer: C
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by jube » Tue Jun 08, 2010 1:25 am
C - pretty much the same logic as used by akhp77
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by crackinggmat » Tue Jun 08, 2010 7:04 am
Hi akhilesh ....i,m not getting what u said ...can u pls elaborate ur reasoning......

thanks
akhp77 wrote:Statement 1:
Q, R, and S occur at 2 places. So these cannot be even

But P can be even prime or just even number.
P =2, Q and R prime, S = 15
P = 4; Q, R, and S are prime

Insufficient

Statement 2:
S = 6x + 3 = 3(2x+1) = ODD number multiple of 3
P, Q, and R must be prime but any one of these can be even prime.
Insufficient

Statement 1 and 2
P, Q, and R ==> Prime
S ==> ODD number multiple of 3

P =2, Q and R prime, S = 15

Sufficient

Answer: C
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by Aman verma » Tue Jun 08, 2010 9:06 am
OAC
800. Arjun's-Bird-Eye
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by lunarpower » Tue Jun 08, 2010 11:14 pm
akhp77 wrote:Statement 1:
Q, R, and S occur at 2 places. So these cannot be even
yeah.

in more detail:

a product of integers will be even if any of the integers is/are even.
the product is odd only if all of the integers are odd.


so, it's impossible for q, r, or s to be even:
* if q were even, then pq, qr, and qs would all be even.
* if r were even, then both qr and rs would be even.
* if s were even, then both rs and qs would be even.

so the only way to satisfy statement (1) is to make 'p' even. (if p were also odd, then none of the products would be even.)

as akhilesh says, this statement gives no clues as to whether p is prime.

--

statement 2:

says nothing about p.

--

together:

examine statement 2.
you could do this algebraically, as done by akhilesh, or you could just make a list:
15, 21, 27, 33, 39, ...
if you make a list, it becomes pretty clear that all of these numbers are divisible by 3, so none of them will be prime.
therefore, S is the non-prime number.
so, from statement 1, p is even.
from statement 2, p is prime.
sufficient.
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