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If x is a prime number, what is the value of x ? (1) x <

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If x is a prime number, what is the value of x ? (1) x < [#permalink] New post 25 Aug 2004, 11:51
If x is a prime number, what is the value of x ?

(1) x < 15
(2) (x – 2) is a multiple of 5.

I think the answer is C (x can be 7 or 17). Kaplan says its B.

Can somebody help on this as well.
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Re: Another KAPAN Question [#permalink] New post 25 Aug 2004, 11:56
amit_drummer wrote:
If x is a prime number, what is the value of x ?

(1) x < 15
(2) (x – 2) is a multiple of 5.

I think the answer is C (x can be 7 or 17). Kaplan says its B.

Can somebody help on this as well.


1) Insufficient. X could be 1 2 3 5 7 11 13
2) X is a multiple of seven, X could be 7, or 17

C. Overlap in sets is 7. Sufficient with Both.
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 [#permalink] New post 25 Aug 2004, 14:25
7 is the answer , combining both you get x = 7 .

1 . x = 2,3,5,7,9,11,13

2 . x = 7 , 37

combining both you have only 7 .
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 [#permalink] New post 26 Aug 2004, 06:01
SigEpUCI,

Just for you information: 1 is not an prime number. Prime number starts with 2 (the only even prime number).

For the question itself I am a little confused.

St.1 is insufficient.
X could be 2, 3, 5, 7, 11 and 13

St.2 is insufficient.
X could be 2, 7 or 17

Combine both statements doesnt solve the answer as well.
X could be 2 or 7

I think 2 - 2 = 0, and 0 is also an multiple of 5

Therefore I think the answer is E.
However I am not sure if 0 is an multiple of 5

Correct me if I am wrong.

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 [#permalink] New post 26 Aug 2004, 06:12
I could not help it.
After I have posted my previous reply I have checked in my Kaplan 2003 book.

The correct answer given is E.

Kaplan:
From the question stem: All we know is that x is a prime number. We want enough information to determine which prime number x is. Our method, then, is to try to find more than one prime that fits with whatever information we're given. If we can, the information is insufficient; if we can't - if we can find only one prime that fits with the information - then the information is sufficient.

1) Insufficient: if x < 15, x could be 2, 3, 5, 7, 11 or 13. Eliminate A and D.
2) Insufficient: if (x-2) is a multiple of five, then x is 2 more than a multiple of 5. So the question is: Can we find more than one prime number that is 2 more than a multiple of five? Yes. Multiples of 5 are 0, 5, 10, 15, 20, and so on. Two more than 5 is 7 - a prime number. While 2 more than 10 is 12, which isn't a prime number, 2 more than 15 is 17, which is prime. Eliminate B.

In combination: Statement 1 narrowed down the possible values of x to 2, 3, 5, 7, 11 and 13. Remember that 0 is a multiple of 5 as well. So both 2 and 7 are more than a multiple of 5, so we cannot find a single answer to the question using both statements. Choose E.

Amit_drummer, I am not sure which Kaplan Book you use. This is the info from the 2003 book.


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 [#permalink] New post 27 Aug 2004, 06:39
Good one. I have seen these too (actually I missed out 2 and chose C, when I first solved this problem).

When we think of multiples of 5, all we think of is 5, 10, 15, ...
It can be ..., -15, -10, -5, 0, 5, 10, 15, ...
(though negative is not relavent to this problem)
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 [#permalink] New post 30 Aug 2004, 07:00
0 is not a multiple of 5 or for that matter any integer.

in that case the LCM of any 2 numbers, will be a 0.

Multiples mean more than ZERO. 1x, 2x, 3x,....etc.
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 [#permalink] New post 30 Aug 2004, 13:06
amit_drummer wrote:
0 is not a multiple of 5 or for that matter any integer.

in that case the LCM of any 2 numbers, will be a 0.

Multiples mean more than ZERO. 1x, 2x, 3x,....etc.


Zero is considered to be a multiple. I believe the LCM is defined to be a positive number. If not, the LCM would be negative infinity since there are an unlimited number of negative numbers that are multiples of two numbers.
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 [#permalink] New post 30 Aug 2004, 14:07
AkamaiBrah wrote:
amit_drummer wrote:
0 is not a multiple of 5 or for that matter any integer.

in that case the LCM of any 2 numbers, will be a 0.

Multiples mean more than ZERO. 1x, 2x, 3x,....etc.


Zero is considered to be a multiple. I believe the LCM is defined to be a positive number. If not, the LCM would be negative infinity since there are an unlimited number of negative numbers that are multiples of two numbers.


in other words, the answer is E
  [#permalink] 30 Aug 2004, 14:07
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