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If you have done a variety of questions, you have come [#permalink] New post 03 Mar 2011, 19:45
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A
B
C
D
E

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0% (00:00) correct 100% (00:50) wrong based on 7 sessions
If you have done a variety of questions, you have come across this concept somewhere before. The point is to recognize which concept I am talking about.

Ques. Given that x and y are positive integers, is x prime?

I. (y + 1)! <= x <= (y + 1)(y! + 1)

II. (y + 1)! + 1 has five factors
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Re: Number Properties [#permalink] New post 03 Mar 2011, 20:43
this question is really tough and time-consuming.
my answer is E and it took me > 5 mins to solve it without knowing if i am right. :oops:

statement 2 is insufficient cos we dont know x
statement 1:
(y+1)!=1*2*...*y*(y+1) =1.2.....y.(y+1)=
=1.2.....y.y +1.2.3....y=y.y!+y!
(y+1)*(y!+1)=y.y!+y!+y+1=(y+1)!+(y+1)
=> (y+1)!<=x<=(y+1)!+(y+1)
whatever y is, prime or not prime, (y+1)! is never a prime
if y=1 so 2!=2<x<4 > x=3
y=2 so 3!=6<x<6+2+1=9 so x coule be 7,8,9 so insufficient

statement 1+2
(y+1)!+1<=x<=(y+1)!+1+(y+1)
A<=x<=A+y+1 ( consider A to be (y+1)!+1)
insufficient
E

( still confused) :-D
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Re: Number Properties [#permalink] New post 03 Mar 2011, 20:44
I think the answer is C.
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Re: Number Properties [#permalink] New post 03 Mar 2011, 21:06
Looks like this is pretty time consuming. On the D Day, I'll move on to the next question after spending 30 seconds and understanding the complexity.

The answer looks like E.
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VeritasPrepKarishma wrote:
If you have done a variety of questions, you have come across this concept somewhere before. The point is to recognize which concept I am talking about.

Ques. Given that x and y are positive integers, is x prime?

I. (y + 1)! <= x <= (y + 1)(y! + 1)

II. (y + 1)! + 1 has five factors
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Re: Number Properties [#permalink] New post 03 Mar 2011, 22:19
(1) implies
(y+1)! <= x <= (y+1)! + (y+1)
Lets replace y+1 by a

=> a! <= x <= a! + a

For all, a! + i where i<=a, there will be a common factor (a+i) in a! + a
=>x is not prime.

(2) No result derivable.

Answer should be A.

OA please.
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Re: Number Properties [#permalink] New post 03 Mar 2011, 22:54
Agreed ! 100% it has to be A.
a will have factors between a! + a and a!. x is NOT prime.

I think this pattern is called "LONE WOLF" trap. it should be A.

http://gmatclub.com/wiki/GMAT_math_tips

IndigoIntentions wrote:
(1) implies
(y+1)! <= x <= (y+1)! + (y+1)
Lets replace y+1 by a

=> a! <= x <= a! + a

For all, a! + i where i<=a, there will be a common factor (a+i) in a! + a
=>x is not prime.

(2) No result derivable.

Answer should be A.

OA please.
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Re: Number Properties [#permalink] New post 03 Mar 2011, 23:59
IndigoIntentions wrote:
(1) implies
(y+1)! <= x <= (y+1)! + (y+1)
Lets replace y+1 by a

=> a! <= x <= a! + a

For all, a! + i where i<=a, there will be a common factor (a+i) in a! + a
=>x is not prime.

(2) No result derivable.

Answer should be A.

OA please.


The statement above is not true. For e.g. for a = 2 x lies between 2 and 4, lets say 3 and 3 is prime
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Re: Number Properties [#permalink] New post 04 Mar 2011, 02:08
From statement 2) can I infer that (y+1)! + 1 is perfect square. some power of 4. ie. z^4 where z > 0.

So far, I am not able to find a single integer z, that satisfies this criteria. I wonder if there is a problem with second statement.
II. (y + 1)! + 1 has five factors

beyondgmatscore wrote:

The statement above is not true. For e.g. for a = 2 x lies between 2 and 4, lets say 3 and 3 is prime
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Re: Number Properties [#permalink] New post 04 Mar 2011, 05:05
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VeritasPrepKarishma wrote:
If you have done a variety of questions, you have come across this concept somewhere before. The point is to recognize which concept I am talking about.

Ques. Given that x and y are positive integers, is x prime?

I. (y + 1)! <= x <= (y + 1)(y! + 1)

II. (y + 1)! + 1 has five factors


Let's analyze the question.

The question stem just tells us that x and y are positive integers. The information was provided mainly to rule out a decimal value for x and since we are using factorials for y.
Question: Is x prime?

I. (y + 1)! <= x <= (y + 1)(y! + 1)
This needs to be modified a little. Why? because right now, there is no apparent relation between (y+1)! and (y + 1)(y! + 1). (y+1)! makes sense to me, (y! +1) does not. Can I bring everything in terms of (y+1)?
I see that I have a (y+1) multiplied with y! and with 1. Recognize that (y+1)*y! = (y+1)!
So, (y + 1)(y! + 1) = (y+1)y! + (y+1) = (y+1)! + (y+1)
How will you know that this is how you would like to split it? Use (y+1)! on left side as a clue. The right side should make sense with respect to the left side which is in its simplest form.

(y + 1)! <= x <= (y+1)! + (y+1)

Now think about it. x can take any of the following values (general case):
(y+1)! - Not prime except if y is 1
(y+1)! + 1 - Cannot say whether it is prime or not. If y = 1, this is prime. If y is 2, this is prime. If y is 3 it is not.
(y+1)! + 2 - Has 2 as a factor. Not prime
(y+1)! + 3 - Has 3 as a factor. Not prime
.
.
(y+1)! + (y+1) - Has (y+1) as a factor. Not prime

Hence x may or may not be prime.

II. (y + 1)! + 1 has five factors
Not sufficient on its own. No mention of x.

Together: There were two exceptions we found above.
1. If y = 1, then (y+1)! is prime.
From stmnt II, since (1+1)! + 1 = 3 does not have 5 factors (It only has 2), y is not 1. Hence y is not 1 and (y+1)! is not prime
2. We don't know whether (y+1)! + 1 is prime
Since (y+1)! + 1 has 5 factors, it is definitely not prime. Prime numbers have only 2 factors.
Since both exceptions have been dealt with using both statements together, we can say that in every case, x is not prime.
Answer (C).

The question is not time consuming if you quickly see the pattern. It's all about getting exposed to the various concepts. The explanation seems long but only because I have written out everything my mind thinks in a few seconds. This is a relatively tough question but not beyond GMAT. Expect such questions if you are shooting for 49-51 in Quant.
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Re: Number Properties [#permalink] New post 04 Mar 2011, 05:20
gmat1220 wrote:
From statement 2) can I infer that (y+1)! + 1 is perfect square. some power of 4. ie. z^4 where z > 0.

So far, I am not able to find a single integer z, that satisfies this criteria. I wonder if there is a problem with second statement.
II. (y + 1)! + 1 has five factors

beyondgmatscore wrote:

The statement above is not true. For e.g. for a = 2 x lies between 2 and 4, lets say 3 and 3 is prime


I now think that above conclusion can also be reached as any number that has more than 2 factors can be written in the form of a^p*b^q.. and so on where a and b are distinct prime numbers and the number of factors for such a number would be given by (p+1)*(q+1). Therefore, any number that has 5 factors would necessarily have the form Z^4 where z is a prime.
However, the second statement in the question is a red herring, as we would get really confused if we try and grapple with the problem of finding a value for y for which (y + 1)! + 1 has five factors. All we need to be concerned here with is the fact that because (y + 1)! + 1 has five factors (y + 1)! + 1 is not prime.

Superb question Karishma +1
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Re: Number Properties [#permalink] New post 04 Mar 2011, 08:19
Sorry cant read those y+1s. Lets substitute y+1 with a.
I a! <= x <= a!+a
II a! + 1 has five factors.

1) Insufficient. We dont know if a is prime. We also dont know if a! is prime.

2) Insufficient. a!+1 has five factors. We can infer a! is not prime. And a is not prime. We don't know x.

Combine 1) + 2) Sufficient.

The pattern is [a! + Integer] has atleast 3 factors namely 1, (a! + Integer) and F such that 2 <= F <= a
Here Integer <= a

Agree?

beyondgmatscore wrote:
I now think that above conclusion can also be reached as any number that has more than 2 factors can be written in the form of a^p*b^q.. and so on where a and b are distinct prime numbers and the number of factors for such a number would be given by (p+1)*(q+1). Therefore, any number that has 5 factors would necessarily have the form Z^4 where z is a prime.
However, the second statement in the question is a red herring, as we would get really confused if we try and grapple with the problem of finding a value for y for which (y + 1)! + 1 has five factors. All we need to be concerned here with is the fact that because (y + 1)! + 1 has five factors (y + 1)! + 1 is not prime.

Superb question Karishma +1
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Re: Number Properties [#permalink] New post 04 Mar 2011, 09:18
tough one :(
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Re: Number Properties [#permalink] New post 04 Mar 2011, 17:59
Brilliant ! Makes all the sense now.
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Re: Number Properties [#permalink] New post 05 Mar 2011, 04:12
Hi Karishma

How is :

(y+1)! + 3 - Has 3 as a factor. Not prime

if y = 1, then (2! +3) = 5, right ?

Regards,
Subhash
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Re: Number Properties [#permalink] New post 05 Mar 2011, 05:00
Expert's post
subhashghosh wrote:
Hi Karishma

How is :

(y+1)! + 3 - Has 3 as a factor. Not prime

if y = 1, then (2! +3) = 5, right ?

Regards,
Subhash


We are considering all terms from (y+1)! to (y+1)! + (y+1)
If y = 1, then we are only considering terms 2!, 2! + 1 and 2! + 2.

I have given the general case above where y is some greater number say 6.
In that case
6!
6! + 1
6! + 2 - factor 2
6! + 3 - factor 3
6! + 4 - factor 4
6! + 5 - factor 5
6! + 6 - factor 6
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Re: Number Properties   [#permalink] 05 Mar 2011, 05:00
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