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

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Veritas Prep GMAT Instructor
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Given that x and y are positive integers, is x prime?  [#permalink]

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03 Mar 2011, 20:45
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Question Stats:

41% (02:30) correct 59% (02:24) wrong based on 131 sessions

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

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

(2) $$(y + 1)! + 1$$ has five factors

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.

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Karishma
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Re: Given that x and y are positive integers, is x prime?  [#permalink]

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03 Mar 2011, 21:43
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.

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)
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Re: Given that x and y are positive integers, is x prime?  [#permalink]

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03 Mar 2011, 21:44
I think the answer is C.
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Re: Given that x and y are positive integers, is x prime?  [#permalink]

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03 Mar 2011, 22:06
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.

Experts ?

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: Given that x and y are positive integers, is x prime?  [#permalink]

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03 Mar 2011, 23:19
(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.

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Re: Given that x and y are positive integers, is x prime?  [#permalink]

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03 Mar 2011, 23:54
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.

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Posts: 166
Re: Given that x and y are positive integers, is x prime?  [#permalink]

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04 Mar 2011, 00:59
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.

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: Given that x and y are positive integers, is x prime?  [#permalink]

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04 Mar 2011, 03:08
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: Given that x and y are positive integers, is x prime?  [#permalink]

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04 Mar 2011, 06:05
3
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.

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: Given that x and y are positive integers, is x prime?  [#permalink]

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04 Mar 2011, 06:20
1
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: Given that x and y are positive integers, is x prime?  [#permalink]

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04 Mar 2011, 09:19
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: Given that x and y are positive integers, is x prime?  [#permalink]

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04 Mar 2011, 18:59
Brilliant ! Makes all the sense now.
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Re: Given that x and y are positive integers, is x prime?  [#permalink]

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05 Mar 2011, 05:12
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: Given that x and y are positive integers, is x prime?  [#permalink]

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05 Mar 2011, 06:00
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: Given that x and y are positive integers, is x prime?  [#permalink]

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29 Sep 2017, 09:11
VeritasPrepKarishma wrote:
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.

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.

Experts help me understand the combination part.
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Given that x and y are positive integers, is x prime?  [#permalink]

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18 Oct 2018, 08:19
Given that x and y are positive integers, is x prime?

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

(2) $$(y + 1)! + 1$$ has five positive factors

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.

Dear Karishma,

Very beautiful problem, congrats!

Let me contribute with a different wording.

I recognize the famous result: for any integer n greater than 1, we don´t have primes in the interval [ n!+2 , n!+n ] (*) .
(Because j is a factor of n!+j , where j is 2,3, ..., n.)

$$x,y\,\, \ge 1\,\,{\rm{ints}}$$

$$x\,\,\mathop {\rm{ = }}\limits^? \,\,{\rm{prime}}$$

$$\left( 1 \right)\,\,\,n!\,\, \le x\,\, \le \,\,n!\, + n\,\,\,,\,\,{\rm{where}}\,\,\,n = y + 1\,\,\,\,\,\,\,\left[ {\,n!\, + n\,\, = \,\,\left( {y + 1} \right) \cdot y!\, + \left( {y + 1} \right) = \left( {y + 1} \right)\left( {y!\, + 1} \right)\,} \right]$$

$$\left\{ \matrix{ \,{\rm{Take}}\,\,y = 1\,\,\left( {n = 2} \right)\,\,{\rm{and}}\,\,x\,{\rm{ = }}\,{\rm{2}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr \,{\rm{Take}}\,\,y = 1\,\,\left( {n = 2} \right)\,\,{\rm{and}}\,\,x\,{\rm{ = }}\,4\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.$$

$$\left( 2 \right)\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,x = 1\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \,{\rm{Take}}\,\,x = 2\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.$$

$$\left( {1 + 2} \right)\,\,\,\,\,\left\{ \matrix{ \,n!\,\, \le x\,\, \le \,\,n!\, + n \hfill \cr \,n!\,\, + \,\,1\,\,{\rm{not}}\,\,{\rm{prime}}\,\,\, \Rightarrow \,\,n \ge 3\,\,\left( {y \ge 2} \right)\,\,\, \hfill \cr} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\left\{ \matrix{ \,x = n!\,\,\,{\rm{not}}\,\,{\rm{prime}} \,\,\, (n \ge 3) \hfill \cr \,x = n!\,\, + \,\,1\,\,\,{\rm{not}}\,\,{\rm{prime}}\,\,\,\left( {{\rm{statement}}\,\,\left( 2 \right)} \right) \hfill \cr \,n!\, + 2\,\, \le x\,\, \le \,\,n!\, + n\,\,\,,\,\,x\,\,\,{\rm{not}}\,\,{\rm{prime}}\,\,\,\left( * \right) \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle$$

Kind Regards,
Fabio.
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Re: Given that x and y are positive integers, is x prime?  [#permalink]

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29 Apr 2019, 08:52
Given that x and y are positive integers, is x prime?

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

(2) $$(y + 1)! + 1$$ has five factors

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.

Responding to a pm:

Quote:
I had a question regarding this problem. I am probably not seeing something correctly but isn't statement 2 inherently impossible because having 5 factors means that the number is a perfect square? I cannot think of any situation in which the factorial of a number generates a perfect square.

You are right that a number with 5 factors will be a perfect square.

(2) $$(y + 1)! + 1$$ has five factors

Note that we are NOT given that (y + 1)! has 5 factors. We are not given that a factorial has 5 factors. We are given that (y+1)! + 1 has 5 factors.
We are given that 1 more than a factorial is a perfect square. That is possible.

e.g.
4! + 1 = 25 (a perfect square)
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Re: Given that x and y are positive integers, is x prime?   [#permalink] 29 Apr 2019, 08:52
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