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If x, y and z are positive integers and x < y, what are the number of
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16 Oct 2017, 08:18
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If x, y and z are positive integers and x < y, what are the number of common factors between x and z? (1) \(z = y! + 1\) (2) \(x = 11\) and \(y = 12\) self made : challenge Q Kudos for first few correct solutions
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1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
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Re: If x, y and z are positive integers and x < y, what are the number of
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16 Oct 2017, 10:14
Answer is A as x,y,z all are positive and x<y, least value of y is 2 and x is 1
I did the following hit and trial if y=2 Z=3 and x=1...Common factor..1
if y=3 Z=7 and x=1,2...Common factor 1
if y=4 z=25, x=1,2,3..common factor 1
if y=5 z=121..x=1,2,3,4..
if y=6 z=721..x=1,2,3,4,5....CF=1.
Seeing this pattern it is A... I am not able to deduce an algebraic solution but saw this pattern



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If x, y and z are positive integers and x < y, what are the number of
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Updated on: 17 Oct 2017, 08:49
chetan2u wrote: If x, y and z are positive integers and x < y, what are the number of common factors between x and z?
(1) \(z = y! + 1\) (2) \(x = 11\) and \(y = 12\)
self made : challenge Q Kudos for first few correct solutions Just trying to answer based on my understanding :  Statement (1) and (2) IMO not sufficient because just talks only about z&m (1) and x&y (2).  So, basically this is C and E split. Combine the statement :  x=11, y=12, so z=(12!)+1.  x just have 2 factor  this is prime number.  So, the question is simply : do z has 11 as a factor? If they HAVE, so the common factors are 2 (1 and 11)  If they DON'T HAVE, so the common factor just 1 (only 1).  So, just analyze these things we can conclude that the answer is C. Whatever the answer, we CAN get how many common factors they have. By the way, z does not have 11 as a factor. (12!) has the 11 as a factor, but when you add 1 into (12!) so you cannot get 11 as a factor. Am I right?
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Originally posted by septwibowo on 17 Oct 2017, 02:03.
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If x, y and z are positive integers and x < y, what are the number of
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Updated on: 17 Oct 2017, 10:01
chetan2u wrote: If x, y and z are positive integers and x < y, what are the number of common factors between x and z?
(1) \(z = y! + 1\) (2) \(x = 11\) and \(y = 12\)
self made : challenge Q Kudos for first few correct solutions Correction: The answer should be A. Given that x < y. Statement 1: since z = y!+1. There is no common factor between z and x beside 1. This statement is sufficient. Statement 2: Does not give any information about z. Hence this is insufficient.
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Originally posted by TheMechanic on 17 Oct 2017, 06:18.
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Re: If x, y and z are positive integers and x < y, what are the number of
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17 Oct 2017, 08:07
chetan2u wrote: If x, y and z are positive integers and x < y, what are the number of common factors between x and z?
(1) \(z = y! + 1\) (2) \(x = 11\) and \(y = 12\)
self made : challenge Q Kudos for first few correct solutions Request u to pls explain this in detail



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If x, y and z are positive integers and x < y, what are the number of
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17 Oct 2017, 09:53
chetan2u wrote: If x, y and z are positive integers and x < y, what are the number of common factors between x and z?
(1) \(z = y! + 1\) (2) \(x = 11\) and \(y = 12\)
self made : challenge Q Kudos for first few correct solutions Statement 1: implies that \(z=y!+1\) & \(y!\) are consecutive and hence coprime. Two consecutive integers are coprime, which means that only common factor is \(1\). For example \(2\) and \(3\), \(40\) & \(41\) etc. are consecutive integers, thus only common factor they share is \(1\). so \(z\) will have \(1\) as a common factor with any number less than \(y!\) and since \(x<y\), so \(z\) will have only one common factor with \(x\) i.e \(1\) you can test this through some examples. for ex. \(x=2\) & \(y=3\) then \(z=3!+1=7\). between \(x\) & \(z\) only \(1\) is the common factor another one, \(x=3\) & \(y=5\), \(z=5!+1=120+1=121=11^2\). only \(1\) is the common factor here. SufficientStatement 2: Nothing mentioned about z. InsufficientOption A



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Re: If x, y and z are positive integers and x < y, what are the number of
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18 Oct 2017, 03:33
chetan2u wrote: If x, y and z are positive integers and x < y, what are the number of common factors between x and z?
(1) \(z = y! + 1\) (2) \(x = 11\) and \(y = 12\)
self made : challenge Q Kudos for first few correct solutions It is a tough Q and the solution would go like this... (1) \(z = y! + 1\) z=y!+1... first thing first  All the numbers below y are factors of y! as \(y! = 1*2*3*....(y1)*y\), so when we add 1 to y!, y!+1 will be coprime to all the numbers below ythis is what the Q is asking... y!+1 is NOTHING but z, and any value < y is NOTHING but x, so x and z are coprime..Only common factor between x and z is 1 sufficient (2) \(x = 11\) and \(y = 12\) Nothing about z Insufficient A
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1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
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Re: If x, y and z are positive integers and x < y, what are the number of
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18 Oct 2017, 18:41
niks18 wrote: chetan2u wrote: If x, y and z are positive integers and x < y, what are the number of common factors between x and z?
(1) \(z = y! + 1\) (2) \(x = 11\) and \(y = 12\)
self made : challenge Q Kudos for first few correct solutions Statement 1: implies that \(z=y!+1\) & \(y!\) are consecutive and hence coprime. Two consecutive integers are coprime, which means that only common factor is \(1\). For example \(2\) and \(3\), \(40\) & \(41\) etc. are consecutive integers, thus only common factor they share is \(1\). so \(z\) will have \(1\) as a common factor with any number less than \(y!\) and since \(x<y\), so \(z\) will have only one common factor with \(x\) i.e \(1\) you can test this through some examples. for ex. \(x=2\) & \(y=3\) then \(z=3!+1=7\). between \(x\) & \(z\) only \(1\) is the common factor another one, \(x=3\) & \(y=5\), \(z=5!+1=120+1=121=11^2\). only \(1\) is the common factor here. SufficientStatement 2: Nothing mentioned about z. InsufficientOption AThanks a lot for the reply What does this mean? so z will have 1 as a common factor with any number less than y!



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If x, y and z are positive integers and x < y, what are the number of
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18 Oct 2017, 19:04
zanaik89 wrote: niks18 wrote: chetan2u wrote: If x, y and z are positive integers and x < y, what are the number of common factors between x and z?
(1) \(z = y! + 1\) (2) \(x = 11\) and \(y = 12\)
self made : challenge Q Kudos for first few correct solutions Statement 1: implies that \(z=y!+1\) & \(y!\) are consecutive and hence coprime. Two consecutive integers are coprime, which means that only common factor is \(1\). For example \(2\) and \(3\), \(40\) & \(41\) etc. are consecutive integers, thus only common factor they share is \(1\). so \(z\) will have \(1\) as a common factor with any number less than \(y!\) and since \(x<y\), so \(z\) will have only one common factor with \(x\) i.e \(1\) you can test this through some examples. for ex. \(x=2\) & \(y=3\) then \(z=3!+1=7\). between \(x\) & \(z\) only \(1\) is the common factor another one, \(x=3\) & \(y=5\), \(z=5!+1=120+1=121=11^2\). only \(1\) is the common factor here. SufficientStatement 2: Nothing mentioned about z. InsufficientOption AThanks a lot for the reply What does this mean? so z will have 1 as a common factor with any number less than y! Hi zanaik89, The important concept to note here is that Two consecutive numbers are CoPrime, that is between themselves they will not have any common factor except 1. So let's take an example of 24 & 25, 24 has factors 1,2,3,4,6,8,12,24 and 25 has factors 1,5,25. As you can see only 1 is the common factor between 24 & 25 because both are consecutive integers and hence coprime. In this question z and y! are consecutive, hence z and y! will be coprime so we can say that any factor of y! except 1 will not be present in z. Since x<y and as y! has all integers that are less than y as a factor so x must be a factor of y! so x cannot be a factor of y!+1 i.e.z. For eg. if x=2, y=3 then y! = 3* 2*1, as you can see x is a factor of y!




If x, y and z are positive integers and x < y, what are the number of &nbs
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