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

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If x, y and z are positive integers and x < y, what are the number of [#permalink]

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New post 16 Oct 2017, 09:18
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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
[Reveal] Spoiler: OA

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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Re: If x, y and z are positive integers and x < y, what are the number of [#permalink]

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New post 16 Oct 2017, 11:14
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 [#permalink]

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New post 17 Oct 2017, 03:03
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? :lol: :lol: :lol:
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Last edited by septwibowo on 17 Oct 2017, 09:49, edited 1 time in total.

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If x, y and z are positive integers and x < y, what are the number of [#permalink]

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New post 17 Oct 2017, 07:18
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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Last edited by TheMechanic on 17 Oct 2017, 11:01, edited 1 time in total.

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Re: If x, y and z are positive integers and x < y, what are the number of [#permalink]

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New post 17 Oct 2017, 09:07
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 [#permalink]

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New post 17 Oct 2017, 10:53
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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 co-prime. Two consecutive integers are co-prime, 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.

Sufficient

Statement 2: Nothing mentioned about z. Insufficient

Option A

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Re: If x, y and z are positive integers and x < y, what are the number of [#permalink]

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New post 18 Oct 2017, 04:33
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 \(y! = 1*2*3*....(y-1)*y\), so when we add 1 to y!, y!+1 will be co-prime to all the numbers below y
this 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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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Re: If x, y and z are positive integers and x < y, what are the number of [#permalink]

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New post 18 Oct 2017, 19:41
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 co-prime. Two consecutive integers are co-prime, 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.

Sufficient

Statement 2: Nothing mentioned about z. Insufficient

Option A


Thanks 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 [#permalink]

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New post 18 Oct 2017, 20:04
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 co-prime. Two consecutive integers are co-prime, 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.

Sufficient

Statement 2: Nothing mentioned about z. Insufficient

Option A


Thanks 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 Co-Prime, 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 co-prime.

In this question z and y! are consecutive, hence z and y! will be co-prime 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!

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If x, y and z are positive integers and x < y, what are the number of   [#permalink] 18 Oct 2017, 20:04
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