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

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Math Expert
Joined: 02 Aug 2009
Posts: 7036
If x, y and z are positive integers and x < y, what are the number of  [#permalink]

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16 Oct 2017, 08:18
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65% (hard)

Question Stats:

30% (01:31) correct 70% (01:28) wrong based on 44 sessions

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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 $$y = 12$$

Kudos for first few correct solutions

_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.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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16 Oct 2017, 10:14
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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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 $$y = 12$$

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.
Last edited by septwibowo on 17 Oct 2017, 08:49, 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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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 $$y = 12$$

Kudos for first few correct solutions

Correction:

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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Citius, Altius, Fortius

Originally posted by TheMechanic on 17 Oct 2017, 06:18.
Last edited by TheMechanic on 17 Oct 2017, 10:01, 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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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 $$y = 12$$

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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17 Oct 2017, 09:53
1
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 $$y = 12$$

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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Joined: 02 Aug 2009
Posts: 7036
Re: If x, y and z are positive integers and x < y, what are the number of  [#permalink]

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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 $$y = 12$$

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 $$y = 12$$
Insufficient

A
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.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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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 $$y = 12$$

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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18 Oct 2017, 19:04
1
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 $$y = 12$$

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!
If x, y and z are positive integers and x < y, what are the number of &nbs [#permalink] 18 Oct 2017, 19:04
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