Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

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.
_________________

Citius, Altius, Fortius

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

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

Show Tags

17 Oct 2017, 10:53

1

This post received KUDOS

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

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

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

Show Tags

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!

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

Show Tags

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!

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!