If x, y and z are positive integers and x

Question

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

deepudiscover · Answer

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

septwibowo · Answer

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?

TheMechanic · Answer

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.

zanaik89 · Answer

Request u to pls explain this in detail

niks18 · Answer

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

chetan2u · Answer

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

zanaik89 · Answer

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!

niks18 · Answer

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

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