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

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.

Request u to pls explain this in detail

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

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

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!