If x is an integer and |1 x| < 2 then which of the following must be

Responding to pm.

D. Number of distinct positive factors of x+2 is a prime number. x+2 is 2, 3, or 4.

2 has 2 factors 1 and 2.
3 has 2 factors 1 and 3.
4 has 3 factors 1, 2 and 4.

The number of factors of each number is a prime number.

Hope it's clear.

Bunuel,

Why does 2 have 2 factors? I thought the smallest factor of any number was 2.. Is there a definition difference that I am missing?

Thanks for the reply!

A factor is a positive integer which divides some integer without a remainder. Thus the smallest factor of any integer is 1.

The factors of 2 are 1 and 2.

I think E ist also correct. as number 1 is not considered prime, as it has only one factor (itself).

Except if 0 is a multiple of every number!

Yes, 1 is NOT prime but it has nothing to do with option E.

E says: x is not a multiple of an odd prime number. IF x=0, then this option is not always true because 0 is a multiple of every integer, hence it's a multiple of all odd primes: 3, 5, 7, ....

This might be a naive question and also highlights a gap in my understand but can you please explain how |1−x|<2 translates into "-2<(1-x)<2". Thank you.

|1 − x| < 2:

(1 - x) < 2;
-(1 - x) < 2 --> -2 < 1 - x

Hence -2 < 1 - x < 2.



Hope this helps.

Bunuel - Thank you very much. It absolutely helps (no pun intended); and like I said there is a gap in my understanding since I believed that the absolute value of anything is always positive, hence I was viewing |x-1| as simply (x-1) and did not consider -(x-1). Thanks again.

Absolute value of any number, expression, is more than or equal to zero but the expression in the modulus can be negative as well as positive. So, \(|x-1|\geq{0}\) but x-1 can be positive negative or 0.

Bunnel, where do the multiples start for an integer? Say for 3
Do they start at 0 or should the negative multiples be considered too?
0 being the multilple of every integer is certainly a revelation to me. Thanks for that!
Btw, zero has no multiples then?

Yes, no number is a multiple of 0.

As for negative multiples: multiples of 3 are: ..., -6, -3, 0, 3, 6, ... But you should not worry about it since every GMAT divisibility question will tell you in advance that any unknowns represent positive integers, which means that ALL GMAT divisibility questions are limited to positive integers only.

Hi again Bunuel,

Just one question. I know by trial and error that the below process is wrong. But why does the algebra not match the intuitive way of solving??? Could you pls point out where you think I am making an error? TIA.

Given |1-x| < 2

(a) If x>0: 1-x < 2 -> x > -1

But this is true only for x>=0 which is a more limiting condition than x > -1. So shouldn't the result of opening the modulus be x>=0?

(b) If x<0: -1+x < 2 -> x<3

But this is true only for x<0 which is a more limiting condition that x<3. So shouldn't the result of opening the modulus be x<0?

By the above logic x = 0. But I can clearly see that x = 1 and x =2 will also work - why the discrepancy :

Hi avgroh,

You are not getting the right result because you are considering the zero points of x when deciding the sign of the expression when it comes out of the modulus. We consider the zero points of the expression inside the modulus to decide the sign of the expression when it comes out of the modulus i.e. in this case zero points of (1-x). So we can solve this modulus as

1. When 1- x > 0
1 - x < 2 i.e. x > -1

2. When 1 - x < 0
-(1-x) < 2 i.e. x < 3

Combining the above we get the range as -1 < x < 3. Since x is an integer it can take values of only 0,1 and 2.

Just to add on to the concept, we write |x| = x if x > 0 and -x if x < 0. On the same line, we would write |x -1| = x-1 if x-1 > 0 and -(x-1) if x-1 < 0

Hope it helps

Regards
Harsh

Hi,

Would you mind having a look? This is what I did.

I 1-x I<2
-2<1-x<2
Subtracted -1 ie -3<-x<1
Multiplied by - 3>x>-1 Now I understand option D, zero is an integer correct? It is just not a positive integer?

Can you explain option D three distinct positive values ie in my case 2,1,0? Just a little confused or do I have it all wrong?

|1-x|<2 means -1<x<3. Since given that x is an integer then x can be 0, 1 or 2.

D says: Number of distinct positive factors of x+2 is a prime number. x+2 is 2, 3, or 4.

2 has 2 factors 1 and 2.
3 has 2 factors 1 and 3.
4 has 3 factors 1, 2 and 4.

The number of factors of each number is a prime number.

Hope it's clear.

Hi Brent BrentGMATPrepNow, could you help with this question please especially D as 4 is not a prime number? Thanks Brent

Two properties involving absolute value inequalities:
Property #1: If |something| < k, then –k < something < k
Property #2: If |something| > k, then EITHER something > k OR something < -k
Note: these rules assume that k is positive

Since the given inequality, |1 − x| < 2, is in the form |something| < k, we know we need to apply Property #1

When we apply Property #1, we get: -2 < 1 - x < 2
Subtract 1 from all sides to get: -3 < -x < 1
Multiply all sides by -1 to get: 3 > x > -1 [ since we multiplied all sides of the inequality by a negative value, we reversed the direction of the inequality symbols]

If x is an INTEGER and if 3 > x > -1, then there are only three possible values of x: 0, 1 or 2

Now let's examine the five answer choices:
(A) x is not a prime number
Since x COULD equal 2, x COULD be prime.

(B) x^2 + x is not a prime number
If x = 1, then x^2 + x = 1^2 + 1 = 2, and 2 IS prime.

(C) x is positive
If x = 0, then x is NOT positive.

(D) Number of distinct positive factors of x + 2 is a prime number
If x = 0, then x + 2 = 2. 2 has TWO positive factors (1 and 2), and TWO IS a prime number.
If x = 1, then x + 2 = 3. 3 has TWO positive factors (1 and 3), and TWO IS a prime number.
If x = 2, then x + 2 = 4. 4 has THREE positive factors (1, 2 and 4), and THREE IS a prime number.

(E) x is not a multiple of an odd prime number
If x = 0, then x IS a multiple of an odd prime number

Answer: D