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.

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:

Re: If x, y, and z are 3 different prime numbers, which of the [#permalink]
25 May 2013, 02:51

1

This post received KUDOS

Expert's post

karishmatandon wrote:

If x, y, and z are 3 different prime numbers, which of the following CANNOT be a multiple of any of x, y, and z?

A. x + y B. y – z C. xy + 1 D. xyz + 1 E. x^2 + y^2

xyz and xyx+1 are are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1.

Since xyz is a multiple of each x, y, and z, then xyx+1 cannot be a multiple of any of them.

Re: If x, y, and z are 3 different prime numbers, which of the [#permalink]
25 May 2013, 02:52

1

This post received KUDOS

If x, y, and z are 3 different prime numbers, which of the following CANNOT be a multiple of any of x, y, and z?

pick: 2,3,5

A) x + y, 3+5=8 multiple of 2 B) y-z, 5-2=2 multiple of 2 C) xy + 1, 3*5+1=16 multiple of 2 D) xyz + 1, CORRECT E) x^2 + y^23^2+5^2=36 multiple of 2

Why D? Every prime number except 2 is odd. In case you pick 2, xyz + 1=E*O*O +1=Odd and an ODD number cannot be a multiple of 2 In case you DO NOT pick 2 (you have only odd numbers), xyz + 1=O*O*O +1=Even and en EVEN number cannot be a multiple of an ODD number.
_________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: If x, y, and z are 3 different prime numbers, which of the [#permalink]
25 May 2013, 02:57

Expert's post

Zarrolou wrote:

If x, y, and z are 3 different prime numbers, which of the following CANNOT be a multiple of any of x, y, and z?

pick: 2,3,5

A) x + y, 3+5=8 multiple of 2 B) y-z, 5-2=2 multiple of 2 C) xy + 1, 3*5+1=16 multiple of 2 D) xyz + 1, CORRECT E) x^2 + y^23^2+5^2=36 multiple of 2

Why D? Every prime number except 2 is odd. In case you pick 2, xyz + 1=E*O*O +1=Odd and an ODD number cannot be a multiple of 2 In case you DO NOT pick 2 (you have only odd numbers), xyz + 1=O*O*O +1=Even and en EVEN number cannot be a multiple of an ODD number.

Your logic to discard D is not correct.

Yes, if one of the primes is 2, then xyz + 1=E*O*O +1=Odd and an odd number cannot be a multiple of an even number but it could be a multiple of remaining odd primes.

Re: If x, y, and z are 3 different prime numbers, which of the [#permalink]
25 May 2013, 03:01

Zarrolou wrote:

If x, y, and z are 3 different prime numbers, which of the following CANNOT be a multiple of any of x, y, and z?

pick: 2,3,5

A) x + y, 3+5=8 multiple of 2 B) y-z, 5-2=2 multiple of 2 C) xy + 1, 3*5+1=16 multiple of 2 D) xyz + 1, CORRECT E) x^2 + y^23^2+5^2=36 multiple of 2

I used picking numbers strategy and was down to D and E..couldn't figure out a way from there!

Bunuel wrote:

xyz and xyx+1 are are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1.

Since xyz is a multiple of each x, y, and z, then xyx+1 cannot be a multiple of any of them.

Answer: D.

This is a much easier way..Thanks for the explanation
_________________

Consider giving +1 Kudo when my post helps you. Also, Good Questions deserve Kudos..!

Re: If x, y, and z are 3 different prime numbers, which of the [#permalink]
25 May 2013, 03:03

Expert's post

karishmatandon wrote:

Zarrolou wrote:

Why D? Every prime number except 2 is odd. In case you pick 2, xyz + 1=E*O*O +1=Odd and an ODD number cannot be a multiple of 2 In case you DO NOT pick 2 (you have only odd numbers), xyz + 1=O*O*O +1=Even and en EVEN number cannot be a multiple of an ODD number.

I used picking numbers strategy and was down to D and E..couldn't figure out a way from there! Thanks this helps..

Bunuel wrote:

xyz and xyx+1 are are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1.

Since xyz is a multiple of each x, y, and z, then xyx+1 cannot be a multiple of any of them.

Answer: D.

This is a much easier way..Thanks for the explanation

Re: If x, y, and z are 3 different prime numbers, which of the [#permalink]
25 May 2013, 03:07

Bunuel wrote:

Your logic to discard D is not correct.

Yes, if one of the primes is 2, then xyz + 1=E*O*O +1=Odd and an odd number cannot be a multiple of an even number but it could be a multiple of remaining odd primes.

Hope it's clear.

Maybe I misunderstood the question...

If I pick E, O, O, I get as result of D an ODD number. So an ODD number can be a multiple of the two odds (of course it can), but since the question asks CANNOT be a multiple of any, if one of the values is Even, then an odd number cannot a multiple of an even one.

Is this reasoning flawed?
_________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: If x, y, and z are 3 different prime numbers, which of the [#permalink]
25 May 2013, 03:11

1

This post received KUDOS

Expert's post

Zarrolou wrote:

Bunuel wrote:

Your logic to discard D is not correct.

Yes, if one of the primes is 2, then xyz + 1=E*O*O +1=Odd and an odd number cannot be a multiple of an even number but it could be a multiple of remaining odd primes.

Hope it's clear.

Maybe I misunderstood the question...

If I pick E, O, O, I get as result of D an ODD number. So an ODD number can be a multiple of the two odds (of course it can), but since the question asks CANNOT be a multiple of any, if one of the values is Even, then an odd number cannot a multiple of an even one.

Is this reasoning flawed?

The correct option must not be a multiple of ANY of the three variables, so if it could be a multiple of some of them, then it's not the right choice.
_________________

Re: If x, y, and z are 3 different prime numbers, which of the [#permalink]
05 Sep 2013, 22:10

Bunuel wrote:

Zarrolou wrote:

Bunuel wrote:

Your logic to discard D is not correct.

Yes, if one of the primes is 2, then xyz + 1=E*O*O +1=Odd and an odd number cannot be a multiple of an even number but it could be a multiple of remaining odd primes.

Hope it's clear.

Maybe I misunderstood the question...

If I pick E, O, O, I get as result of D an ODD number. So an ODD number can be a multiple of the two odds (of course it can), but since the question asks CANNOT be a multiple of any, if one of the values is Even, then an odd number cannot a multiple of an even one.

Is this reasoning flawed?

The correct option must not be a multiple of ANY of the three variables, so if it could be a multiple of some of them, then it's not the right choice.

Hi,

If i pick the numbers as x=3 y=5 z=7

then to discard 1) x+y = 3+5=8 this is not divisible by any of 3 numbers

Re: If x, y, and z are 3 different prime numbers, which of the [#permalink]
05 Sep 2013, 22:33

Expert's post

rrsnathan wrote:

Bunuel wrote:

Zarrolou wrote:

Maybe I misunderstood the question...

If I pick E, O, O, I get as result of D an ODD number. So an ODD number can be a multiple of the two odds (of course it can), but since the question asks CANNOT be a multiple of any, if one of the values is Even, then an odd number cannot a multiple of an even one.

Is this reasoning flawed?

The correct option must not be a multiple of ANY of the three variables, so if it could be a multiple of some of them, then it's not the right choice.

Hi,

If i pick the numbers as x=3 y=5 z=7

then to discard 1) x+y = 3+5=8 this is not divisible by any of 3 numbers

Same applies for other options too.

Pls tell me where i am wrong.

Thanks in Advance, Rrsnathan.

The question asks which of the options CANNOT be a multiple of any of x, y, and z in ANY case. Four options will not be multiples in SOME cases (not all) and only one of the options CANNOT be a multiple in ANY case.

Re: If x, y, and z are 3 different prime numbers, which of the [#permalink]
05 Sep 2013, 23:44

Quote:

The question asks which of the options CANNOT be a multiple of any of x, y, and z in ANY case. Four options will not be multiples in SOME cases (not all) and only one of the options CANNOT be a multiple in ANY case.

Does this make sense?

Yeah Bunuel i got it. It CAN be a multiple of any of X,Y and Z but we need the option that is CANNOT be a multiple in ANY case(ANY Prime number). Thanks a lot

Re: If x, y, and z are 3 different prime numbers, which of the [#permalink]
11 Nov 2013, 01:35

A prime number cannot be a multiple of another prime number. How can i apply this logic in this case? Odd number cannot be a multiple of an even number. Is it always true? 3 is multiple of 6...
_________________

Practice Makes a Man Perfect. Practice. Practice. Practice......Perfectly

Helpful Geometry formula sheet:best-geometry-93676.html I hope these will help to understand the basic concepts & strategies. Please Click ON KUDOS Button.

Re: If x, y, and z are 3 different prime numbers, which of the [#permalink]
11 Nov 2013, 01:42

Expert's post

monirjewel wrote:

A prime number cannot be a multiple of another prime number. How can i apply this logic in this case? Odd number cannot be a multiple of an even number. Is it always true? 3 is multiple of 6...

3 is a factor of 6, not a multiple.

An odd number is not divisible by 2, thus it cannot be a multiple of even number.
_________________

(e.g.) If x= 2, y = 3, z = 5 (E): If x = 2 & y = 5, → (x2 + y2) = (4 + 25) = 29 & z could be 29 → Wrong (D): If x = 2 & y = 3 & z = 7, → (xyz + 1) = (2 x 3 x 7 + 1) = (42 + 1) = 43 Since used up all variables (x, y, z) to plug-in, cannot be a multiple of x, y or z. → Answer (C): If x = 2 & y = 3, → (xy + 1) = (2 x 3 + 1) = (6 + 1) = 7, z can be 7. → Wrong (B): If y = 5 & z = 2, → (y – z) = (5 – 2) = 3. x could be 3 → Wrong (A): If x = 2 & y = 3, → (x + y) = (2 + 3) = 5. z could be 5 → Wrong

Hi, Can we say, because only (D) has all 3 variables (x, y, z), this is the answer?

(e.g.) If x= 2, y = 3, z = 5 (E): If x = 2 & y = 5, → (x2 + y2) = (4 + 25) = 29 & z could be 29 → Wrong (D): If x = 2 & y = 3 & z = 7, → (xyz + 1) = (2 x 3 x 7 + 1) = (42 + 1) = 43 Since used up all variables (x, y, z) to plug-in, cannot be a multiple of x, y or z. → Answer (C): If x = 2 & y = 3, → (xy + 1) = (2 x 3 + 1) = (6 + 1) = 7, z can be 7. → Wrong (B): If y = 5 & z = 2, → (y – z) = (5 – 2) = 3. x could be 3 → Wrong (A): If x = 2 & y = 3, → (x + y) = (2 + 3) = 5. z could be 5 → Wrong

Hi, Can we say, because only (D) has all 3 variables (x, y, z), this is the answer?

Merging similar topics. Please refer to the solutions above.
_________________