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

2

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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

1

This post received KUDOS

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, 03:11

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]
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]
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

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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. _________________