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Re: If x, y, and z are 3 different prime numbers, which of the [#permalink]
25 May 2013, 02:51
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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) \(x^2 + y^2\) \(3^2+5^2=36\) 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) \(x^2 + y^2\) \(3^2+5^2=36\) 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) \(x^2 + y^2\) \(3^2+5^2=36\) 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. _________________
Re: If x, y, and z are 3 different prime numbers, which of the [#permalink]
27 Mar 2015, 19:22
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Re: If x, y, and z are 3 different prime numbers, which of the [#permalink]
26 Jun 2015, 06:24
The question was: If we pick any three different prime numbers, which expression gives a result which cannot be a multiple of the initial three prime numbers.
So, if we pick 3,5 and 7 and we try answer A (x+y), the result (8) is not a multiple of any of the three primes chosen.
Re: If x, y, and z are 3 different prime numbers, which of the [#permalink]
26 Jun 2015, 06:34
Expert's post
alefal wrote:
The question was: If we pick any three different prime numbers, which expression gives a result which cannot be a multiple of the initial three prime numbers.
So, if we pick 3,5 and 7 and we try answer A (x+y), the result (8) is not a multiple of any of the three primes chosen.
Cannot there means in any case. Meaning that for ANY 3 distinct primes (not only for some particular triplet), correct option must not be a multiple of ANY of x, y, and z.
If 3 primes are 2, 3, and 5, then 2+3=5 IS a multiple of 5, so x+y CAN be a multiple of one of the primes. _________________
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