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If x, y, and z are 3 different prime numbers, which of the

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If x, y, and z are 3 different prime numbers, which of the [#permalink] New post 25 May 2013, 02:45
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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
[Reveal] Spoiler: OA

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Re: If x, y, and z are 3 different prime numbers, which of the [#permalink] New post 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.

Answer: D.

Similar question to practice: if-x-and-y-are-two-different-prime-numbers-which-of-the-134177.html

Hope it helps.
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Re: If x, y, and z are 3 different prime numbers, which of the [#permalink] New post 25 May 2013, 02:52
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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.
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Re: If x, y, and z are 3 different prime numbers, which of the [#permalink] New post 25 May 2013, 02:57
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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.

Hope it's clear.
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Re: If x, y, and z are 3 different prime numbers, which of the [#permalink] New post 25 May 2013, 03:11
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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.
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Re: If x, y, and z are 3 different prime numbers, which of the [#permalink] New post 25 May 2013, 02:53
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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


To discard other options, consider z=2, x=3, and y=5, in this case A, B, C, and E will be a multiple of z=2.

Hope it helps.
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Re: If x, y, and z are 3 different prime numbers, which of the [#permalink] New post 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 :-D
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Re: If x, y, and z are 3 different prime numbers, which of the [#permalink] New post 25 May 2013, 03:03
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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 :-D


Note that Zarrolou's way to discard D is not 100% correct: if-x-y-and-z-are-3-different-prime-numbers-which-of-the-153338.html#p1229032
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Re: If x, y, and z are 3 different prime numbers, which of the [#permalink] New post 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?
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Re: If x, y, and z are 3 different prime numbers, which of the [#permalink] New post 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

Same applies for other options too.

Pls tell me where i am wrong.

Thanks in Advance,
Rrsnathan.
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Re: If x, y, and z are 3 different prime numbers, which of the [#permalink] New post 05 Sep 2013, 22:33
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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.

Does this make sense?
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Re: If x, y, and z are 3 different prime numbers, which of the [#permalink] New post 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
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Re: If x, y, and z are 3 different prime numbers, which of the [#permalink] New post 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...
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Re: If x, y, and z are 3 different prime numbers, which of the [#permalink] New post 11 Nov 2013, 01:42
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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.
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Re: If x, y, and z are 3 different prime numbers, which of the [#permalink] New post 15 Jan 2014, 22:41
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


Let us say x = 2, y = 3 and z = 5

A. 5 (ELIMINATED)
B. -2 (ELIMINATED)
C. 7
D. 31
E. 4 + 9 = 13

Let us say x = 7, y = 11 and z = 2
C. 77 + 1 = 78 (ELIMINATED)
D. 155
E. 49 + 121 = 170 (ELIMINATED)

Answer is D
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Re: If x, y, and z are 3 different prime numbers, which of the [#permalink] New post 27 Mar 2015, 19:22
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Re: If x, y, and z are 3 different prime numbers, which of the [#permalink] New post 19 Jun 2015, 11:54
If we pick 3,5 & 7

xyz =105
xyz+1 =106 which is a multiple of 2 how do we accept D?
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Re: If x, y, and z are 3 different prime numbers, which of the [#permalink] New post 20 Jun 2015, 00:38
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Re: If x, y, and z are 3 different prime numbers, which of the [#permalink] New post 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.
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Re: If x, y, and z are 3 different prime numbers, which of the [#permalink] New post 26 Jun 2015, 06:34
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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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Re: If x, y, and z are 3 different prime numbers, which of the   [#permalink] 26 Jun 2015, 06:34

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