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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

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]

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25 May 2013, 03:52

1

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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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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.

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]

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25 May 2013, 04:01

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
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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]

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25 May 2013, 04:07

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]

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05 Sep 2013, 23:10

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

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]

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06 Sep 2013, 00:44

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]

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11 Nov 2013, 02:35

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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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]

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15 Jan 2014, 23:41

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]

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27 Mar 2015, 20:22

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

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26 Jun 2015, 07:24

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.

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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