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If x, y, and z are 3 different prime numbers, which of the [#permalink]
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25 May 2013, 03:45
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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
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Re: If x, y, and z are 3 different prime numbers, which of the [#permalink]
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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 coprime, 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: ifxandyaretwodifferentprimenumberswhichofthe134177.htmlHope it helps.
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Re: If x, y, and z are 3 different prime numbers, which of the [#permalink]
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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) \(yz\), \(52=2\) 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]
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25 May 2013, 03:53
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]
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25 May 2013, 03:57
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) \(yz\), \(52=2\) 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]
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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) \(yz\), \(52=2\) 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 coprime, 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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Re: If x, y, and z are 3 different prime numbers, which of the [#permalink]
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25 May 2013, 04:03
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 coprime, 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 Note that Zarrolou's way to discard D is not 100% correct: ifxyandzare3differentprimenumberswhichofthe153338.html#p1229032
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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, 04:07
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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25 May 2013, 04:11
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]
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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 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]
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05 Sep 2013, 23:33
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]
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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



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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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Re: If x, y, and z are 3 different prime numbers, which of the [#permalink]
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Re: If x, y, and z are 3 different prime numbers, which of the [#permalink]
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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 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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19 Jun 2015, 12:54
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]
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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.



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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, 07:34
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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