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# If x, y, z are positive integers, is xyz divisible by 6? 1) x, y, z ar

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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GMAT 1: 760 Q51 V42
GPA: 3.82
If x, y, z are positive integers, is xyz divisible by 6? 1) x, y, z ar  [#permalink]

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28 Jun 2017, 01:12
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Difficulty:

25% (medium)

Question Stats:

75% (00:55) correct 25% (01:33) wrong based on 51 sessions

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If x, y, z are positive integers, is xyz divisible by 6?

1) x, y, z are consecutive.
2) x+y+z is a multiple of 3.

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MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Director Joined: 04 Dec 2015 Posts: 743 Location: India Concentration: Technology, Strategy WE: Information Technology (Consulting) If x, y, z are positive integers, is xyz divisible by 6? 1) x, y, z ar [#permalink] ### Show Tags 28 Jun 2017, 01:30 MathRevolution wrote: If x, y, z are positive integers, is xyz divisible by 6? 1) x, y, z are consecutive. 2) x+y+z is a multiple of 3. 1) $$x, y, z$$ are consecutive. Rule : Product of $$3$$ consecutive numbers are always divisible by $$6$$. Lets check this rule. Let $$x, y , z = 1,2,3$$ $$xyz = 1*2*3 = 6$$. $$6$$ is divisible by $$6$$. Therefore $$xyz$$ is divisible by $$6$$. Hence I is Sufficient. 2) $$x+y+z$$ is a multiple of $$3$$. Multiples of $$3$$ are $$= 3,6,12,15,18$$ etc... From above $$3, 15$$ are not divisible by $$6$$. Hence Not all multiples of $$3$$ are divisible by $$6$$. Hence II is Not Sufficient. Answer (A)... Senior PS Moderator Joined: 26 Feb 2016 Posts: 3335 Location: India GPA: 3.12 Re: If x, y, z are positive integers, is xyz divisible by 6? 1) x, y, z ar [#permalink] ### Show Tags 28 Jun 2017, 07:44 1) x, y, z are consecutive Since we are talking about positive integers, lets assume the lowest set of consecutive positive integers(1,2 and 3) possible The product of these numbers is divisible by 6. Hence, the sum of all three positive consecutive numbers is always divisible by 6 (Sufficient) 2) x+y+z is a multiple of 3. Consider x+y+z = 6( a multiple of 3) which is divisible by 6 Similarly x+y+z = 21(a multiple of 3) which is not divisible by 6. (Insufficient)(Option A) _________________ You've got what it takes, but it will take everything you've got Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8017 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If x, y, z are positive integers, is xyz divisible by 6? 1) x, y, z ar [#permalink] ### Show Tags 30 Jun 2017, 01:00 ==> The product of 3 consecutive integers is always a multiple of 6. For con 1), it is yes and sufficient, and for con 2), (x,y,z)=(1,2,3) yes but (x,y,z)=(2,2,2) no, hence it is not sufficient. Therefore, the answer is A. Answer: A _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Re: If x, y, z are positive integers, is xyz divisible by 6? 1) x, y, z ar   [#permalink] 30 Jun 2017, 01:00
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