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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## If x, y and z are positive integers, then... tagged by: swerve ##### This topic has 1 expert reply and 0 member replies ### Top Member ## If x, y and z are positive integers, then... If x, y and z are positive integers, then $$x^2y^3z^4$$ must be divisible by which of the following? $$I.\ \ x^2+y^3+z^4$$ $$II.\ \ xy+xz$$ $$III.\ \ xyz+z$$ A. None B. I only C. I and II only D. I and III only E. II and III only The OA is A. I don't understand this PS question. Please, can any expert help me to solve it? Thanks. ### GMAT/MBA Expert Legendary Member Joined 14 Jan 2015 Posted: 2667 messages Followed by: 122 members Upvotes: 1153 GMAT Score: 770 swerve wrote: If x, y and z are positive integers, then $$x^2y^3z^4$$ must be divisible by which of the following? $$I.\ \ x^2+y^3+z^4$$ $$II.\ \ xy+xz$$ $$III.\ \ xyz+z$$ A. None B. I only C. I and II only D. I and III only E. II and III only The OA is A. I don't understand this PS question. Please, can any expert help me to solve it? Thanks. Take the easiest possible case: x = 1, y = 1, and z = 1. $$x^2y^3z^4$$= 1. $$I.\ \ x^2+y^3+z^4$$ --> 1 + 1 + 1 = 3. 1 is not divisible by 3. I is out. $$II.\ \ xy+xz$$ --> 1 + 1 = 2. 1 is not divisible by 2. II is out. $$III.\ \ xyz+z$$ --> 1 + 1 = 2. 1 is not divisible by 2. III is out. So none of these statements needs to be true. The answer is A _________________ Veritas Prep | GMAT Instructor Veritas Prep Reviews Save$100 off any live Veritas Prep GMAT Course

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