text: It is currently 21 Sep 2017, 19:57
  
  ### GMAT Club Daily Prep
  
  #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
  
  Customized
  for You
  
  we will pick new questions that match your level based on your Timer History
  
  Track
  Your Progress
  
  every week, we’ll send you an estimated GMAT score based on your performance
  
  Practice
  Pays
  
  we will pick new questions that match your level based on your Timer History
  
  # Events & Promotions
  
  ###### Events & Promotions in June
  Open Detailed Calendar
  
  # m03 Q 22
  
   new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
  Author Message
  Intern
  Joined: 12 Feb 2010
  Posts: 26
  
  Kudos [?]: 1 [0], given: 5
  
  m03 Q 22 [#permalink]
  
  ### Show Tags
  
  16 Mar 2010, 13:52
  Is the product of three integers P, Q, and R even?
  
  1. (p-1) (r+1) is odd
  2. Square of (q-r) is odd
  
  a. Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
  b. Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
  c. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
  d. EACH statement ALONE is sufficient
  e. Statements (1) and (2) TOGETHER are NOT sufficient
  
  [Reveal] Spoiler: OA
  D
  
  Let me know your reasoning too.
  
  Kudos [?]: 1 [0], given: 5
  
  Intern
  Joined: 14 Mar 2010
  Posts: 49
  
  Kudos [?]: 4 [0], given: 0
  
  Re: m03 Q 22 [#permalink]
  
  ### Show Tags
  
  16 Mar 2010, 21:42
  PQR is even if any of P, Q, or R is even and none of them are 0.
  
  A: (p-1)*(r+1) = odd either if p and r are both even (since only even+-odd = odd and only odd*odd = odd) or if p and/or r is 0. NOT SUFFICIENT
  
  B: (q-r)^2 = odd only if either q or r is even and the other is odd or if one of them is 0 and the other is odd. NOT SUFFICIENT
  
  Together: r is restricted to 0 or even. Can't determine PQR.
  
  Not sure why answer is C
  
  Kudos [?]: 4 [0], given: 0
  
  Intern
  Joined: 12 Feb 2010
  Posts: 26
  
  Kudos [?]: 1 [0], given: 5
  
  Re: m03 Q 22 [#permalink]
  
  ### Show Tags
  
  17 Mar 2010, 08:25
  I chose e too with 0 value as deciding factor for my choice but it turns out the 0 is considered an even number. I just googled it and found it out. Hope the understanding is correct.
  
  Kudos [?]: 1 [0], given: 5
  
  Math Expert
  Joined: 02 Sep 2009
  Posts: 41666
  
  Kudos [?]: 124373 [0], given: 12077
  
  Re: m03 Q 22 [#permalink]
  
  ### Show Tags
  
  17 Mar 2010, 09:18
  Ironduke wrote:
  Is the product of three integers P, Q, and R even?
  
  1. (p-1) (r+1) is odd
  2. Square of (q-r) is odd
  
  [Reveal] Spoiler:
  OA c
  
  Let me know your reasoning too.
  
  First of all zero is an even number.
  Second answer can not be C, it should be D.
  
  The product of three integers P, Q, and R is even if either of them is even.
  
  (1) $$(p-1)(r+1)=odd$$. Product of two integers is odd only if both are odd. Hence $$p-1=odd$$ and $$r+1=odd$$, which means $$p$$ and $$r$$ are even. Sufficient.
  
  (2) $$(q-r)^2=odd$$ --> $$q-r=odd$$. Difference of two integers is odd only if one of them is odd and another is even. Hence $$p$$ or $$r$$ is even. Sufficient.
  
  Answer: D.
  _________________
  
  Kudos [?]: 124373 [0], given: 12077
  
  Intern
  Joined: 01 Mar 2010
  Posts: 12
  
  Kudos [?]: 2 [0], given: 0
  
  Re: m03 Q 22 [#permalink]
  
  ### Show Tags
  
  17 Mar 2010, 10:43
  Excellent. Good reasoning.
  
  Bunuel wrote:
  Ironduke wrote:
  Is the product of three integers P, Q, and R even?
  
  1. (p-1) (r+1) is odd
  2. Square of (q-r) is odd
  
  [Reveal] Spoiler:
  OA c
  
  Let me know your reasoning too.
  
  First of all zero is an even number.
  Second answer can not be C, it should be D.
  
  The product of three integers P, Q, and R is even if either of them is even.
  
  (1) $$(p-1)(r+1)=odd$$. Product of two integers is odd only if both are odd. Hence $$p-1=odd$$ and $$r+1=odd$$, which means $$p$$ and $$r$$ are even. Sufficient.
  
  (2) $$(q-r)^2=odd$$ --> $$q-r=odd$$. Difference of two integers is odd only if one of them is odd and another is even. Hence $$p$$ or $$r$$ is even. Sufficient.
  
  Answer: D.
  
  _________________
  
  A beginner. I will post my progress here.
  
  Kudos [?]: 2 [0], given: 0
  
  Intern
  Joined: 14 Mar 2010
  Posts: 49
  
  Kudos [?]: 4 [0], given: 0
  
  Re: m03 Q 22 [#permalink]
  
  ### Show Tags
  
  17 Mar 2010, 13:21
  Ironduke wrote:
  I chose e too with 0 value as deciding factor for my choice but it turns out the 0 is considered an even number. I just googled it and found it out. Hope the understanding is correct.
  
  Then it's D. Dang math rules
  
  Kudos [?]: 4 [0], given: 0
  
  CIO
  Joined: 02 Oct 2007
  Posts: 1218
  
  Kudos [?]: 970 [0], given: 334
  
  Re: m03 Q 22 [#permalink]
  
  ### Show Tags
  
  18 Mar 2010, 03:32
  The OA is D, not C as was indicated in the first post. I've edited the first post to include a correct OA.
  _________________
  
  Welcome to GMAT Club!
  
  Want to solve GMAT questions on the go? GMAT Club iPhone app will help.
  Please read this before posting in GMAT Club Tests forum
  Result correlation between real GMAT and GMAT Club Tests
  Are GMAT Club Test sets ordered in any way?
  
  Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas.
  
  GMAT Club Premium Membership - big benefits and savings
  
  Kudos [?]: 970 [0], given: 334
  
  Manager
  Joined: 13 Dec 2009
  Posts: 249
  
  Kudos [?]: 233 [0], given: 13
  
  Re: m03 Q 22 [#permalink]
  
  ### Show Tags
  
  19 Mar 2010, 07:36
  Ironduke wrote:
  Is the product of three integers P, Q, and R even?
  
  1. (p-1) (r+1) is odd
  2. Square of (q-r) is odd
  
  a. Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
  b. Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
  c. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
  d. EACH statement ALONE is sufficient
  e. Statements (1) and (2) TOGETHER are NOT sufficient
  
  [Reveal] Spoiler: OA
  D
  
  Let me know your reasoning too.
  
  product P Q R can be even only if one of them is even.
  1. (p-1)(r+1)
  both p-1 and r+1 are odd
  means p and r are even => pqr is even suff.
  
  2. (q-r)^2 is odd => q-r is odd.
  there can be following cases for q-r to be odd
  q is odd and r is even or q is even and r is odd
  in any case one of either q or r are even so pqr will be even
  hence d
  _________________
  
  My debrief: done-and-dusted-730-q49-v40
  
  Kudos [?]: 233 [0], given: 13
  
  Manager
  Joined: 02 May 2012
  Posts: 108
  
  Kudos [?]: 71 [0], given: 34
  
  Location: United Kingdom
  WE: Account Management (Other)
  Re: m03 Q 22 [#permalink]
  
  ### Show Tags
  
  18 Sep 2012, 08:16
  The key here is any multiplication by an even number results in an even.
  
  So therefore we're on the hunt for even numbers:
  
  1) (p-1)(r+1) is odd. Only odd x odd gives = odd, so both p-1 and r+1 are odd and hence both are even. An even number in qpr will given an even results
  
  Sufficient
  
  2)Similar to above, odd x odd = odd. So q-r must be odd, and therefore one of q or r must be even. An even number in qpr makes it eve
  
  Sufficient
  _________________
  
  In the study cave!
  
  Kudos [?]: 71 [0], given: 34
  
  Re: m03 Q 22   [#permalink] 18 Sep 2012, 08:16
  Similar topics Replies Last post
  Similar
  Topics:
  14 M03 Q27 31 11 Dec 2013, 05:57
  8 M03 Q14 22 08 May 2014, 03:16
  20 M03 Q33 15 09 Oct 2012, 08:16
  3 M03 Q 36 20 18 Sep 2012, 08:54
  15 m03#22 12 04 Jul 2014, 06:53
  Display posts from previous: Sort by
  
  # m03 Q 22
  
   new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
  
  Moderator: Bunuel
  
   Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.
metadata: {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}