text: If k and t are integer and k^2 - t^2 is an odd integer, : Quant Question Archive [LOCKED]
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  # If k and t are integer and k^2 - t^2 is an odd integer,
  
  Author Message
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  If k and t are integer and k^2 - t^2 is an odd integer, [#permalink]
  
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  21 Apr 2008, 02:58
  This topic is locked. If you want to discuss this question please re-post it in the respective forum.
  
  If k and t are integer and k^2 - t^2 is an odd integer, which of the following must be an even integer ?
  
  i)k+t+2
  ii)k^2=2kt+t^2
  iii)k^2+t^2
  
  1)None
  2)i only
  3)ii only
  4)iii only
  5)i,ii and iii
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  21 Apr 2008, 05:17
  patrannn wrote:
  If k and t are integer and k^2 - t^2 is an odd integer, which of the following must be an even integer ?
  
  i)k+t+2
  ii)k^2=2kt+t^2
  iii)k^2+t^2
  
  1)None
  2)i only
  3)ii only
  4)iii only
  5)i,ii and iii
  
  Given : (k+t)(k-t)= odd, i.e k+t=odd k-t=odd i.e either k is odd and t is even OR k is even and t is odd.
  
  (i) k+t is odd => (k+t) + 2 => odd
  
  (ii) (k^2 - t^2) which is odd = even (as anything multiplied by 2)...??? WTH
  
  (iii) even square + odd square (or vice versa) = even
  
  So Answer is 4 (iii) only
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  21 Apr 2008, 06:32
  alpha_plus_gamma wrote:
  patrannn wrote:
  If k and t are integer and k^2 - t^2 is an odd integer, which of the following must be an even integer ?
  
  i)k+t+2
  ii)k^2=2kt+t^2
  iii)k^2+t^2
  
  1)None
  2)i only
  3)ii only
  4)iii only
  5)i,ii and iii
  
  Given : (k+t)(k-t)= odd, i.e k+t=odd k-t=odd i.e either k is odd and t is even OR k is even and t is odd.
  
  (i) k+t is odd => (k+t) + 2 => odd
  
  (ii) (k^2 - t^2) which is odd = even (as anything multiplied by 2)...??? WTH
  
  (iii) even square + odd square (or vice versa) = even
  
  So Answer is 4 (iii) only
  
  Even + odd is odd.
  
  I think there is some problem with the option ii as odd can't be equal to even.
  i think 'none' is the answer.
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  21 Apr 2008, 14:45
  None. Plug in k = 3, t = 2. 3^2 - 2^2 = 5 = odd. None of the options in the list will come up as even.
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  21 Apr 2008, 19:32
  vshaunak@gmail.com wrote:
  alpha_plus_gamma wrote:
  patrannn wrote:
  If k and t are integer and k^2 - t^2 is an odd integer, which of the following must be an even integer ?
  
  i)k+t+2
  ii)k^2=2kt+t^2
  iii)k^2+t^2
  
  1)None
  2)i only
  3)ii only
  4)iii only
  5)i,ii and iii
  
  Given : (k+t)(k-t)= odd, i.e k+t=odd k-t=odd i.e either k is odd and t is even OR k is even and t is odd.
  
  (i) k+t is odd => (k+t) + 2 => odd
  
  (ii) (k^2 - t^2) which is odd = even (as anything multiplied by 2)...??? WTH
  
  (iii) even square + odd square (or vice versa) = even
  
  So Answer is 4 (iii) only
  
  Even + odd is odd.
  
  I think there is some problem with the option ii as odd can't be equal to even.
  i think 'none' is the answer.
  
  Thanks for pointing that out..I don't know when will I avoid such mistakes [:(]
  VP
  Joined: 10 Jun 2007
  Posts: 1459
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  ### Show Tags
  
  21 Apr 2008, 20:48
  patrannn wrote:
  If k and t are integer and k^2 - t^2 is an odd integer, which of the following must be an even integer ?
  
  i)k+t+2
  ii)k^2=2kt+t^2
  iii)k^2+t^2
  
  1)None
  2)i only
  3)ii only
  4)iii only
  5)i,ii and iii
  
  We know that k^2 - t^2 = odd = (k+t)*(k-t)
  We also know that only
  odd*odd = odd
  
  This means:
  (k+t) = odd & (k-t) = odd
  Now, Eliminate "i" because odd + 2 will always be odd.
  
  In "ii", I think this is what you mean:
  k^2 - 2kt + t^2 = (k-t)*(k-t) = odd*odd = odd
  Eliminate "ii"
  
  In "iii", we need to work out the solution further...
  Knowing that
  (k+t) = odd
  (k-t) = odd
  There are two cases here:
  odd + even = odd
  even + odd = odd
  This means k and t will never be odd&odd or even&even pair. They will always be either even&odd or odd&even. In other word, they will always be different.
  
  This means
  k^2 + t^2 = even^2 + odd^2 = odd
  Eliminate "iii"
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  21 Apr 2008, 23:29
  1- None
  
  Picked by plugging numbers 3,2
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  Re: number properties   [#permalink] 21 Apr 2008, 23:29
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explanation: The provided text extract contains a math question and no answer. The question is about the properties of integers and their squares, specifically when the difference of two squares is an odd integer. The task is to determine which of the given expressions must be an even integer.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: If k and t are integer and k^2 - t^2 is an odd integer, : Quant Question Archive [LOCKED]
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  # If k and t are integer and k^2 - t^2 is an odd integer,
  
  Author Message
  Intern
  Joined: 30 Dec 2003
  Posts: 33
  Location: Danbury
  Followers: 0
  
  Kudos [?]: 0 [0], given: 0
  
  If k and t are integer and k^2 - t^2 is an odd integer, [#permalink]
  
  ### Show Tags
  
  21 Apr 2008, 02:58
  This topic is locked. If you want to discuss this question please re-post it in the respective forum.
  
  If k and t are integer and k^2 - t^2 is an odd integer, which of the following must be an even integer ?
  
  i)k+t+2
  ii)k^2=2kt+t^2
  iii)k^2+t^2
  
  1)None
  2)i only
  3)ii only
  4)iii only
  5)i,ii and iii
  Director
  Joined: 14 Aug 2007
  Posts: 733
  Followers: 10
  
  Kudos [?]: 182 [0], given: 0
  
  ### Show Tags
  
  21 Apr 2008, 05:17
  patrannn wrote:
  If k and t are integer and k^2 - t^2 is an odd integer, which of the following must be an even integer ?
  
  i)k+t+2
  ii)k^2=2kt+t^2
  iii)k^2+t^2
  
  1)None
  2)i only
  3)ii only
  4)iii only
  5)i,ii and iii
  
  Given : (k+t)(k-t)= odd, i.e k+t=odd k-t=odd i.e either k is odd and t is even OR k is even and t is odd.
  
  (i) k+t is odd => (k+t) + 2 => odd
  
  (ii) (k^2 - t^2) which is odd = even (as anything multiplied by 2)...??? WTH
  
  (iii) even square + odd square (or vice versa) = even
  
  So Answer is 4 (iii) only
  Director
  Joined: 14 Jan 2007
  Posts: 777
  Followers: 2
  
  Kudos [?]: 136 [0], given: 0
  
  ### Show Tags
  
  21 Apr 2008, 06:32
  alpha_plus_gamma wrote:
  patrannn wrote:
  If k and t are integer and k^2 - t^2 is an odd integer, which of the following must be an even integer ?
  
  i)k+t+2
  ii)k^2=2kt+t^2
  iii)k^2+t^2
  
  1)None
  2)i only
  3)ii only
  4)iii only
  5)i,ii and iii
  
  Given : (k+t)(k-t)= odd, i.e k+t=odd k-t=odd i.e either k is odd and t is even OR k is even and t is odd.
  
  (i) k+t is odd => (k+t) + 2 => odd
  
  (ii) (k^2 - t^2) which is odd = even (as anything multiplied by 2)...??? WTH
  
  (iii) even square + odd square (or vice versa) = even
  
  So Answer is 4 (iii) only
  
  Even + odd is odd.
  
  I think there is some problem with the option ii as odd can't be equal to even.
  i think 'none' is the answer.
  SVP
  Joined: 11 Mar 2008
  Posts: 1634
  Location: Southern California
  Schools: Chicago (dinged), Tuck (November), Columbia (RD)
  Followers: 9
  
  Kudos [?]: 201 [0], given: 0
  
  ### Show Tags
  
  21 Apr 2008, 14:45
  None. Plug in k = 3, t = 2. 3^2 - 2^2 = 5 = odd. None of the options in the list will come up as even.
  _________________
  
  Check out the new Career Forum
  http://gmatclub.com/forum/133
  
  Director
  Joined: 14 Aug 2007
  Posts: 733
  Followers: 10
  
  Kudos [?]: 182 [0], given: 0
  
  ### Show Tags
  
  21 Apr 2008, 19:32
  vshaunak@gmail.com wrote:
  alpha_plus_gamma wrote:
  patrannn wrote:
  If k and t are integer and k^2 - t^2 is an odd integer, which of the following must be an even integer ?
  
  i)k+t+2
  ii)k^2=2kt+t^2
  iii)k^2+t^2
  
  1)None
  2)i only
  3)ii only
  4)iii only
  5)i,ii and iii
  
  Given : (k+t)(k-t)= odd, i.e k+t=odd k-t=odd i.e either k is odd and t is even OR k is even and t is odd.
  
  (i) k+t is odd => (k+t) + 2 => odd
  
  (ii) (k^2 - t^2) which is odd = even (as anything multiplied by 2)...??? WTH
  
  (iii) even square + odd square (or vice versa) = even
  
  So Answer is 4 (iii) only
  
  Even + odd is odd.
  
  I think there is some problem with the option ii as odd can't be equal to even.
  i think 'none' is the answer.
  
  Thanks for pointing that out..I don't know when will I avoid such mistakes [:(]
  VP
  Joined: 10 Jun 2007
  Posts: 1459
  Followers: 7
  
  Kudos [?]: 255 [0], given: 0
  
  ### Show Tags
  
  21 Apr 2008, 20:48
  patrannn wrote:
  If k and t are integer and k^2 - t^2 is an odd integer, which of the following must be an even integer ?
  
  i)k+t+2
  ii)k^2=2kt+t^2
  iii)k^2+t^2
  
  1)None
  2)i only
  3)ii only
  4)iii only
  5)i,ii and iii
  
  We know that k^2 - t^2 = odd = (k+t)*(k-t)
  We also know that only
  odd*odd = odd
  
  This means:
  (k+t) = odd & (k-t) = odd
  Now, Eliminate "i" because odd + 2 will always be odd.
  
  In "ii", I think this is what you mean:
  k^2 - 2kt + t^2 = (k-t)*(k-t) = odd*odd = odd
  Eliminate "ii"
  
  In "iii", we need to work out the solution further...
  Knowing that
  (k+t) = odd
  (k-t) = odd
  There are two cases here:
  odd + even = odd
  even + odd = odd
  This means k and t will never be odd&odd or even&even pair. They will always be either even&odd or odd&even. In other word, they will always be different.
  
  This means
  k^2 + t^2 = even^2 + odd^2 = odd
  Eliminate "iii"
  Director
  Joined: 05 Jan 2008
  Posts: 701
  Followers: 5
  
  Kudos [?]: 416 [0], given: 0
  
  ### Show Tags
  
  21 Apr 2008, 23:29
  1- None
  
  Picked by plugging numbers 3,2
  _________________
  
  Persistence+Patience+Persistence+Patience=G...O...A...L
  
  Re: number properties   [#permalink] 21 Apr 2008, 23:29
  Display posts from previous: Sort by
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The provided text extract contains a math question and no answer. The question is about the properties of integers and their squares, specifically when the difference of two squares is an odd integer. The task is to determine which of the given expressions must be an even integer.
  
  Label: question