If a, b, and c are consecutive integers such that a < b : GMAT Data Sufficiency (DS)
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If a, b, and c are consecutive integers such that a < b

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If a, b, and c are consecutive integers such that a < b [#permalink]

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24 Apr 2010, 07:19
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If a, b, and c are consecutive integers such that a < b < c, then which of the following must be true?
I. c – a = 2
II. b = a + 1 = c – 1
III. abc is an even integer

A I only
B II only
C III only
D I and III only
E I, II, and III

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Re: Seems easy, but is it really??? [#permalink]

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24 Apr 2010, 07:29
hardnstrong wrote:
If a, b, and c are consecutive integers such that a < b < c, then which of the following must be true?
I. c – a = 2 -> a, b, c are consecutive integers and c > a. So in any case this is true.
II. b = a + 1 = c – 1 -> same as above. a, b, c are consecutive Integers and b is between a and c.
III. abc is an even integer -> 0 is an even integer. Also, multiplication of odd and even is even. So in any case this would be true.

A I only
B II only
C III only
D I and III only
E I, II, and III

IMO E. Please feel free to correct if my explainations are wrong.
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Re: Seems easy, but is it really??? [#permalink]

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24 Apr 2010, 07:31
hardnstrong wrote:
If a, b, and c are consecutive integers such that a < b < c, then which of the following must be true?
I. c – a = 2
II. b = a + 1 = c – 1
III. abc is an even integer

A I only
B II only
C III only
D I and III only
E I, II, and III

given
b= a+1 & c = a+2 & c=b+1
I is true
II is true
III: atleast one of a,b,c is even and odd * even is always even hence it is also true.
so its E.
Whats the OA
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Re: Seems easy, but is it really??? [#permalink]

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24 Apr 2010, 07:35
Yes E only.

Take a = n-1 . b = n and c = n+1

substitute in all the equations, you will see its satisfying all of them.
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Re: Seems easy, but is it really??? [#permalink]

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24 Apr 2010, 21:59
thanks guys ...........i got confused with the third option
Got it now
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Re: Seems easy, but is it really??? [#permalink]

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15 Jun 2010, 15:51
what if one of is zero, then it can not satisfy the third option.
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Re: Seems easy, but is it really??? [#permalink]

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15 Jun 2010, 16:08
what if one of is zero, then it can not satisfy the third option.

Zero is considered to be an even integer.
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Re: Seems easy, but is it really??? [#permalink]

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15 Jul 2010, 04:56
nsp007 wrote:
what if one of is zero, then it can not satisfy the third option.

Zero is considered to be an even integer.

really? I didn't know that... thanks for the info.
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Re: If a, b, and c are consecutive integers such that a < b [#permalink]

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28 May 2016, 07:28
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Re: If a, b, and c are consecutive integers such that a < b [#permalink]

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28 May 2016, 07:40
hardnstrong wrote:
If a, b, and c are consecutive integers such that a < b < c, then which of the following must be true?
I. c – a = 2
II. b = a + 1 = c – 1
III. abc is an even integer

A I only
B II only
C III only
D I and III only
E I, II, and III

Any 3 consecutive integers are of the form n, n+1, n+2. (n+2)-n will always be (2) so statement I is always right. This facts means B and C are out of contention.

Similarly, b= n+1 and a=n then b=a+1; also b=c-1. This statement is always valid. So A is out of contention. Also D is out of contention.

By elimination E remains.

(By the way III statement is always right. Whenever 3 consecutive integers are there there will be at least one even integer. So the product will always be even.)
Re: If a, b, and c are consecutive integers such that a < b   [#permalink] 28 May 2016, 07:40
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