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# If x and y are consecutive integers and x < y, then y^2 - x^2 must be

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Senior RC Moderator
Joined: 02 Nov 2016
Posts: 4106
GPA: 3.39
If x and y are consecutive integers and x < y, then y^2 - x^2 must be  [#permalink]

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04 Mar 2019, 14:13
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25% (medium)

Question Stats:

80% (01:16) correct 20% (01:21) wrong based on 40 sessions

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If x and y are consecutive integers and $$x < y$$, then $$y^2 - x^2$$ must be

A. a prime number.
B. an odd number.
C. an even number.
D. the square of an integer.
E. $$(y-x)^2$$.

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e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3074
Re: If x and y are consecutive integers and x < y, then y^2 - x^2 must be  [#permalink]

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04 Mar 2019, 22:55

Solution

Given:
• The numbers x and y are consecutive integers
• Also x < y

To find:
• Which of the given options is always true, regarding the value of $$y^2 – x^2$$

Approach and Working:
As x < y and x, y are consecutive integers, we can say that
• If x = n, then y = n + 1
• Therefore, $$y^2 – x^2 = (n + 1)^2 – n^2 = n^2 + 2n + 1 – n^2 = 2n + 1$$, which is always an odd integer.

Alternatively, we can assume values also to determine the answer.
• If x = 2 and y = 3, then $$y^2 – x^2 = 9 – 4 = 5$$
o Therefore, $$y^2 – x^2$$ can be either a prime or an odd number, we can rule out the other possibilities.

• If x = 7 and y = 8, then $$y^2 – x^2 = 64 – 49 = 15$$, which is odd but not a prime number.

Hence, the correct answer is option B.

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Joined: 31 Oct 2013
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Concentration: Accounting, Finance
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Re: If x and y are consecutive integers and x < y, then y^2 - x^2 must be  [#permalink]

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04 Mar 2019, 14:23
If x and y are consecutive integers and $$x < y$$, then $$y^2 - x^2$$ must be

A. a prime number.
B. an odd number.
C. an even number.
D. the square of an integer.
E. $$(y-x)^2$$.

x and y are consecutive integers. Thus, one of them must be odd and one of them must be even.

even - odd = odd.

Result must be an odd integer.

Manager
Joined: 21 Feb 2019
Posts: 125
Location: Italy
Re: If x and y are consecutive integers and x < y, then y^2 - x^2 must be  [#permalink]

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04 Mar 2019, 17:05
1
$$x < y$$ and $$x$$ and $$y$$ are consecutive.

Hence we can rewrite $$y$$ as: $$y = x +1$$

Thus: $$y^2 - x^2 = (y - x) (y + x) = (x + 1 - x) (x + 1 + x) = 2x + 1$$,

which is an odd number.

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Re: If x and y are consecutive integers and x < y, then y^2 - x^2 must be  [#permalink]

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06 Mar 2019, 02:19
difference of square of two consective no will always result in an odd no
IMO B

If x and y are consecutive integers and $$x < y$$, then $$y^2 - x^2$$ must be

A. a prime number.
B. an odd number.
C. an even number.
D. the square of an integer.
E. $$(y-x)^2$$.
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8069
Location: United States (CA)
Re: If x and y are consecutive integers and x < y, then y^2 - x^2 must be  [#permalink]

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08 Mar 2019, 08:00
If x and y are consecutive integers and $$x < y$$, then $$y^2 - x^2$$ must be

A. a prime number.
B. an odd number.
C. an even number.
D. the square of an integer.
E. $$(y-x)^2$$.

Since x and y are consecutive integers and x < y, then y = x + 1. Thus, y^2 - x^2 = (x + 1)^2 - x^2 = x^2 + 2x + 1 - x^2 = 2x + 1, which is always an odd number, regardless of what integer x is.

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Re: If x and y are consecutive integers and x < y, then y^2 - x^2 must be   [#permalink] 08 Mar 2019, 08:00
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