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# If a, b, and c are consecutive integers and a<b<c,

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VP
Joined: 18 May 2008
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If a, b, and c are consecutive integers and a<b<c, [#permalink]

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27 Jun 2008, 00:31
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If a, b, and c are consecutive integers and a<b<c, which of the following CANNOT be the value of c^2-a^2-b^2?
(A) -12
(B) 0
(C) 3
(D) 4
(E) 5

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Intern
Joined: 25 Jun 2008
Posts: 13

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27 Jun 2008, 01:19
A is the ciorrect Ans?
Can someone tell the correct approach

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SVP
Joined: 04 May 2006
Posts: 1886

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Schools: CBS, Kellogg

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27 Jun 2008, 01:21
ritula wrote:
If a, b, and c are consecutive integers and a<b<c, which of the following CANNOT be the value of c^2-a^2-b^2?
(A) -12
(B) 0
(C) 3
(D) 4
(E) 5

E for me!

a=a
b=a+1
c=a+2

c^2-a^2-b^2 = -a^2+2a+3
a=0, c^2-a^2-b^2 =3
a=1, c^2-a^2-b^2 =4
...
_________________

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Current Student
Joined: 12 Jun 2008
Posts: 287

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Schools: INSEAD Class of July '10

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27 Jun 2008, 02:38

We can write a=b-1 and c=b+1

Then c^2-a^2-b^2 = (b+1)^2 - (b-1)^2 - b^2 = 4b - b^2 = (4-b)*b

(4-b)*b = -12 for b=-2
(4-b)*b = 0 for b=0 or b=4
(4-b)*b = 3 for b=1
(4-b)*b = 4 for b=2

but we cannot write (4-b)*b = 5 with b integer

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Intern
Joined: 26 May 2008
Posts: 40

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27 Jun 2008, 06:21
a<b<c

a=b-1
c=b+1

c^2-a^2-b^2 = (b+1)^2 -(b-1)^2 - b^2
= b(4-b)
Solve this.

Ans - E

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CEO
Joined: 29 Mar 2007
Posts: 2554

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27 Jun 2008, 06:31
ritula wrote:
If a, b, and c are consecutive integers and a<b<c, which of the following CANNOT be the value of c^2-a^2-b^2?
(A) -12
(B) 0
(C) 3
(D) 4
(E) 5

E.

(b-1)<b<b+1

(b+1)^2-(b-1)^2-b^2

--> b(4-b) = (A,B,C,D,E)

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Current Student
Joined: 28 Dec 2004
Posts: 3351

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Location: New York City
Schools: Wharton'11 HBS'12

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27 Jun 2008, 09:16
ritula wrote:
If a, b, and c are consecutive integers and a<b<c, which of the following CANNOT be the value of c^2-a^2-b^2?
(A) -12
(B) 0
(C) 3
(D) 4
(E) 5

a<b<c
b-1<b<b+1

(b+1)^2 - (b-1)^2 - b^2

b(4-b)=-12, doesnt =0 doesnt, =3 doesnt work, =4 doesnt work..=5 WORKS..

i still had to try all the numbers..i wonder if there is a more short cut to solve this..took about 1.5 min to do this

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VP
Joined: 18 May 2008
Posts: 1259

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29 Jun 2008, 22:32
OA is E indeed. Thanks to all!

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Re: consecutive integers   [#permalink] 29 Jun 2008, 22:32
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