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

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

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26 Jun 2008, 23:31
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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27 Jun 2008, 00:19
A is the ciorrect Ans?
Can someone tell the correct approach
SVP
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27 Jun 2008, 00:21
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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27 Jun 2008, 01:38

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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27 Jun 2008, 05:21
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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27 Jun 2008, 05:31
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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27 Jun 2008, 08:16
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
VP
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29 Jun 2008, 21:32
OA is E indeed. Thanks to all!
Re: consecutive integers   [#permalink] 29 Jun 2008, 21:32
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