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If a, b, and c are consecutive integers and a<b<c, [#permalink ]
26 Jun 2008, 23:31

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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Re: consecutive integers [#permalink ]
27 Jun 2008, 00:19

A is the ciorrect Ans? Can someone tell the correct approach

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Re: consecutive integers [#permalink ]
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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Re: consecutive integers [#permalink ]
27 Jun 2008, 01:38

Answer is (E) 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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Re: consecutive integers [#permalink ]
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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Re: consecutive integers [#permalink ]
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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Re: consecutive integers [#permalink ]
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

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Re: consecutive integers [#permalink ]
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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