It is currently 26 Jun 2017, 09:20

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# a,b, and c are consecutive integers where a<b<c. which

Author Message
VP
Joined: 22 Nov 2007
Posts: 1080
a,b, and c are consecutive integers where a<b<c. which [#permalink]

### Show Tags

29 Dec 2007, 10:20
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

a,b, and c are consecutive integers where a<b<c. which of the following could not be the value of c^2-b^2-a^2?

a. -21
b. -12
c. -6
d. 0
e. 3
Senior Manager
Joined: 01 Sep 2006
Posts: 301
Location: Phoenix, AZ, USA

### Show Tags

29 Dec 2007, 11:21
[quote="marcodonzelli"]a,b, and c are consecutive integers where a<b<c>B(4-B)

B=0 ===> 0 possible
B=1 ===> 3 possible
B=5==> -5
B=6==> -12
B=7=>-21

-6 is not possible

Intern
Joined: 24 Feb 2006
Posts: 48

### Show Tags

29 Dec 2007, 11:38
Its C.
Let a =x, then b = x+1, c= x+2

c^2-b^2-a^2
=>(x+2)^2 - (x+1)^2 -x^2
=> -x^2 +2x + 3

x can't be an integer when -x^2 +2x + 3 = -6
Intern
Joined: 24 Feb 2006
Posts: 48

### Show Tags

29 Dec 2007, 11:50
marcodonzelli wrote:
nope OA is 3

Hmm OA is 3 ?? Not convinced ..
c^2-b^2-a^2
(0, 1,2) gives a value of 3 and 2,3,4 also gives a value of 3..
CEO
Joined: 17 Nov 2007
Posts: 3584
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

### Show Tags

29 Dec 2007, 12:24
Manager
Joined: 11 Dec 2007
Posts: 121

### Show Tags

29 Dec 2007, 12:37
OA cant be 3. Damager is right.

If A=2, B=3, C=4, then the answer is 3.
If 3,4,5, then answer is 0.
If 5,6,7, then -12.
If 6,7,8, then -21.

Senior Manager
Joined: 19 Nov 2007
Posts: 459

### Show Tags

29 Dec 2007, 18:05
mkl_in_2001 wrote:
Its C.
Let a =x, then b = x+1, c= x+2

c^2-b^2-a^2
=>(x+2)^2 - (x+1)^2 -x^2
=> -x^2 +2x + 3

x can't be an integer when -x^2 +2x + 3 = -6

thats a good way of solving it.
good job!
Re: consecutive integers   [#permalink] 29 Dec 2007, 18:05
Display posts from previous: Sort by