Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Sep 1507:00 PM EDT
-09:00 PM EST
Elite scores are possible when you enroll in a powerful GMAT course, taught live online + 6 months access to TTP OnDemand video courses included! Class starts Tues/Thurs Sept. 15, 2026 - Nov. 15, 2027, 7:00pm-9:00pm EST
Sep 2107:00 PM EDT
-10:00 PM EDT
Boost your GMAT score in less than one month in a live online class + 6 months access to TTP OnDemand video courses included! Class starts Mon, Tues, Wed, Thur, Fri Sept. 21, 2026 - Oct. 9, 2027, 7:00pm-10:00pm EST
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
61% (03:02) correct
39%
(02:19)
wrong
based on 510
sessions
History
Date
Time
Result
Not Attempted Yet
If a, b, and c are consecutive integers, where \(a < b < c\), which of the following cannot be the value of \(c^2 - (a^2 + b^2)\)?
(A) -77
(B) -32
(C) -21
(D) -10
(E) 0
Are You Up For the Challenge: 700 Level Questions
_________________
New to the GMAT Club?
- • Questions: Question Bank | Bunuel's Signature Collection
• Tests: GMAT Club Tests | Forum Quiz
• Guides: Quantitative | Critical Reasoning | Reading Comprehension | Ultimate Quant Guide | All You Need for Quant
• Rules: Quantitative | Verbal
Bunuel
Convert \(c^2 - (a^2 + b^2)\) in terms of b......
\((b+1)^2 - ((b-1)^2 + b^2)=b^2+2b+1-(b^2-2b+1+b^2)=4b-b^2=b(4-b)=-(b)(b-4)\)
Look for the choices that can be converted in the form \(-(b)(b-4)\), that is product of numbers that differ by 4 *(-)
(A) -77 = \(-(11)(7)\)
(B) -32 = \(-(8)(4)\)
(C) -21 = \(-(7)(3)\)
(D) -10 = \(-(10)(1)=-(5)(2)\)...NO
(E) 0 = \(-(4)(0)\)
D
_________________
General Discussion
CareerGeek
Joined: 20 Jul 2017
Last visit: 11 Sep 2026
Posts: 1,285
Location: India
Concentration: Entrepreneurship, Marketing
GMAT 1: 690 Q51 V30
WE:Education (Education)
GMAT 1: 690 Q51 V30
Posts: 1,285
If a, b, and c are consecutive integers, where a<b<c, which of the following cannot be the value of c2−(a2+b2)?
(A) -77
(B) -32
(C) -21
(D) -10
(E) 0
Let the values of a = n - 1, b = n & c = n + 1
--> c^2−(a^2+b^2) = (n + 1)^2 - [(n - 1)^2 + n^2]
--> n^2 + 2n + 1 - 2n^2 + 2n - 1
--> -n^2 + 4n
--> -(n^2 - 4n + 4 - 4)
--> -(n - 2)^2 + 4
(A) -77 = -(9)^2 + 4 --> Possible
(B) -32 = -(6)^2 + 4 --> Possible
(C) -21 = -(5)^2 + 4 --> Possible
(D) -10 --> Not Possible
(E) 0 = -(2)^2 + 4 --> Possible
IMO Option D
(A) -77
(B) -32
(C) -21
(D) -10
(E) 0
Let the values of a = n - 1, b = n & c = n + 1
--> c^2−(a^2+b^2) = (n + 1)^2 - [(n - 1)^2 + n^2]
--> n^2 + 2n + 1 - 2n^2 + 2n - 1
--> -n^2 + 4n
--> -(n^2 - 4n + 4 - 4)
--> -(n - 2)^2 + 4
(A) -77 = -(9)^2 + 4 --> Possible
(B) -32 = -(6)^2 + 4 --> Possible
(C) -21 = -(5)^2 + 4 --> Possible
(D) -10 --> Not Possible
(E) 0 = -(2)^2 + 4 --> Possible
IMO Option D
Quote:
\(c^2−(a^2+b^2)\)
\(c^2=(a+2)^2=(a^2+4+4a)\)
\(a^2+b^2=a^2+(a+1)^2=a^2+a^2+1+2a=2a^2+2a+1\)
\(c^2−(a^2+b^2)=(a^2+4+4a)-(2a^2+2a+1)=-a^2+2a+3\)
\((A)-a^2+2a+3=-77…a^2-2a-80=0…(a-10)(a+8)=0:a=integer=valid\)
\((B)-a^2+2a+3=-32…a^2-2a-35=0…(a-7)(a+5)=0:a=integer=valid\)
\((C)-a^2+2a+3=-21…a^2-2a-24=0…(a-6)(a+4)=0:a=integer=valid\)
\((E)-a^2+2a+3=0…a^2-2a-3=0…(a-3)(a+1)=0:a=integer=valid\)
\((D)-a^2+2a+3=-10…a^2-2a-13=0…13=prime:a=not.integer=invalid\)
Ans (D)
If a, b, and c are consecutive integers, where a<b<c, which of the following cannot be the value of \(c^2−(a^2+b^2)\)?
(A) -77
(B) -32
(C) -21
(D) -10
(E) 0
\(c^2−(a^2+b^2)\) suggests that a,b and c can take any integer values where b = a + 1 and c = a + 2
Now, converting the equation in terms of lowest variable i.e. 'a'
\(c^2−(a^2+b^2)\) = \((a+2)^2 - (a^2 + (a + 1)^2)\)
= \(3 + 2a - a^2\)
For sake of ease checking values starting with a = 1 gives \(3 + 2a - a^2\) = 4
if a = 3, \(3 + 2a - a^2\) = 0
a = 6, \(3 + 2a - a^2\) = -21
a = 7, \(3 + 2a - a^2\) = -32
a = 10, \(3 + 2a - a^2\) = -77
Though by observing we can see that \(3 + 2a - a^2\) would give a negative value after giving a value of '0' for a = 3.
We can stop at a = 4 which gives -21 thus -10 cannot be the value of \(3 + 2a - a^2\).
IMO Answer D.
_________________
(A) -77
(B) -32
(C) -21
(D) -10
(E) 0
\(c^2−(a^2+b^2)\) suggests that a,b and c can take any integer values where b = a + 1 and c = a + 2
Now, converting the equation in terms of lowest variable i.e. 'a'
\(c^2−(a^2+b^2)\) = \((a+2)^2 - (a^2 + (a + 1)^2)\)
= \(3 + 2a - a^2\)
For sake of ease checking values starting with a = 1 gives \(3 + 2a - a^2\) = 4
if a = 3, \(3 + 2a - a^2\) = 0
a = 6, \(3 + 2a - a^2\) = -21
a = 7, \(3 + 2a - a^2\) = -32
a = 10, \(3 + 2a - a^2\) = -77
Though by observing we can see that \(3 + 2a - a^2\) would give a negative value after giving a value of '0' for a = 3.
We can stop at a = 4 which gives -21 thus -10 cannot be the value of \(3 + 2a - a^2\).
IMO Answer D.
_________________
Pain + Reflection = Progress | Ray Dalio
Good Books to read prior to MBA
Good Books to read prior to MBA
Given that a,b, and c are consecutive integers such that a<b<c. We are to determine which of the following cannot be the value of c^2 - (a^2 + b^2)
Since a,b, and c are consecutive numbers, and b is the number in the middle, then c=b+1 and a=b-1
so c^2 - (a^2 + b^2) = (b+1)^2 - [(b^2) + (b-1)^2]
b^2 + 2b +1 -[b^2 + b^2 -2b + 1]
4b - b^2 = b(4-b)
from this we know that when b=4, then b(4-b)=0, hence option E is out.
when b=5, 5(-1) = -5
when b=6, 6(-2)=-12
Since -10 lies between -5 and -12, there is no way of getting an integer value of b which yields -10 in the given expression above.
The answer, therefore, has to be -10.
A further check is to equate b(4-b) to -10 and find the determinant in order to confirm if it leads to a perfect square.
4b-b^2=-10
b^2-4b-10=0
The determinant of the above quadratic equation is [16-4(1)(-10)] = 56
Since 56 is not a perfect square to begin with, there is no way we can get integral roots of the above equation.
The answer is, therefore, option D.
Since a,b, and c are consecutive numbers, and b is the number in the middle, then c=b+1 and a=b-1
so c^2 - (a^2 + b^2) = (b+1)^2 - [(b^2) + (b-1)^2]
b^2 + 2b +1 -[b^2 + b^2 -2b + 1]
4b - b^2 = b(4-b)
from this we know that when b=4, then b(4-b)=0, hence option E is out.
when b=5, 5(-1) = -5
when b=6, 6(-2)=-12
Since -10 lies between -5 and -12, there is no way of getting an integer value of b which yields -10 in the given expression above.
The answer, therefore, has to be -10.
A further check is to equate b(4-b) to -10 and find the determinant in order to confirm if it leads to a perfect square.
4b-b^2=-10
b^2-4b-10=0
The determinant of the above quadratic equation is [16-4(1)(-10)] = 56
Since 56 is not a perfect square to begin with, there is no way we can get integral roots of the above equation.
The answer is, therefore, option D.
D
C^2-(b^2+a^2)=(a+2)^2-(a+1)^2-a^2
=4-(a-1)^2
=4-(int)^2 is never -10
Posted from my mobile device
C^2-(b^2+a^2)=(a+2)^2-(a+1)^2-a^2
=4-(a-1)^2
=4-(int)^2 is never -10
Posted from my mobile device
Answer is -10 . Option D
Consider the integers : 10, 11, 12
12^2-(10^2+11^2) = 144-221 = -77
Consider integers : 7,8,9
9^2-(8^2+7^2) = 81-113 = -32
Consider integers : 6,7,8
8^2-(7^2 + 6^2 ) = 64-85 = -21
Consider integers : 3,4,5
5^2-(3^2+4^2) =0
Even if the numbers are negative , since we are squaring the values it will become positive.
Hence, answer is -10
Posted from my mobile device
Consider the integers : 10, 11, 12
12^2-(10^2+11^2) = 144-221 = -77
Consider integers : 7,8,9
9^2-(8^2+7^2) = 81-113 = -32
Consider integers : 6,7,8
8^2-(7^2 + 6^2 ) = 64-85 = -21
Consider integers : 3,4,5
5^2-(3^2+4^2) =0
Even if the numbers are negative , since we are squaring the values it will become positive.
Hence, answer is -10
Posted from my mobile device
Ans.is D
C^2-(b^2+a^2)=(a+2)^2-(a+1)^2-a^2
=4-(a-1)^2
=4-(int)^2 is never -10
Posted from my mobile device
C^2-(b^2+a^2)=(a+2)^2-(a+1)^2-a^2
=4-(a-1)^2
=4-(int)^2 is never -10
Posted from my mobile device
The answer is D.
Using pattern, we figure out that as the numbers increase , the difference in the squares also increase.
For numbers 3,4,5, The given expression is zero
When you try 4,5,6 , the given expression is 6
When you try 5,6,7, the given expression is -12
As you can see -10 is not possible
Therefore D
Posted from GMAT ToolKit
Using pattern, we figure out that as the numbers increase , the difference in the squares also increase.
For numbers 3,4,5, The given expression is zero
When you try 4,5,6 , the given expression is 6
When you try 5,6,7, the given expression is -12
As you can see -10 is not possible
Therefore D
Posted from GMAT ToolKit 
Posts: 31
Bunuel
a, b, c are consecutive integer. So a = k, b = k+1, c= k+2
c^2 - a^2 - b^2 = (k+2)^2 - (k+1)^2 - k^2
= (2k+3) - k^2
= - [(k^2 - 2k +1) -4]
= - [(k-1)^2 - 2^2]
= -(k-3)(k+1)
So this equation can written as the -1* [product of two number a & b, in which a-b = 4]
I. -77 = -1 * 7 * 11. 11-7 =4. SUFF
II. -32 = -1 * 4 * 8. 8-4 =4. SUFF
III. -21 = -1 * 3 * 7. 7-3=4. SUFF
III. -10. INSUFF
IV. 0. SUFF
IMO D.
Automated notice from GMAT Club BumpBot:
A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.
This post was generated automatically.
_________________
A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.
This post was generated automatically.
_________________
Moderator:
Announcements
How to Tell If Your GMAT Prep Resources Are Actually Helping You
Subscribe to GMAT Club for New Podcast Episodes, Debriefs, & Insightful Sessions on GMAT Preparation
Subscribe to GMAT Club for New Podcast Episodes, Debriefs, & Insightful Sessions on GMAT Preparation
Tuck at Dartmouth
AI at Dartmouth Tuck: How the School Is Preparing Leaders for the Age of Artificial Intelligence
Inside the courses, research, and hands-on learning defining wise, decisive leadership in an AI-driven world.
Read More
Inside the courses, research, and hands-on learning defining wise, decisive leadership in an AI-driven world.
Read More
Top 5 GMAT Debrief Videos
Success stories and strategies from high-scoring candidates.