GRE Prep 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.
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
If a, b, and c are consecutive integers such that a < b < c,
[#permalink]
23 Jan 2020, 07:23
23 Jan 2020, 07:23
2
7
Bookmarks
00:00
Question Stats:
59% (03:34) correct
40% (02:45) wrong based on 37 sessions
Hide Show timer Statistics
If a, b, and c are consecutive integers such that a < b < c, which of the following CANNOT be the value of (b² – a²)(c² – b²)?
(A) 323
(B) 483
(C) 575
(D) 613
(E) 899
(A) 323
(B) 483
(C) 575
(D) 613
(E) 899
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: If a, b, and c are consecutive integers such that a < b < c,
[#permalink]
24 Jan 2020, 08:14
24 Jan 2020, 08:14
2
1
Bookmarks
GreenlightTestPrep wrote:
If a, b, and c are consecutive integers such that a < b < c, which of the following CANNOT be the value of (b² – a²)(c² – b²)?
(A) 323
(B) 483
(C) 575
(D) 613
(E) 899
(A) 323
(B) 483
(C) 575
(D) 613
(E) 899
Since a, b, and c are consecutive integers and a < b < c, we know that...
a = a
b = a + 1
c = a + 2
So, (b² – a²)(c² – b²) = [(a + 1)² - a²][(a + 2)² - (a + 1)²]
= [a² + 2a + 1 - a²][(a² + 4a + 4) - (a² + 2a + 1)]
= [2a + 1][2a + 3]
= 4a² + 8a + 3
= 4(a² + 2a) + 3
= 3 greater than some multiple of 4
Now let's check the answer choices....
(A) 323
320 is a multiple of 4, which means 323 is 3 greater than a multiple of 4
Keep A
(B) 483
480 is a multiple of 4, which means 483 is 3 greater than a multiple of 4
Keep B
(C) 575
572 is a multiple of 4, which means 575 is 3 greater than a multiple of 4
Keep C
(D) 613
610 is NOT a multiple of 4, which means 613 is NOT 3 greater than a multiple of 4
Voila!
Answer: D
Cheers,
Brent
GRE Instructor
Joined: 19 Jan 2020
Status:Entrepreneur | GMAT, GRE, CAT, SAT, ACT coach & mentor | Founder @CUBIX | Edu-consulting | Content creator
Posts: 117
Given Kudos: 0
GPA: 3.72
Re: If a, b, and c are consecutive integers such that a < b < c,
[#permalink]
24 Jan 2020, 13:08
24 Jan 2020, 13:08
1
Question:
If a, b, and c are consecutive integers such that a < b < c, which of the following CANNOT be the value of (b² – a²)(c² – b²)?
(A) 323
(B) 483
(C) 575
(D) 613
(E) 899
Solution:
a, b, and c are consecutive integers such that a < b < c
=> b = a + 1; c = b + 1
(b² – a²)(c² – b²)
= (b - a)(b + a)(c - b)(c + b)
= (1)(b + a)(1)(c + b)
= (b + a)(c + b)
= (b + b - 1)(b + 1 + b)
= (2b - 1)(2b + 1)
= \(4b^2 - 1\)
= \((2b)^2 - 1\)
Thus, if 1 is added to the number (the numbers given in the options), the resulting number should be definitely the square of an even number
Let us try the options:
A: 323 + 1 = 324 = \(18^2\) --- it is the square of an even number
B: 483 + 1 = 484 = \(22^2\) --- it is the square of an even number
C: 575 + 1 = 576 = \(24^2\) --- it is the square of an even number
D: 613 + 1 = 614 --- it is not the square of an even number, hence definitely NOT possible
E: 899 + 1 = 900 = \(30^2\) --- it is the square of an even number
Answer D
Alternate approach:
a, b, and c are consecutive integers such that a < b < c
=> b = a + 1; c = b + 1
=> c = a + 2
(b² – a²)(c² – b²)
= (b - a)(b + a)(c - b)(c + b)
= (1)(b + a)(1)(c + b)
= (a + 1 + a)(a + 2 + a + 1)
= (2a + 1)(2a + 3)
= \(4a^2 + 8a + 3\)
Thus, if 3 is removed from the number (the numbers given in the options), the resulting number should be definitely divisible by 4
Let us try the options:
A: 323 - 3 = 320 ---- divisible by 4, hence maybe possible
B: 483 - 3 = 480 ---- divisible by 4, hence maybe possible
C: 575 - 3 = 572 ---- divisible by 4, hence maybe possible
D: 613 - 3 = 610 ---- not divisible by 4, hence definitely NOT possible
E: 899 - 3 = 896 ---- divisible by 4, hence maybe possible
Answer D
If a, b, and c are consecutive integers such that a < b < c, which of the following CANNOT be the value of (b² – a²)(c² – b²)?
(A) 323
(B) 483
(C) 575
(D) 613
(E) 899
Solution:
a, b, and c are consecutive integers such that a < b < c
=> b = a + 1; c = b + 1
(b² – a²)(c² – b²)
= (b - a)(b + a)(c - b)(c + b)
= (1)(b + a)(1)(c + b)
= (b + a)(c + b)
= (b + b - 1)(b + 1 + b)
= (2b - 1)(2b + 1)
= \(4b^2 - 1\)
= \((2b)^2 - 1\)
Thus, if 1 is added to the number (the numbers given in the options), the resulting number should be definitely the square of an even number
Let us try the options:
A: 323 + 1 = 324 = \(18^2\) --- it is the square of an even number
B: 483 + 1 = 484 = \(22^2\) --- it is the square of an even number
C: 575 + 1 = 576 = \(24^2\) --- it is the square of an even number
D: 613 + 1 = 614 --- it is not the square of an even number, hence definitely NOT possible
E: 899 + 1 = 900 = \(30^2\) --- it is the square of an even number
Answer D
Alternate approach:
a, b, and c are consecutive integers such that a < b < c
=> b = a + 1; c = b + 1
=> c = a + 2
(b² – a²)(c² – b²)
= (b - a)(b + a)(c - b)(c + b)
= (1)(b + a)(1)(c + b)
= (a + 1 + a)(a + 2 + a + 1)
= (2a + 1)(2a + 3)
= \(4a^2 + 8a + 3\)
Thus, if 3 is removed from the number (the numbers given in the options), the resulting number should be definitely divisible by 4
Let us try the options:
A: 323 - 3 = 320 ---- divisible by 4, hence maybe possible
B: 483 - 3 = 480 ---- divisible by 4, hence maybe possible
C: 575 - 3 = 572 ---- divisible by 4, hence maybe possible
D: 613 - 3 = 610 ---- not divisible by 4, hence definitely NOT possible
E: 899 - 3 = 896 ---- divisible by 4, hence maybe possible
Answer D
