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If a, b, and c are consecutive integers such that a < b < c,
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23 Jan 2020, 07:23
23 Jan 2020, 07:23
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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,
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24 Jan 2020, 08:14
24 Jan 2020, 08:14
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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
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Re: If a, b, and c are consecutive integers such that a < b < c,
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24 Jan 2020, 13:08
24 Jan 2020, 13:08
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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
