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If a, b, and c are consecutive integers such that a < b < c

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If a, b, and c are consecutive integers such that a < b < c  [#permalink]

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New post 22 Jan 2020, 06:44
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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

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Re: If a, b, and c are consecutive integers such that a < b < c  [#permalink]

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New post 22 Jan 2020, 07:02
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2
GMATPrepNow 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, b, and c are consecutive integers such that a < b < c

i.e. b-a = 1 and c-b = 1 and c-a = 2

(b² – a²)(c² – b²) = (b+a)*(b-a)*(c-b)*(c+b) = (b+a)*(c+b) = Product of two consecutive ODD integers

(A) 323 = 17*19
(B) 483 = 21*23
(C) 575 = 23*25
(D) 613 = CAN NOT be product of two integers
(E) 899 = 29*31

Answer: Option D
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Re: If a, b, and c are consecutive integers such that a < b < c  [#permalink]

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New post 22 Jan 2020, 16:00
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While GMATinsight provides a valid approach, there's another (easier/faster) approach that we can use.
Anyone care to try to find that approach?

Cheers,
Brent
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Re: If a, b, and c are consecutive integers such that a < b < c  [#permalink]

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New post 22 Jan 2020, 16:42
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(b² – a²)(c² – b²) = (b+a)*(c+b) = (2b-1) * (2b+1) = 4b^2 -1

So the answer plus 1 must divided by 4
(A) 323 + 1 = 324 ok
(B) 483 + 1 ok
(C) 575 + 1 ok
(D) 613 + 1 = 614 nope
(E) 899 + 1 ok

So imo D.

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Re: If a, b, and c are consecutive integers such that a < b < c  [#permalink]

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New post 22 Jan 2020, 17:47
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GMATPrepNow 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


Great solution, maxx0811!
My solution is similar to yours, but I didn't notice that I could save a bit of time by assigning variables similar to the way in which you assigned to the variables.

Here's how I approached the question:

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  [#permalink]

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New post 31 Jan 2020, 04:28
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This question can be solved using basic algebra and some logic. First of all, (b^2-a^2)(c^2-b^2)= (b-a)(b+a)(c-b)(c+b). Since a,b and c are consequtive, both b-a and c-b are equal to 1. So the product is equal to (a+b)(b+c). Since a b and c are consequtive, either a odd b even and c odd or a even b odd and c even. So both a+b and b+c are odd and they are consequtive odd integers. Hence we have the product of two consecutive odd integers.
Now let's see the answer choices
323 <400=20^2 the unit is 3 which has to be 1*3 or 9*7. So let's try 19*17 which gives 323
483>400 which is 20^2. If we try 21*23 it is 483
575<625 which is 25^2. Since unit digit is 5 let's try 23*25 which gives 575
613<625 which is 25^2 but we have already seen 21*23 as 483 and 29*27 will be greater than 625 so this is the answer
899<900 which is 30^2 so since unit digit is 9 if we try 29*31 we get 899.
So D

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Re: If a, b, and c are consecutive integers such that a < b < c   [#permalink] 31 Jan 2020, 04:28
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