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

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

We have the consecutive integers: A < B < C

Step 1: factor the difference of squares given and you get

(b + a) (b - a) (c + b) (c - a)

Step 2: since a < b < c are consecutive integers, the difference between each variable will be +1

(c - a) = +1

And

(b - a) = +1

Since multiplying by +1 will not make a difference in the ultimate product, we can focus on the other 2 factors

(a + b) (b + c)

Step 3: substitute the consecutive integer pattern for a < b < c such that the middle term B = X

Let

a = X - 1

b = X

c = X + 1

(a + b) (b + c) = (X - 1 + X) (X + X + 1) = (2X - 1) (2X + 1)

This is a Difference of Squares again:

4(X)^2 - 1

Thus, since X stands for the Integer b, the answer must be -1 less than a Multiple of 4

The only answer that does NOT satisfy this is

D 613

Adding +1 to all the other answers makes them multiples of 4

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Re: If a, b, and c are consecutive integers such that a < b < c [#permalink]
Top Contributor
Given that a, b, and c are consecutive integers such that a < b < c and we need to find which of the following CANNOT be the value of (b² – a²)(c² – b²)

a, b and c are consecutive integers
=> b = a + 1
=> c = b + 1 = a+ 2

We know that $$A^2 - B^2$$ = (A-B) *(A+B)

=> (b² – a²)(c² – b²) = (b - a) * (b + a) * (c - b) * (c + b) = (a + 1 - a) * (a + 1 + a) * (a+ 2 - (a + 1)) * (a+ 2 + (a + 1))
= 1 * (2a + 1) * 1 * (2a + 3) = (2a + 1) * (2a + 3)
= Product of two consecutive odd numbers as (2a is even making 2a +1 and 2a + 3 two consecutive odd numbers)

Let's try the answer choices into product of two consecutive odd numbers

(A) 323 = 17 * 19 => POSSIBLE

(B) 483 = 21 * 23 => POSSIBLE

(C) 575 = 23 * 25 => POSSIBLE

(D) 613 = Cannot be split into product of two consecutive odd numbers => NOT POSSIBLE

(E) 899 = 29 * 31 => POSSIBLE