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For Consecutive integers x, y and z, where x > y > z, which of the [#permalink]
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11 Sep 2016, 19:10
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For Consecutive integers x, y and z, where x > y > z, which of the following CANNOT be the value of \((x^2  y^2)*(y^2  z^2)\) ? A) 63 B) 99 C) 195 D) 276 E) 323
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Last edited by Bunuel on 12 Sep 2016, 22:50, edited 2 times in total.
Renamed the topic and edited the question.



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Re: For Consecutive integers x, y and z, where x > y > z, which of the [#permalink]
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11 Sep 2016, 21:46
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[quote="accountrail"]Plz let me know,method to solve these kind of questions
The question mentions that the numbers are consecutive and \(x >y > z\) so first simplify \((x^2−y^2)∗(y^2−z^2) = (x  y) * (x + y) * (y  z) * (y + z)\) and since \(x > y > z , x  y = 1\) and \(y  z = 1\) So what remains is \((x +y ) * (y + z)\) and since the numbers are consecutive, \(x +y\) and \(y + z\) has to odd, so only option which is even is D CANNOT be value.



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Re: For Consecutive integers x, y and z, where x > y > z, which of the [#permalink]
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12 Sep 2016, 02:06
Senthil1981 wrote: accountrail wrote: Plz let me know,method to solve these kind of questions
The question mentions that the numbers are consecutive and \(x >y > z\) so first simplify \((x^2−y^2)∗(y^2−z^2) = (x  y) * (x + y) * (y  z) * (y + z)\) and since \(x > y > z , x  y = 1\) and \(y  z = 1\) So what remains is \((x +y ) * (y + z)\) and since the numbers are consecutive, \(x +y\) and \(y + z\) has to odd, so only option which is even is D CANNOT be value. This is too simple....even n odd Say if I gave you D as 209.. and A as 35 and rest all same What will be the answer
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Re: For Consecutive integers x, y and z, where x > y > z, which of the [#permalink]
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12 Sep 2016, 22:32
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OE: Perhaps we should list a few of our squares here: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169. OK, that should take us far enough. Now, since we just need the differences between consecutive squares, we can just consider the intervals between values listed above: 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25. Upon noticing that these intervals are only odd, one must conclude that a product of two of them must be odd and the answer is clearly D.
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Re: For Consecutive integers x, y and z, where x > y > z, which of the [#permalink]
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12 Sep 2016, 22:47
Nevernevergiveup wrote: OE:
Perhaps we should list a few of our squares here: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169. OK, that should take us far enough. Now, since we just need the differences between consecutive squares, we can just consider the intervals between values listed above: 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25. Upon noticing that these intervals are only odd, one must conclude that a product of two of them must be odd and the answer is clearly D. The main logic to this Q is that the answer has to be a multiple of CONSECUTIVE odd integers.... Incase no choice was EVEN, we would be looking into this aspect
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For Consecutive integers x, y and z, where x > y > z, which of the [#permalink]
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12 Sep 2016, 23:28
Senthil1981 wrote: accountrail wrote: Plz let me know,method to solve these kind of questions
The question mentions that the numbers are consecutive and \(x >y > z\) so first simplify \((x^2−y^2)∗(y^2−z^2) = (x  y) * (x + y) * (y  z) * (y + z)\) and since \(x > y > z , x  y = 1\) and \(y  z = 1\) So what remains is \((x +y ) * (y + z)\) and since the numbers are consecutive, \(x +y\) and \(y + z\) has to odd, so only option which is even is D CANNOT be value. Also we can derive whole wuqation \((x +y ) * (y + z)\) in terms of x as \((2x1)(2x3)\) since \(y=x1\) and \(z=y1=x2\) we can easily note that (2n+k) where k is any odd number indicates an odd number. Hence product of two odd numbers is an odd number.
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Re: For Consecutive integers x, y and z, where x > y > z, which of the [#permalink]
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13 Sep 2016, 07:04
Senthil1981 wrote: For Consecutive integers x, y and z, where x > y > z, which of the following CANNOT be the value of \((x^2  y^2)*(y^2  z^2)\) ?
A) 63 B) 99 C) 195 D) 276 E) 323 2 ways: 1) xodd yeven zodd \((odd^2even^2)^(even^2odd^2) = even\) 2) xeven yodd zeven \((even^2odd^2)^(odd^2even^2) =even\) The only even answer we have is D) 276



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Re: For Consecutive integers x, y and z, where x > y > z, which of the [#permalink]
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04 Feb 2017, 01:06
Continuing from Senthil1981's solution, (x+y)(y+z) can be represented in terms of y as (y+1+y)(y+y1)>(2y+1)(2y1)>4y^21 Now equate 4y^21 to each answer option and check if you are getting a valid integer value for y. It will work for all options excluding 276.Hence, D is the answer.



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Re: For Consecutive integers x, y and z, where x > y > z, which of the [#permalink]
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11 Jan 2018, 15:00
Hi All, GMAT questions are almost always built around patterns  even if you don't realize that the pattern is there, you can probably do a bit of 'brute force' work and define the pattern. By extension, if you know the pattern, then you should be able to use that knowledge to your advantage to either answer the question immediately (or do another step or two of work to get the answer). Here, we're given some specific facts to work with: 1) X, Y and Z are CONSECUTIVE integers 2) X > Y > Z We're asked for what CANNOT be the value of (X^2  Y^2)(Y^2  Z^2). Let's TEST VALUES and see if a pattern emerges... IF... X = 3, Y = 2, Z = 1.... (9  4)(4  1) = (5)(3) = 15 So "15" is a possible answer. Also note that we ended up multiplying two ODD numbers together... Let's try another TEST.... IF... X = 4, Y = 3, Z = 2.... (16  9)(9  4) = (7)(5) = 35 So "35" is a possible answer. Notice that we again ended up multiplying two ODD numbers together... That looks like a pattern. If the end result is just going to be an ODD number every time, then there's clearly an answer that CANNOT be the value... If you're not convinced yet, then try another example (and feel free to try as many as you like  as the numbers increase, you'll eventually hit all 4 of the possible answers, at which point you'll know which answer is NOT possible. Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: For Consecutive integers x, y and z, where x > y > z, which of the
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