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Re: For positive integers x and y, which of the following can [#permalink]

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11 Jun 2013, 06:12

Bunuel wrote:

emmak wrote:

For positive integers x and y, which of the following can be written as y^2?

A. (x+1)! B. (x+9)! C. x^2−9 D. x^2+1 E. ((x+1)^2)!

Plug values: if x=5, then x^2-9=16=4^2.

Answer: C.

How the solution can be made easy? I was plugging in smaller numbers than 5 and in such cases even (C) fails. As the time was running out, I made an educated guess as (C) and it turned out to be right !!!!

Re: For positive integers x and y, which of the following can [#permalink]

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12 Jun 2013, 17:35

The question is asking "Which of these equations could equal a perfect square?"

If you're familiar with number properties regarding perfect squares, you should be able to rule out everything except the correct answer.

the fact that 5^2 - 4^2 is equal to 3^2 should be a pattern that you have memorized (esp. with regards to right triangles) also, another one you might see often is some form of 13^2-5^2=12^2

Re: For positive integers x and y, which of the following can [#permalink]

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12 Jun 2013, 22:57

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The way I solved this is by process of elimination.

First, all factorials should be written off. A factorial would never be a perfect square since the last prime in the series would be the only single factor of its kind in the factorial. (i.e. 11! would have an 11 as one of its factor, but would not have another 11 as another factor, hence making it impossible for it to be a perfect square). A, B and E are out.

Second, D is never going to be a perfect square, since a perfect square plus one is never going to be a perfect square.

Re: For positive integers x and y, which of the following can [#permalink]

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13 Apr 2015, 10:41

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For positive integers x and y, which of the following can [#permalink]

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24 Jul 2016, 23:01

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For positive integers x and y, which of the following can be written as y^2? A. (x+1)! B. (x+9)! C. x^2−9 D. x^2+1 E. ((x+1)^2)!

It is worth while to remember that the only factorial that are perfect square is 0 and 1. {0!=1 and 1!=1 and \(1=1^2}\) If you know this fact then you can figure out that the answer is C But just to reconfirm I will solve this question. Now lets attack the problem.

A. (x+1)! = OHHHH ! so close, If x= 0 then this could have been our answer, \({0+1=1^2}\)but question stem tells us that x and y are +ve integer. 0 is not a positive integer.

B. (x+9)! = Cannot be a perfect square. No factorial except 1 is a perfect square

C. x^2−9 = Yes ! this can be ; take \(x= 3; 3^2-9 = 0-0 = 0 ; y= 0^2\) ||||| take \(x=5; 5^2-9=16 ; Y^2=4^2\)

D. x^2+1 = NO ! if you add 1 to a perfect square you cannot another perfect square; exception is 0 ; \(0^2+1 = 1 = 1^2\)

E. ((x+1)^2)!= Again so close if x= 0 then we could have a perfect square but x cannot be zero; so this expression cannot be a perfect square.

emmak wrote:

For positive integers x and y, which of the following can be written as y^2?

A. (x+1)! B. (x+9)! C. x^2−9 D. x^2+1 E. ((x+1)^2)!

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Posting an answer without an explanation is "GOD COMPLEX". The world doesn't need any more gods. Please explain you answers properly. FINAL GOODBYE :- 17th SEPTEMBER 2016. .. 16 March 2017 - I am back but for all purposes please consider me semi-retired.

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