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For any integer x, x^2 − x

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Bunuel
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For any integer x, x^2 − x [#permalink] New post   05 Jun 2018, 00:29
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86% (00:52) correct 14% (01:38) wrong based on 77 sessions
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For any integer x, x^2 − x

A. is always even.
B. always odd.
C. odd only when x is negative.
D. even only when x is even.
E. even only when x is positive.
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A
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DavidTutorexamPAL
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Re: For any integer x, x^2 − x [#permalink] New post   05 Jun 2018, 02:17
Bunuel wrote:
For any integer x, x^2 − x

A. is always even.
B. always odd.
C. odd only when x is negative.
D. even only when x is even.
E. even only when x is positive.


Questions dealing with number properties can often be solved logically, without explicitly calculating.
We'll look for such a solution, a Logical approach.

If x is an even integer then x^2 - x = (even)^2 - even = even*even - even = even - even = even.
If x is an odd integer then x^2 - x = (odd)^2 - odd = odd*odd - odd = odd - odd = even.
Then x^2 - x is always even.

(A) is our answer.

Notice that all we did is write things out clearly, use the basic properties of even/odd and the question became very simple.
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For any integer x, x^2 − x [#permalink] New post   05 Jun 2018, 05:46
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Bunuel wrote:
For any integer x, x^2 − x

A. is always even.
B. always odd.
C. odd only when x is negative.
D. even only when x is even.
E. even only when x is positive.


\(x^2-x=x(x-1)\)

Note that, here we have two consecutive integers: \(x-1\) and \(x\). One of this integers is always even and therefore the product is always even as well. Product of two consecutive integers is always even.

Answer: A
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Re: For any integer x, x^2 − x [#permalink] New post   05 Jun 2018, 08:08
I also picked A but I have a question:

If x=1 then \((x^2)-x=0\)

0 is neither even nor negative. Considering this, the rule that \((x^2)-x\) doesn't seem always true. Can someone please explain?
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generis
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For any integer x, x^2 − x [#permalink] New post   05 Jun 2018, 17:32
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Masterscorp wrote:
I also picked A but I have a question:

If x=1 then \((x^2)-x=0\)

0 is neither even nor negative. Considering this, the rule that \((x^2)-x\) doesn't seem always true. Can someone please explain?

Masterscorp , 0 is even.

The test for whether an integer is even: is it evenly divisible by 2? That is, can you divide the integer by 2 and get an integer quotient with no remainder?

0/2 = 0
0 is an integer. No remainder.

Zero a little strange, I grant you. Hope that helps. :-)
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ScottTargetTestPrep
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Re: For any integer x, x^2 − x [#permalink] New post   06 Jun 2018, 16:37
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Bunuel wrote:
For any integer x, x^2 − x

A. is always even.
B. always odd.
C. odd only when x is negative.
D. even only when x is even.
E. even only when x is positive.


If x is even, then x^2 - x = even - even = even. If x is odd, then x^2 - x = odd - odd = even. So x^2 − x is always even.

Alternate solution:

If x is an integer, then x^2 - x = x(x - 1) is a product of two consecutive integers. When we have two consecutive integers, one must be even and the other one must be odd. Since even x odd = even, we see that x^2 - x must be even.

Answer: A
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Re: For any integer x, x^2 − x [#permalink]
06 Jun 2018, 16:37
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