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PS: The operation # is defined by x#y =

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by Yaj » Thu Sep 12, 2013 2:46 am
The operation # is defined by x#y = (2x+y)^2+(x+2y)^2 for all integers x and y. If z is a integer, which of the following must be true?
I. x#x is even
II. x#x >x
III. x#(-2x)=0
(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III

[spoiler]OA:A[/spoiler]

Thanks in advance!
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Source: — Problem Solving |

by ganeshrkamath » Thu Sep 12, 2013 3:03 am
Yaj wrote:The operation # is defined by x#y = (2x+y)^2+(x+2y)^2 for all integers x and y. If z is a integer, which of the following must be true?
I. x#x is even
II. x#x >x
III. x#(-2x)=0
(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III

[spoiler]OA:A[/spoiler]

Thanks in advance!
Statement 1:
x#x = (2x+x)^2+(x+2x)^2 = (3x)^2 + (3x)^2
= 2*(3x)^2
Since even*(any integer) = even, x#x is even.
True.

Statement 2:
x#x = 2*(3x)^2 = 2*(9x^2) = 18*x^2
Put x = 0, x#x = 0
False.

Statement 3:
x#(-2x) = (2x-2x)^2+(x-4x)^2 = 0 + (-3x)^2
= 9x^2
False.

Choose A

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by [email protected] » Thu Sep 12, 2013 11:42 am
Hi Yaj,

This question appears to have some extra wording that shouldn't be there: the phrase "if Z is an integer" appears irrelevant to the rest of the prompt and the answer choices.

Regardless, you can solve this question in a number of ways: with algebra, Number Properties or TESTing values.

Since ganeshrkamath has shown you the algebra, I'll show you a combination of Number Properties and TESTing values.

This question asks us to prove what MUST be true. In Roman Numeral questions, it's usually easier to prove when something is NOT true, so I will use THAT tactic whenever possible.

I. x#x

We're told that x and y are integers. Since we're plugging in the SAME integer for x and y, we can use a Number Property against the given equation....

Each of the calculations inside the parentheses will be the same: (2x + x)^2 = (x + 2x)^2 regardless of what integer we plug in. The Number property is that whenever you add 2 integers together that are the SAME, the result is ALWAYS EVEN. So Roman Numeral I is TRUE.

II. x#x > x

Here, I see a real easy, fast way to disprove this. Try plugging in x = 0

0#0 creates a big calculation of 0s which equals 0

0 is NOT > 0. Roman Numeral II is NOT TRUE.

III. x#(-2x) = 0

Any number of values will disprove this. Try plugging in x = 1

1#(-2) will give you (0)^2 + (-3)^2 = 9

9 is NOT = 0. Roman Numeral III is NOT TRUE

Final Answer: A

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Rich
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