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Re: If x is a positive integer, is x^2 + 6x +10 odd? [#permalink]

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23 Jun 2015, 04:34

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If x is a positive integer, is x^2 + 6x +10 odd?

In the expression x^2 + 6x + 10, 6x + 10 is always even. So we need to check whether x^2 is either even or odd to decide if the expression x^2 + 6x +10 is odd. We have 2 conditions: 1)If x^2 is even then x^2 + 6x +10 is even 2)If x^2 is odd then x^2 + 6x +10 is odd

(1) x^2 + 4x + 5 is odd

In the above expression 4x + 5 is always odd. Since the expression x^2 + 4x + 5 is odd, x^2 has to be even. As x^2 is even, x is also even. Referring to the above 2 conditions we can say x^2 + 6x +10 is even.

Sufficient

(2) x^2 + 3x + 4 is even

For the above expression to be even then x^2 + 3x has to be even. If x is odd then x^2 + 3x is even If x is even then x^2 + 3x is even

So the expression x^2 + 6x +10 can be either odd or even.

Since x is a Positive Integer therefore 6x+10 will always be even

so for x^2 + 6x +10 to be odd, x^2 MUST be odd i.e. the question is asking whether x is an odd integer or not

QUESTION REDEFINED: is x an odd Integer?

Statement 1: x^2 + 4x + 5 is odd

Since 4x is even therefore we infer that, x^2 + 5 is odd But Since EVEN + ODD = ODD therefore, x^2 must be even for x^2+5 to be odd i.e. x is certainly ODD Hence, SUFFICIENT

Statement 2: x^2 + 3x + 4 is even

EVE + EVE = EVEN Since, 4 is even, therefore x^2 + 3x MUST be even as well but, ODD + ODD = EVEN also, EVEN + EVEN = EVEN therefore x may be odd or even Hence, NOT SUFFICIENT

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the Q basically asks us "is x odd?"(as all terms except x^2 are even).. 1) stat 1 tells us that x^2 + 4x + 5 is odd... here x^2 has to be even... suff 2) stat 2 tells us that x^2 + 3x + 4 is even... here x can take even or odd value to satisfy the eq... insuff ans A
_________________

The question can first be simplified by noting that, if x is even, x^2 + 6x + 10 will be even, and if x is odd, x^2 + 6x+ 10 will be odd.

Thus, you can simplify this question: “Is x odd or even?” (A couple of shortcuts to save time in reaching that conclusion: the exponent on the first term can be ignored, since an even squared is still even and an odd squared is still odd. 6x will be even no matter what, since 6 is even, and obviously 10 is even no matter what. So, an even plus two evens is even, and an odd plus two evens is odd.)

(1) SUFFICIENT: You can plug in numbers or simply use number theory. If x is even, you get even + even + odd = odd, and if x is odd, you get odd + even + odd = even. Thus, since x^2 + 4x + 5 is odd, x is even.

(2) INSUFFICIENT: x^2 + 3x + 4 is actually even regardless of what integer is plugged in for x. If x is even, you get even + even + even = even, and if x is odd, you get odd + odd + even = even. Thus, x could be odd or even. Plugging in numbers will yield the same conclusion - x could be any integer.

Note that you should not factor any of the expressions above. If you wasted time factoring, remember: factoring is meaningless if you don’t have an equation set equal to zero! This problem was about number theory (or number testing), not factoring.

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