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If x is an integer, is (x^2 +1) (x+5) an even number?

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by phoenixhazard » Tue Oct 19, 2010 7:10 pm
If x is an integer, is (x^2 +1) (x+5) an even number?
(1) x is an odd number
(2) Each prime factor of x^2 is greater than 7

OA = Each statement ALONE is sufficient


spoiler to answer: (my question)

I understand (1), that definitely gives you always an odd number with the formula. Not sure how (2) works out though?
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by uwhusky » Tue Oct 19, 2010 7:22 pm
My first question for you is, do you know what a prime factor is?

It's the smallest factor that cannot be divided by another integer, or in another words, it is a "prime number" factor.

All of the prime numbers except 2 are odd, and an odd number plus another odd number results in an even number: 3 + 5 = 8; 3 + 9 = 12; 5 + 13 = 18.

An even number has at least one factor of 2 in it, and that any integer multiplies by 2 will result in an even number.

Statement (2) says that whatever prime factor is in x^2 is greater than 7, and that means x must be a multiple of only odd numbers, and that x must be odd.

Odd number + odd number = even number, and therefore (x + 5) must be even, and (x^2 + 1) (x + 5) must also be even.

I did a horrible job explaining this one, but hopefully it makes a little bit of sense.
Yep.
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by Rahul@gurome » Tue Oct 19, 2010 8:05 pm
phoenixhazard wrote:If x is an integer, is (x^2 +1) (x+5) an even number?
(1) x is an odd number
(2) Each prime factor of x^2 is greater than 7

OA = Each statement ALONE is sufficient


spoiler to answer: (my question)

I understand (1), that definitely gives you always an odd number with the formula. Not sure how (2) works out though?
Though it is not needed in the context of DS questions, (1) does not always gives an odd number with the formula. In fact, the expression gives an even number whenever x is odd. We can proceed as follows...

If x is an odd number, x^2 must be odd => (x^2+1) must be even.
By the same logic, (x+5) is even. As addition of two odd numbers results in an even number.
Multiplication of two even numbers results in an even number. Thus (x^2+1)(x+5) is an even number.

Satetment (1) is SUFFICIENT.


And for (2), each prime factor of x^2 is greater than 7 implies that 2 is not a prime factor of x^2 and in turn 2 is not a prime factor of x. (As all the prime factors of x^2 are also prime factors of x)

That means x is composed of 11, 13, 17, 19, 23 etc (Prime numbers greater than 7).
Therefore x is an odd number => Equivalent to statement (1).

Statement (2) is SUFFICIENT.

The correct answer is D.
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by phoenixhazard » Tue Oct 19, 2010 8:46 pm
Rahul@gurome wrote:
phoenixhazard wrote:If x is an integer, is (x^2 +1) (x+5) an even number?
(1) x is an odd number
(2) Each prime factor of x^2 is greater than 7

OA = Each statement ALONE is sufficient


spoiler to answer: (my question)

I understand (1), that definitely gives you always an odd number with the formula. Not sure how (2) works out though?
Though it is not needed in the context of DS questions, (1) does not always gives an odd number with the formula. In fact, the expression gives an even number whenever x is odd. We can proceed as follows...

If x is an odd number, x^2 must be odd => (x^2+1) must be even.
By the same logic, (x+5) is even. As addition of two odd numbers results in an even number.
Multiplication of two even numbers results in an even number. Thus (x^2+1)(x+5) is an even number.

Satetment (1) is SUFFICIENT.


And for (2), each prime factor of x^2 is greater than 7 implies that 2 is not a prime factor of x^2 and in turn 2 is not a prime factor of x. (As all the prime factors of x^2 are also prime factors of x)

That means x is composed of 11, 13, 17, 19, 23 etc (Prime numbers greater than 7).
Therefore x is an odd number => Equivalent to statement (1).

Statement (2) is SUFFICIENT.

The correct answer is D.
Sorry it was a typo I meant (1) is always even. Thanks for the clarification though
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by [email protected] » Fri Oct 22, 2010 3:28 am
after reading this post am kinda confused now for some reason.

Even though x is an odd integer; when substituted in (x^2+1)/x+5, it can give me numbers other than an integer

For eg. when x=1. then 2/6=.333 wel how can this be odd or a even number.

pls clarify some1.. god.. ds sucks..

I perfectly agree with al the explanations.
i went in with the same logic onl
but i was thinkin after solvin it and found out that why cant this b a possibility
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by Rahul@gurome » Fri Oct 22, 2010 3:40 am
[email protected] wrote:after reading this post am kinda confused now for some reason.

Even though x is an odd integer; when substituted in (x^2+1)/x+5, it can give me numbers other than an integer

For eg. when x=1. then 2/6=.333 wel how can this be odd or a even number.

pls clarify some1.. god.. ds sucks..

I perfectly agree with al the explanations.
i went in with the same logic onl
but i was thinkin after solvin it and found out that why cant this b a possibility
Read the expression carefully. It is (x^2+1)(x+5) not (x^2+1)/x+5, i.e. (x^2+1) is multiplied with (x+5). The expression will always result in an integer whenever x is an integer. Thus when x = 1, (x^2+1)(x+5) = (1+1)(1+5) = 2*6 = 12, an even number.
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by Testluv » Fri Oct 22, 2010 6:55 am
(Even) * (any integer) is always Even.

(1): if x is odd, then x^2 is also odd (as odd*odd is odd). If x^2 is odd, then x^2 + 1 is even. Even * any integer is Even, thus (1) is always Even.

(2): if every prime factor of x^2 is greater than 7, then every prime factor of x^2 is odd. Odd*Odd is always Odd. Thus, x^2 is odd, and x^2 + 1 is even. Thus, (2) will also be Even.
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by siddhans » Fri Jul 29, 2011 1:29 am
Testluv wrote:(Even) * (any integer) is always Even.

(1): if x is odd, then x^2 is also odd (as odd*odd is odd). If x^2 is odd, then x^2 + 1 is even. Even * any integer is Even, thus (1) is always Even.

(2): if every prime factor of x^2 is greater than 7, then every prime factor of x^2 is odd. Odd*Odd is always Odd. Thus, x^2 is odd, and x^2 + 1 is even. Thus, (2) will also be Even.
I have a question...How can we calculate if x is odd or even from the prime factor....the statement 2 just says the prime factors of x^2 is odd ....that doesnt mean x^2 is odd or x is odd...Please explain...???
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by Dhaswe » Sun Jul 31, 2011 9:50 am
For (1), why are we not considering the value 0. If so, (x^2+1)(x+5) will be odd and (1) will be insufficient.
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by rohanpande » Sun Aug 26, 2012 1:47 am
From the question stem and first statement, we can deduce that x is an integer and it is odd. Zero is not an odd integer; it is an even integer.
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