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(x^3 + 19837) (x^2 + 5) (x – 3) an odd

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(x^3 + 19837) (x^2 + 5) (x – 3) an odd

by sanju09 » Fri Oct 15, 2010 10:50 pm
x is a positive integer greater than 2; is (x^3 + 19837) (x^2 + 5) (x - 3) an odd number?

[1] The sum of any prime factor of x and x is even.

[2] 3 x is an even number.


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by this_time_i_will » Fri Oct 15, 2010 11:02 pm
IMO B.
SInce II gives x to be an even number.
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by ankurmit » Sat Oct 16, 2010 1:44 am
Stem B is sufficient.

I could not solve for A.

Sanju.. Can you post explanation .
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by limestone » Sat Oct 16, 2010 2:54 am
If x is even, then x^3 and x^2 are even too,
then (x^3 + 19837), (x^2 + 5), and (x - 3) are odd.
then the product will be odd.

If x is odd, then x^3 and x^2 are odd too,
then (x^3 + 19837), (x^2 + 5), and (x - 3) are even.
then the product will be even.

Thus if we can define whether x is an even or odd, we can define the property of the product.

1. All prime numbers are odd, except 2. Thus prime numbers can be odd or even
So we cannot define whether x is even or odd.
INSUF.

2. 3x is an even, then x is an even. As even divided by an odd, if divisible, will give out an even.
Now we know that x is even, then the product must be odd.
SUFF.

Pick B.
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by ankurmit » Sat Oct 16, 2010 3:08 am
Its mentioned that X is geater than 2

x is a positive integer greater than 2; is (x^3 + 19837) (x^2 + 5) (x - 3) an odd number?
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by limestone » Sat Oct 16, 2010 8:09 am
For 1.
Let's say x = 8
Prime factor of x: 2
The sum of x and its prime factor : 2+8 = 10 is an even

Let's say x = 9
Prime factor of x: 3
The sum of x and its prime factor : 3+9 = 12 is an even too

So with the given information that the sum of x and its prime factor is even, x can be either odd or even.
The the product : (x^3 + 19837) (x^2 + 5) (x - 3) can be either odd or even too.

Then 1 is INSUFF.
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by Stuart@KaplanGMAT » Sat Oct 16, 2010 10:04 am
ankurmit wrote:Its mentioned that X is geater than 2

x is a positive integer greater than 2; is (x^3 + 19837) (x^2 + 5) (x - 3) an odd number?
Correct - but it doesn't say that all of the prime factors of x are greater than 2. So, from (1) x could be either even or odd.
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