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x is a positive integer greater than two; is (x^3 + 19837)(x^2 + 5)(x

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x is a positive integer greater than two; is (x^3 + 19837)(x^2 + 5)(x [#permalink]

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New post 13 Jun 2011, 09:21
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x is a positive integer greater than two; 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) 3x is an even number

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/x-is-a-posit ... 00569.html
[Reveal] Spoiler: OA

Last edited by agdimple333 on 13 Jun 2011, 10:31, edited 1 time in total.

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Re: x is a positive integer greater than two; is (x^3 + 19837)(x^2 + 5)(x [#permalink]

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New post 13 Jun 2011, 10:20
agdimple333 wrote:
x is a positive integer greater than two; is (x^3 + 19837)(x^2 + 5)(x – 3) an odd number?
--> the sum of any prime factor of x and x is even
--> 3x is an even number


Please denote powers with "^" to avoid ambiguity.

Even \(\pm\) odd=odd
Even \(\pm\) even=even
odd \(\pm\) odd=even
Even*Integer=Even
Odd*odd=odd

If x=even
The expression: (x^3+19837)(x^2+5)(x-3) will be odd
E^3=E
E^2=E

If x=odd
The expression: (x^3+19837)(x^2+5)(x-3) will be even
O^3=O
O^2=O

Question: Is x even?
1. Factor of x and x is even
x=9
Factor=3
3+9=12

x=4
Factor=2
4+2=6

x can be even or odd.
Not Sufficient.

2. 3x is even
3 is odd.
For the expression to be even, x must be even.
Sufficient.

Ans: "B"
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Re: x is a positive integer greater than two; is (x^3 + 19837)(x^2 + 5)(x [#permalink]

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New post 13 Jun 2011, 23:37
a the number can be 2 and powers raised to 2 such as 2,4,8.
or the number can be 3*5*7 and powers of them,with no multiple of 2 present.
thus x can be both even 4,8 or odd 15,21 and so on.

Not sufficient.


b x is even. thus the product is odd. Sufficient.

B
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Re: x is a positive integer greater than two; is (x^3 + 19837)(x^2 + 5)(x [#permalink]

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New post 03 Dec 2017, 05:18
agdimple333 wrote:
x is a positive integer greater than two; 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) 3x is an even number


Platinum GMAT Official Solution:

In order to solve this question efficiently, it is necessary to begin with number properties. For a product of any number of terms to be odd, all the terms must be odd. If there is but one even term, the product will be even. To see this, consider the following examples:
All Terms Odd --> Odd Product
3*7*9*5 = 945
7*9*3 = 189
1*3*5 = 15

But: One or More Even Terms --> Even Product
3*7*9*2 = 378
7*9*4 = 252
1*3*5*6 = 90

In order for (x^3 + 19837)(x^2 + 5)(x – 3) to be an odd number, all the terms must be odd.

To determine under what conditions each term will be odd, it is important to remember the following relationships:
odd + odd = even
odd - odd = even

even + even = even
even - even = even

even + odd = odd
even - odd = odd
odd + even = odd
odd - even = odd

The only way for each term of (x^3 + 19837)(x^2 + 5)(x – 3) to be odd is if an even and an odd number are added or subtracted together within the parenthesis of each term. In other words:
even + odd = odd: For (x^3 + 19837) to be odd, since 19837 is odd, x^3 will need to be even. This will happen only when x is even.

even + odd = odd: For (x^2 + 5) to be odd, since 5 is odd, x^2 will need to be even. This will happen only when x is even.

even - odd = odd: For (x – 3) to be odd, since 3 is odd, x will need to be even.

When combining the results from the analysis of the three terms above, the only way for (x^3 + 19837)(x^2 + 5)(x – 3) to be odd is if each term is odd. This will only happen if x is even. Consequently, the original question can be simplified to: is x even? Another version of the simplified question is: what is the parity of x?

Evaluate Statement (1) alone.

In order for the sum of any prime factor of x and x to be even, it must follow one of two patterns:
Pattern (1): even + even = even
Pattern (2): odd + odd = even

There are two possible cases:

Case (1): x is even. In this case, Pattern (1) must hold. Since x is even in Case (1), any and every prime factor of x must be even (otherwise we could choose an odd prime factor of x and the sum of x and the odd prime factor would be odd). Let's consider some examples:
Let x = 12: However, x cannot equal 12 since one prime factor of 12 is 3 and 12 + 3 = odd number.
Let x = 26: However, x cannot equal 26 since one prime factor of 26 is 13 and 26 + 13 = odd number.
Let x = 14: However, x cannot equal 14 since one prime factor of 14 is 7 and 14 + 7 = odd number.
Let x = 16: x can equal 16 since every prime factor of 16 is even and as a result we know that and 16 + any prime factor = even number.
It is clear that Statement (1) allows x to be even (e.g., 16 is a possible value of x).

Case (2): x is odd. In this case, Pattern (2) must hold. Since x is odd in Case (2), any and every prime factor of x must be odd (otherwise we could choose an even prime factor of x and the sum of x and the even prime factor would be odd). Since all the prime factors of x are odd, x must be odd in Case (2). Let's consider some examples:
Let x = 11: Every prime factor of 11 is odd, so: 11 + prime factor of 11 = even number.
Let x = 15: Every prime factor of 15 is odd, so: 15 + prime factor of 15 = even number.
Since Statement (1) allows x to be either even (e.g., 16) or odd (e.g., 15), we cannot determine the parity of x.

Statement (1) is NOT SUFFICIENT.

Evaluate Statement (2) alone.
3x = Even Number
(odd)(x)=(even)
x must be even because, as shown above, if x were odd, 3x would be odd.

Statement (2) is SUFFICIENT since it definitively tells the parity of x.

Since Statement (1) alone is NOT SUFFICIENT but Statement (2) alone is SUFFICIENT, answer B is correct.

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/x-is-a-posit ... 00569.html
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Re: x is a positive integer greater than two; is (x^3 + 19837)(x^2 + 5)(x   [#permalink] 03 Dec 2017, 05:18
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x is a positive integer greater than two; is (x^3 + 19837)(x^2 + 5)(x

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