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If X is an integer

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Source: — Data Sufficiency |
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by theCodeToGMAT » Sun Mar 09, 2014 7:23 am
kaudes11114 wrote:If X is an integer, is (x2 + 1)(x + 5) an even number?
(1) x is an odd number.
(2) Each prime factor of x^2 is greater than 7.

How to approach this one?
To find: (x^2 + 1) (x+5) is EVEN

Statement 1:
x = ODD
(x^2 + 1) = EVEN
(x + 5) = EVEN
Hence, EVEN * EVEN = EVEN
SUFFICIENT

Statement 2:
x^2 = p * p .. that means all prime factors are ODD
(x^2 + 1) = EVEN
(x + 5) = EVEN
Hence, EVEN * EVEN = EVEN
SUFFICIENT

[spoiler]{D}[/spoiler]
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by Brent@GMATPrepNow » Sun Mar 09, 2014 7:50 am
kaudes11114 wrote:If x is an integer, is (x² + 1)(x + 5) an even number?
(1) x is an odd number.
(2) Each prime factor of x² is greater than 7.
My solution is similar to Rahul's, except I handle statement 2 a little differently

Aside: Don't forget these useful rules:
(ODD)(ODD) = ODD
(EVEN)(ODD) = EVEN
(EVEN)(EVEN) = EVEN

ODD + ODD = EVEN
EVEN + ODD = ODD
EVEN + EVEN = EVEN

Okay, onto the question...

Target question: Is (x² + 1)(x + 5) an even number?

Statement 1: x is an odd number.
If x is ODD, then x² is ODD, which means (x² + 1) is EVEN
If x is ODD, then (x + 5) is EVEN
So, (x² + 1)(x + 5) = (EVEN)(EVEN) = EVEN
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: Each prime factor of x² is greater than 7.
This is a clever way of telling us that the number 2 is NOT a prime factor of x².
In other words, x² is ODD
If x² is ODD, then x is ODD
At this point, we should recognize that statements 1 and 2 are both telling us that x is odd
So, if statement 1 is SUFFICIENT, then statement 2 must be SUFFICIENT.

Answer = D

Cheers,
Brent
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by GMATGuruNY » Sun Mar 09, 2014 12:45 pm
kaudes11114 wrote:If X is an integer, is (x2 + 1)(x + 5) an even number?
(1) x is an odd number.
(2) Each prime factor of x² is greater than 7.
The approaches above are great.
An alternate approach is to TEST CASES.

Statement 1: x is odd
If x=1, then (x²+1)(x+5) = (1² + 1)(1 + 5) = 2*6 = 12.
If x=3, then (x²+1)(x+5) = (3² + 1)(3 + 5) = 10*8 = 80.
If x=5, then (x²+1)(x+5) = (5² + 1)(5 + 5) = 26*10 = 260.
The cases above illustrate that -- if x is odd -- then (x²+1)(x+5) = even.
SUFFICIENT.

Statement 2: Each prime factor of x² is greater than 7.
Since each prime factor of x² is greater than 7, each prime factor of X ITSELF must be greater than 7.
Options for x:
11, 13, 17, 19, 23...
Notice that only ODD VALUES FOR X will satisfy statement 2.
As we saw in statement 1, if x is odd, then (x²+1)(x+5) = even.
SUFFICIENT.

The correct answer is D.
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by Matt@VeritasPrep » Sun Mar 09, 2014 12:52 pm
Just to add two final notes to a great cavalcade of answers:

The first thing to see here is that the two statements are IDENTICAL. (They each say "x is odd".) Given that, the answer is D or E.

The second is that you can always factor the original equation. (x²+1)(x+5) = x³ + 5x² + x + 5. If x is odd, we have ODD + ODD + ODD + ODD, which must be even. Sufficient!
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