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

[Gmat2Go]

This is a DS problem

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

a) x is an odd number
b) Each prime factor of x^2 is greater than 7

I know a by itself will work. So down to A and D.
I don't understand what statement B is saying? ITs only talking about prime factor of x^2 but there could be other number that's no prime factor of x^2 so how do i know whether x is even or odd? Please advise.

[RonPurewal]

The prime factors of x^2 are the same as the prime factors of x. The reason is that x^2 is just x times x; therefore, a 'factor tree' or 'prime box' for x^2 would contain EXACTLY the same prime numbers as for x, but just twice as many of each one.

So (2) now says, 'all the prime factors of x are at least 7'.
So, x is a product of primes 7 or greater. All those are odd, so, x is odd. As you've figured out, odd is sufficient, so this also works by itself. Therefore, answer is D.

Another way of deciphering (2) is to use this logic: 2 is the only even prime. Therefore, ALL even numbers have at least one 2 in their prime factorizations. So, if there are no 2's in the prime factorization, then the number is odd.

Hope that helps.

[pravsr]

Well explained Ron!!

[RonPurewal]

[tekofrlo]

thanks Ron.

[parthatayi]

The prime factors of x^2 are the same as the prime factors of x. The reason is that x^2 is just x times x; therefore, a 'factor tree' or 'prime box' for x^2 would contain EXACTLY the same prime numbers as for x, but just twice as many of each one.

So (2) now says, 'all the prime factors of x are at least 7'.
So, x is a product of primes 7 or greater. All those are odd, so, x is odd. As you've figured out, odd is sufficient, so this also works by itself. Therefore, answer is D.

Another way of deciphering (2) is to use this logic: 2 is the only even prime. Therefore, ALL even numbers have at least one 2 in their prime factorizations. So, if there are no 2's in the prime factorization, then the number is odd.

Hope that helps.

Hi Ron,

I dont think the option A holds good if the x takes the value -5.
The answer would be 0 which is niether even nor odd..

Can you please explain what happens if x= -5

[mundlia]

hey parthatayi

in case of x=-5 the equation wud result in 0
and since 0 is counted among even numbers
A wud be sufficient

[parthatayi]

hey parthatayi

in case of x=-5 the equation wud result in 0
and since 0 is counted among even numbers
A wud be sufficient

hey 0 is not an even number..0 is called a composite number !!

[mundlia]

Parthatayi

i wont argue with u..please get ur basics right...u can google it as well
0 is an even number

[RonPurewal]

0 is even.

parthatayi, "composite" has nothing at all to do with the issue of even/odd; "composite" just means that a number is not prime.

[shobhitdixit]

So Ron,
For the 2nd statement, the solution would hold if they said that x^2 has prime factors bigger than 3 - right?

[RonPurewal]

So Ron,
For the 2nd statement, the solution would hold if they said that x^2 has prime factors bigger than 3 - right?

yep.
or even greater than 2 (since 2 wouldn't be included)

[sachin.w]

So (2) now says, 'all the prime factors of x are at least 7'.

Hi Ron,
I believe, 2 says that each prime factor is greater than or equal to 11?

Please correct me if my understanding is wrong.
Regards,
Sachin

[RonPurewal]

So (2) now says, 'all the prime factors of x are at least 7'.

Hi Ron,
I believe, 2 says that each prime factor is greater than or equal to 11?

Please correct me if my understanding is wrong.
Regards,
Sachin

ya, if the original problem says "greater than 7" then that's how you should interpret it.
thanks, good catch -- i think i just misread it the first time.