Data Sufficiency

If x is an integer. Is (X^2 +1) (X+5) an even number?
1. X is an odd no.
2. Each prime factor of X^2 is greater than 7.

From 1 : x is odd
so, (x^2 +1) = even

and odd+odd = odd; finally, even X odd = even so, sufficient

From 2 :

x is always going to be an odd integer, so sufficient

Answer D.

amitansu wrote:From 1 : x is odd
so, (x^2 +1) = even

and odd+odd = odd; finally, even X odd = even so, sufficient

From 2 :

x is always going to be an odd integer, so sufficient

Answer D.

Can you explain what statement #2 means? Thx.

Each prime factor of x^2 is greater than 7 is the condition :

for example x^2=121 => x = 11
and 121= 11 x 11 (prime factor is 11 here which is greater than 7)
169=13 X 13 (13>7), so x is odd here.

likewise we have to test for different values of x whose each prime factors should be greater than 7.
Prime factors are those which are basically prime numbers as well as factors of x taking different values of it.

amitansu wrote:Each prime factor of x^2 is greater than 7 is the condition :

for example x^2=121 => x = 11
and 121= 11 x 11 (prime factor is 11 here which is greater than 7)
169=13 X 13 (13>7), so x is odd here.

likewise we have to test for different values of x whose each prime factors should be greater than 7.
Prime factors are those which are basically prime numbers as well as factors of x taking different values of it.

Thanks!

Any number can be broken down into powers of primes. The only even prime is 2. All primes of this number are greater than 7. Therefore, this number has all odd prime factors. Since odd * odd = odd, this number must therefore be odd. If this number is odd, we can determine that when plugged into the original equation, we always get an even number, so we can answer the stem definitively.

Great approach guys!!

Just one thing..what if x is -5 then it will be (26)(0)=0 so does this mean that 0 is an even integer??

moneyman wrote:Great approach guys!!

Just one thing..what if x is -5 then it will be (26)(0)=0 so does this mean that 0 is an even integer??

Hey Maxx- Yes, 0 is indeed an even integer. I've been burned more than once by forgetting that.

Thanks mim3 for that info!!

amitansu wrote:From 1 : x is odd
so, (x^2 +1) = even

and odd+odd = odd; finally, even X odd = even so, sufficient

From 2 :

x is always going to be an odd integer, so sufficient

Answer D.

odd +odd = even and even * even = even