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.
1. X is an odd no.
2. Each prime factor of X^2 is greater than 7.
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Can you explain what statement #2 means? Thx.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.
Thanks!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.
Hey Maxx- Yes, 0 is indeed an even integer. I've been burned more than once by forgetting that.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??
odd +odd = even and even * even = evenamitansu 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.
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