stmt 1 : if x is odd then (x + 5) will be even as odd+odd =even
hence
(x^2+1) (x+5) will be even SUFF

stmt 2 : Each prime factor of x^2>7

i.e prime factors of x^ 2 would be same as x only with greater powers

so we can say that x has prime factors greater than 7 viz 7,11,13.. etc

now the only even prime nos is 2 which not its factor hence its prime factors are all odd & product of all odd nos would always be odd

so again x + 5 would be even thus the product would be even

hence SUFF

so ans should be D

samirpandeyit62 wrote:stmt 1 : if x is odd then (x + 5) will be even as odd+odd =even
hence
(x^2+1) (x+5) will be even SUFF

stmt 2 : Each prime factor of x^2>7

i.e prime factors of x^ 2 would be same as x only with greater powers

so we can say that x has prime factors greater than 7 viz 7,11,13.. etc

now the only even prime nos is 2 which not its factor hence its prime factors are all odd & product of all odd nos would always be odd

so again x + 5 would be even thus the product would be even

hence SUFF

so ans should be D

Hi Samir

I dont understand your statement above highlighted in bold "now the only even prime nos is 2 which not its factor "

can you pls explain?

thanks

As would be aware that 2 is the only prime nos that is even rest all of them are odd

Now as I mentioned that x will have prime factors greater than 7 hence all its prime factors will all be odd

I've metioned this to highligh that even X odd =even will not happen here

now odd X odd is always odd hence x will be odd.

ahh, thanks so much Samir. It was quite stupid of me I dint read the question properly.- I though it said Each prime factor of x^2<7 instead of > 7

At this rate I am going to have to thank you at least 10 times a day :lol:

No probs RI, anytime, And u dont need to thank me
I've solved the other question of TT pls check if it is helpful to u.

The one "Again from my practice test"

Thanks Samir. D is the right answer.