If x is an integer, is (x^2+1) (x+5) an even integer?
1. X is odd
2. Each prime factor of x^2>7
From my practice test
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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
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
Regards
Samir
Samir
Hi Samirsamirpandeyit62 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
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
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- Master | Next Rank: 500 Posts
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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.
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.
Regards
Samir
Samir
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:
At this rate I am going to have to thank you at least 10 times a day :lol:
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- Master | Next Rank: 500 Posts
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- Joined: Sun Mar 25, 2007 7:42 am
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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"
I've solved the other question of TT pls check if it is helpful to u.
The one "Again from my practice test"
Regards
Samir
Samir