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Gmat2Go
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Post subject: If x is an integer, is (x^2 +1)(x+5) an even number?  Posted: Mon Nov 05, 2007 9:19 pm |
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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.
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RonPurewal
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Post subject: Posted: Tue Nov 06, 2007 4:40 am |
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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.
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pravsr
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Post subject: Well explained Posted: Thu Nov 08, 2007 2:24 am |
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RonPurewal
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Post subject: Posted: Fri Nov 09, 2007 4:04 am |
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tekofrlo
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Post subject: Re: If x is an integer, is (x^2 +1)(x+5) an even number? Posted: Mon Sep 07, 2009 3:35 pm |
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parthatayi
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Post subject: Re: Posted: Wed Aug 18, 2010 1:06 am |
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RonPurewal wrote: 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
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mundlia
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Post subject: Re: If x is an integer, is (x^2 +1)(x+5) an even number? Posted: Wed Aug 18, 2010 7:14 pm |
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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
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parthatayi
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Post subject: Re: If x is an integer, is (x^2 +1)(x+5) an even number? Posted: Wed Aug 18, 2010 9:39 pm |
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mundlia wrote: 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 !!
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mundlia
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Post subject: Re: If x is an integer, is (x^2 +1)(x+5) an even number? Posted: Thu Aug 19, 2010 8:24 am |
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Parthatayi
i wont argue with u..please get ur basics right...u can google it as well
0 is an even number
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RonPurewal
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Post subject: Re: If x is an integer, is (x^2 +1)(x+5) an even number? Posted: Thu Sep 16, 2010 6:24 am |
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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.
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shobhitdixit
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Post subject: Re: If x is an integer, is (x^2 +1)(x+5) an even number? Posted: Wed Dec 01, 2010 9:43 am |
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So Ron,
For the 2nd statement, the solution would hold if they said that x^2 has prime factors bigger than 3 - right?
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RonPurewal
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Post subject: Re: If x is an integer, is (x^2 +1)(x+5) an even number? Posted: Thu Dec 02, 2010 10:18 am |
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shobhitdixit wrote: 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)
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