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If x is an integer, is (x^2+1)(x+5) an even number? [#permalink]
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If x is an integer, is (x^2 + 1)(x + 5) an even number? (1) x is an odd number (2) Each prime factor of x^2 is greater than 7
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Last edited by Bunuel on 30 Jan 2016, 01:36, edited 2 times in total.
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If x is an integer, is (x^2+1)(x+5) an even number? [#permalink]
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Re: If x is an integer, is (x^2+1)(x+5) an even number? [#permalink]
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19 Dec 2010, 22:45
I could not understand the reason for the explanation of teh second statement... Can you plsss please explain in a elaborative way...



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Re: If x is an integer, is (x^2+1)(x+5) an even number? [#permalink]
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jullysabat wrote: I could not understand the reason for the explanation of teh second statement... Can you plsss please explain in a elaborative way... Exponentiation does not "produce" new prime factors means that the prime factors of a positive integer x and the prime factors of x^n (where n is a positive integer) are the same. For example if x=12=2^2*3 then it has two prime factors 2 and 3 and x^2=144=2^4*3^2 also has the same two prime factors 2 and 3. Now, (2) says: each prime factor of x^2 is greater than 7 > the prime factors of \(x^2\), or which is the same the prime factors of \(x\) can be 11, 13, 17, ... So 2<7 is not a prime factor of \(x\) which means that \(x\) must be odd, and as we concluded if \(x\) is an odd number then \((x^2+1)(x+5)=even*even=even\). Sufficient. Hope it's clear.
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Re: If x is an integer, is (x^2+1)(x+5) an even number? [#permalink]
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20 Dec 2010, 10:43
Bunuel wrote: jullysabat wrote: I could not understand the reason for the explanation of teh second statement... Can you plsss please explain in a elaborative way... Exponentiation does not "produce" new prime factors means that the prime factors of a positive integer x and the prime factors of x^n (where n is a positive integer) are the same. For example if x=12=2^2*3 then it has two prime factors 2 and 3 and x^2=144=2^4*3^2 also has the same two prime factors 2 and 3. Now, (2) says: each prime factor of x^2 is greater than 7 > the prime factors of \(x^2\), or which is the same the prime factors of \(x\) can be 11, 13, 17, ... So 2<7 is not a prime factor of \(x\) which means that \(x\) must be odd, and as we concluded if \(x\) is an odd number then \((x^2+1)(x+5)=even*even=even\). Sufficient. Hope it's clear. Hey thanks now its clear.... I have also given a kudos to you...



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Re: If x is an integer, is (x^2+1)(x+5) an even number? [#permalink]
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agdimple333 wrote: if x is an integer, is (x^2 + 1) (x + 5) an even number?
(1) x is an odd number (2) each prime factor of x ^ 2 is greater than 7 Statement 1  if x is odd then x+5 is even and hence the expression is even, sufficient Statement 2  if each prime factor of x^2 is greater than 7, then all factors are odd and hence x^2 is odd and hence x^2+1 is even making the expression even, sufficient Answer D



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Re: If x is an integer, is (x^2+1)(x+5) an even number? [#permalink]
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24 Apr 2011, 21:04
(1) If x is odd, (X+5) is even. Hence (x^2 + 1) (x + 5) is an even number. (2) If X^2 has each prime factor greater than 7, x^2 is odd, and then x is odd too. (x^2 does not have 2 as factor) By same reasoning as above, (x^2 + 1) (x + 5) is an even number. Answer  D
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Re: If x is an integer, is (x^2+1)(x+5) an even number? [#permalink]
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24 Apr 2011, 22:43
Straight D it is. Only thing to be considered here is each of the prime numbers greater than 3 is either 6x+1 or 6x1. Hence necessarily an odd number.



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Re: If x is an integer, is (x^2+1)(x+5) an even number? [#permalink]
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29 Jul 2011, 01:56
The answer should be (D). We need to state if we have enough information to conclude whether (x^2 +1) (x+5) is an even number. From statement (1), if x is odd, then x+5 will be even. Therefore (x^2 +1) (x+5) will be (odd +1)*(odd+5) = even * even = even From statement (2), we know that each prime factor of x^2 is greater than 7. This rules out 2 as being a factor of x^2. Since 2 is the only prime that is also even, this means that x^2 can therefore be expressed entirely as a product of primes, all of which are odd. Therefore x^2 is odd. Therefore (x^2 + 1) is even. Therefore (x^2 + 1) (x+5) is even. So (D).
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Re: If x is an integer, is (x^2+1)(x+5) an even number? [#permalink]
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siddhans wrote: If x is an integer, is (x^2 +1) (x+5) an even number? (1) x is an odd number (2) Each prime factor of x^2 is greater than 7 Question asks: Is \(x\) even? OR Is \(x\) odd? (1) \(x\) is odd. Sufficient. (2) All prime factors of \(x^2\) is greater than 7. Rule: If a number doesn't have at least one 2 as its prime factor, the number will be odd. Statement 2 is a convoluted way to say that there is NO 2 as a prime factor of \(x\) OR \(x\) is odd. Same as statement 1. Sufficient. Ans: "D"
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Re: If x is an integer, is (x^2+1)(x+5) an even number? [#permalink]
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29 Jul 2011, 03:29
is (x^2 +1) (x+5) an even
To be even, we need one of the brackets to be even. In both brackets, we are adding odd numbers (1 and 5), hence, if x or x^2 is odd, result will be even.
Restate: is x odd?
1) Sufficient. Plug in a couple numbers as well, looks good. 2) x^2 does not have a two as a factor, as 2 < 7. x is odd.
D



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Re: If x is an integer, is (x^2+1)(x+5) an even number? [#permalink]
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29 Jul 2011, 05:11
If x is an integer, is (x^2 +1) (x+5) an even number? (1) x is an odd number (2) Each prime factor of x^2 is greater than 7
Ans: 1. Stmt 1 suff as an odd number squared will also be an odd number and sum of two odd numbers will give an even number. Without checking further, the product will also be even. Suff
Stmt 2: Prime factor greater than 7, then only odd factors, therefore, an odd integer. Following the same logic as derived in stmt1, product will always give an even number. Suff
D



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Re: If x is an integer, is (x^2+1)(x+5) an even number? [#permalink]
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12 Dec 2011, 09:40
I have one question here, what if x = 5? in that case (5+5) = 0 and hence the equation's value ll be 0 right ? is 0 too considered as even ?



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Re: If x is an integer, is (x^2+1)(x+5) an even number? [#permalink]
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12 Dec 2011, 19:02
Yes,0 is also an even integer
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Re: If x is an integer, is (x^2+1)(x+5) an even number? [#permalink]
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If x is an integer, is (x^2+1)(x+5) an even number?
(1) x is an odd number
(2) Each prime factor of x^2 is greater than 7
Last edited by Bunuel on 23 May 2013, 03:53, edited 2 times in total.
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Re: If x is an integer, is (x^2+1)(x+5) an even number? [#permalink]
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Re: If x is an integer, is (x^2+1)(x+5) an even number? [#permalink]
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28 Jan 2012, 05:20
i copied this question from source as it is
but still i am not able to understand this point
2) Each prime factor of x^2 is greater than 7 > 2 is not a prime factor of x^2, so not a prime factor of x as well > x=odd > the same as above. Can u expalin with example ?



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Re: If x is an integer, is (x^2+1)(x+5) an even number? [#permalink]
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28 Jan 2012, 05:33
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pbull78 wrote: i copied this question from source as it is
but still i am not able to understand this point
2) Each prime factor of x^2 is greater than 7 > 2 is not a prime factor of x^2, so not a prime factor of x as well > x=odd > the same as above. Can u expalin with example ? Each prime factor of x^2 is greater than 7 > as 2 is less than 7, then 2 is not a prime factor of x^2, which means that 2 is not a prime of x as well, because if it is a prime of x then x^2 would also have it. Or: Each prime factor of x^2 is greater than 7 > as 2 is less than 7, then 2 is not a prime factor of x^2, which means that x^2 is an odd number > x^2=odd > x^2=x*x=odd > x=odd. Or: Each prime factor of x^2 is greater than 7 > primes more than 7 are odd (in fact the only even prime is 2) > x is a product of some odd primes more than 7 > x^2 is an odd number > x=odd. Hope it's clear.
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Re: If x is an integer, is (x^2+1)(x+5) an even number? [#permalink]
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01 Jan 2017, 16:07
Here We really need to find the even / odd nature of x here Here both the statements are sufficient if x is odd=> sufficient x^2 has all prime factors >7 => x^2 will be odd => x will be odd to hence D
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Re: If x is an integer, is (x^2+1)(x+5) an even number? [#permalink]
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23 Jan 2017, 10:36
(x^2+1)(x+5) an even number 1, x = odd number oddxodd = odd odd + 1 = even odd + odd = even even x even = even number always AD 2, x^2 > 7 All x values are ODD other than 2 D
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