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If x is an integer, is (x^2+1)(x+5) an even number?

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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
[Reveal] Spoiler: OA

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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If x is an integer, is (x^2+1)(x+5) an even number?

The product of 2 integers to be even at least one of them must be even. So if \(x\) is an odd number then \((x^2+1)(x+5)=even*even=even\).

(1) x is an odd number. Sufficient.

(2) Each prime factor of x^2 is greater than 7 --> first of all, prime factors of \(x^2\) are the same as the prime factors of \(x\): exponentiation does not "produce" new prime factors. So, each prime of \(x\) is greater than 7, which means that 2 is not a prime factor of \(x\), thus \(x\) is an odd number. 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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New post 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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New post 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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New post 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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New post 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 6x-1. 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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New post 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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New post 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.

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Re: If x is an integer, is (x^2+1)(x+5) an even number? [#permalink]

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New post 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

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Re: If x is an integer, is (x^2+1)(x+5) an even number? [#permalink]

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New post 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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New post 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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pbull78 wrote:
If x is an integer, is (x2+1)(x+5) an even number?
1). x is an odd number. 2). each prime factor of x2 is greater than 7

although discussed but still need explanation of this not able to understand .......... 2) each prime factor of x2 is greater than 7

answer is D


pbull78 you please format the question properly. Question should read:

If x is an integer, is (x^2+1)(x+5) an even number?

(1) x is an odd number --> (x^2+1)(x+5)=(odd+odd)(odd+odd)=even*even=even. Sufficient.

(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. 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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New post 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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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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New post 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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New post 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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Re: If x is an integer, is (x^2+1)(x+5) an even number?   [#permalink] 23 Jan 2017, 10:36

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