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

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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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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I could not understand the reason for the explanation of teh second statement...
Can you plsss please explain in a elaborative way...

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

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).

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"

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

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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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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If x is an odd number, x^2 must be odd => (x^2+1) must be even.
By the same logic, (x+5) is even. As addition of two odd numbers results in an even number.
Multiplication of two even numbers results in an even number. Thus (x^2+1)(x+5) is an even number.

Satetment (1) is SUFFICIENT.


And for (2), each prime factor of x^2 is greater than 7 implies that 2 is not a prime factor of x^2 and in turn 2 is not a prime factor of x. (As all the prime factors of x^2 are also prime factors of x)

That means x is composed of 11, 13, 17, 19, 23 etc (Prime numbers greater than 7).
Therefore x is an odd number => Equivalent to statement (1).

Statement (2) is SUFFICIENT.

The correct answer is D.

In statement 2. Can X not be 0?
each prime number of X^2 is greater than 7. Understanding that 2 is the only prime number so if the prime factor of x^2 is above 7 then the only numbers we are left with are odd numbers. Since 0 a multiple of every number. Can x be 0?

Maybe it's too dumb of a question but hey we are all here to learn. Thanks

x cannot be 0 because 0 is divisible be primes which are less than 7 (2, 3, and 5).
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If x is an integer, is (x^2 + 1)(x + 5) an even number?

(1) x is an odd number

if x is odd, then x^2 is odd. Furthermore, x + odd = even. sufficient

(2) Each prime factor of x^2 is greater than 7

if the prime factors of x^2 are greater than 7, we know that there is no 2 in the prime factorization.
This means x is odd. sufficient

11 is an odd prime and is a factor of 22 which is an even,

what about that ?

What you are saying is correct.

However, I will point out that the question states: ALL the prime factors of X^2 are larger than 7.

This means that X^2 cannot have a 2, the only even prime, in the prime factorization (neither can X).

If 22 happened to equal X, X^2 would most certainly have a prime factor of a 2. However, statement 2 does not permit that.

(x^2 + 1)(x + 5)

1)
if x is odd, then x^2 = x * x = odd * odd = odd --> x^2 is odd
If x^2 is odd, then x^2 + 1 is even
if x is odd, x + 5 = odd + odd = even --> x + 5 is even
(x^2 + 1)(x + 5) = even * even = even
suff

2)
All prime numbers are odd EXCEPT 2
Each prime factor of x^2 is greater than 7. So that means the prime factorization of x^2 does not contain a 2
If a prime factorization does not contain a 2, it is an odd number --> x^2 is odd
x^2 + 1 = odd + 1 = even --> x^2 + 1 is even

If x^2 is odd, then so is x --> x^2 = odd --> x * x = odd --> odd * odd = odd only --> x is odd
odd + 5 = odd + odd = even --> x + 5 is even
(x^2 + 1)(x + 5) = even * even = even
suff

That Each prime factor of x2 is greater than 7 is different from x has only prime factors. what if x has both prime factors and non prime factors and those non prime factors are even?

ANSWER IS D

(1) If x is odd, then (x^2 + 1) is even and (x + 5) is even, so EVEN * EVEN= EVEN SUFFICIENT

(2) We have to know that exponent does not change the number prime factor, therefore x and x^2 have the same prime factor.
Now we know that only 2 is the even prime number, rest of then are odd, and here Each prime factor of x^2 is greater than 7, then x must be odd number. then (x^2 + 1) is even and (x + 5) is even, so EVEN * EVEN= EVEN SUFFICIENT
Answer is D